How likely is the flotsam of a space battle to be captured by our sun's gravity well?












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A light-year distant from Sol saw the collision of a massive space-born battle fleet against an awe-inspiring dreadnought. At one point in the battle, the flotilla fired rail-guns with 100 metric ton slugs and nuclear missiles.



...And a lot of them missed their target



...And kept on sailing through space, generally toward our beloved Sol.



Eons later, while eating an Italian custard at a French bistro whilst admiring the visas in my American passport, I look up to see an odd reflection in the early afternoon sun. Hours later, Earth becomes collateral damage in a battle fought long before Gelato ice cream was invented.




  • Ignore the scattering effect of objects sailing through space. Assume that the rail-gun slugs and missiles (their fuel long since exhausted) are lined up to pass through our system, parallel with the ecliptic, 15 degrees above the ecliptic referenced from the sun, and 2 AU from the sun.


  • Assume the missiles are similar in size and mass to a Minuteman-III missile.


  • Rail-gun slug: 100 metric tons, muzzle velocity of 2.5 Km/s.



Question: Is it possible for these munitions to become captured in our sun's gravity well, thereby making them a believable catastrophe-in-the-making?






gravity orbital-mechanics







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  • 3




    $begingroup$
    "15 degrees above the ecliptic" does not actually mean anything. Please rephrase this. Do you mean that at closest approach to the sun it will appear to be 15 degrees above or below the plane of the ecliptic when observed from earth? If so, calculate the actual distance involved, and then try to reconcile it with your 2 AU number.
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    – WhatRoughBeast
    Dec 23 '18 at 4:25






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    @WhatRoughBeast, when observed from earth? Odd. At a distance of 2 AU from the sun, passing parallel with the ecliptic, 15 degrees above the ecliptic (which, obviously, is referenced from the sun...) I'll put in the minor clarification.
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    – JBH
    Dec 23 '18 at 6:34






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    "Eons later, ... a battle fought long before Gelato ice cream was invented." Approximately 120,000 years ago. Yep Gelato ice cream wasn't invented in 118 millennium BCE. Certainly a bit later than that.
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    – a4android
    Dec 23 '18 at 8:37










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    In the interests linguistic good practice. "Flotsam" isn't the right word. It's the floating wreckage from wrecks at sea. Jetsam is the stuff that sinks. Floating & sinking wreckage in space doesn't make sense. "Spent munitions" might be a better phrase.
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    – a4android
    Dec 23 '18 at 8:40






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    If the spent munitions are 2 AU from the Sun, how can there be "a believable catastrophe-in-the-making"? It should miss by 150 million kilometres.
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    – a4android
    Dec 23 '18 at 8:42
















6












$begingroup$


A light-year distant from Sol saw the collision of a massive space-born battle fleet against an awe-inspiring dreadnought. At one point in the battle, the flotilla fired rail-guns with 100 metric ton slugs and nuclear missiles.



...And a lot of them missed their target



...And kept on sailing through space, generally toward our beloved Sol.



Eons later, while eating an Italian custard at a French bistro whilst admiring the visas in my American passport, I look up to see an odd reflection in the early afternoon sun. Hours later, Earth becomes collateral damage in a battle fought long before Gelato ice cream was invented.




  • Ignore the scattering effect of objects sailing through space. Assume that the rail-gun slugs and missiles (their fuel long since exhausted) are lined up to pass through our system, parallel with the ecliptic, 15 degrees above the ecliptic referenced from the sun, and 2 AU from the sun.


  • Assume the missiles are similar in size and mass to a Minuteman-III missile.


  • Rail-gun slug: 100 metric tons, muzzle velocity of 2.5 Km/s.



Question: Is it possible for these munitions to become captured in our sun's gravity well, thereby making them a believable catastrophe-in-the-making?






gravity orbital-mechanics







share|improve this question











$endgroup$








  • 3




    $begingroup$
    "15 degrees above the ecliptic" does not actually mean anything. Please rephrase this. Do you mean that at closest approach to the sun it will appear to be 15 degrees above or below the plane of the ecliptic when observed from earth? If so, calculate the actual distance involved, and then try to reconcile it with your 2 AU number.
    $endgroup$
    – WhatRoughBeast
    Dec 23 '18 at 4:25






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    @WhatRoughBeast, when observed from earth? Odd. At a distance of 2 AU from the sun, passing parallel with the ecliptic, 15 degrees above the ecliptic (which, obviously, is referenced from the sun...) I'll put in the minor clarification.
    $endgroup$
    – JBH
    Dec 23 '18 at 6:34






  • 1




    $begingroup$
    "Eons later, ... a battle fought long before Gelato ice cream was invented." Approximately 120,000 years ago. Yep Gelato ice cream wasn't invented in 118 millennium BCE. Certainly a bit later than that.
    $endgroup$
    – a4android
    Dec 23 '18 at 8:37










  • $begingroup$
    In the interests linguistic good practice. "Flotsam" isn't the right word. It's the floating wreckage from wrecks at sea. Jetsam is the stuff that sinks. Floating & sinking wreckage in space doesn't make sense. "Spent munitions" might be a better phrase.
    $endgroup$
    – a4android
    Dec 23 '18 at 8:40






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    $begingroup$
    If the spent munitions are 2 AU from the Sun, how can there be "a believable catastrophe-in-the-making"? It should miss by 150 million kilometres.
    $endgroup$
    – a4android
    Dec 23 '18 at 8:42














6












6








6


1



$begingroup$


A light-year distant from Sol saw the collision of a massive space-born battle fleet against an awe-inspiring dreadnought. At one point in the battle, the flotilla fired rail-guns with 100 metric ton slugs and nuclear missiles.



...And a lot of them missed their target



...And kept on sailing through space, generally toward our beloved Sol.



Eons later, while eating an Italian custard at a French bistro whilst admiring the visas in my American passport, I look up to see an odd reflection in the early afternoon sun. Hours later, Earth becomes collateral damage in a battle fought long before Gelato ice cream was invented.




  • Ignore the scattering effect of objects sailing through space. Assume that the rail-gun slugs and missiles (their fuel long since exhausted) are lined up to pass through our system, parallel with the ecliptic, 15 degrees above the ecliptic referenced from the sun, and 2 AU from the sun.


  • Assume the missiles are similar in size and mass to a Minuteman-III missile.


  • Rail-gun slug: 100 metric tons, muzzle velocity of 2.5 Km/s.



Question: Is it possible for these munitions to become captured in our sun's gravity well, thereby making them a believable catastrophe-in-the-making?






gravity orbital-mechanics







share|improve this question











$endgroup$




A light-year distant from Sol saw the collision of a massive space-born battle fleet against an awe-inspiring dreadnought. At one point in the battle, the flotilla fired rail-guns with 100 metric ton slugs and nuclear missiles.



...And a lot of them missed their target



...And kept on sailing through space, generally toward our beloved Sol.



Eons later, while eating an Italian custard at a French bistro whilst admiring the visas in my American passport, I look up to see an odd reflection in the early afternoon sun. Hours later, Earth becomes collateral damage in a battle fought long before Gelato ice cream was invented.




  • Ignore the scattering effect of objects sailing through space. Assume that the rail-gun slugs and missiles (their fuel long since exhausted) are lined up to pass through our system, parallel with the ecliptic, 15 degrees above the ecliptic referenced from the sun, and 2 AU from the sun.


  • Assume the missiles are similar in size and mass to a Minuteman-III missile.


  • Rail-gun slug: 100 metric tons, muzzle velocity of 2.5 Km/s.



Question: Is it possible for these munitions to become captured in our sun's gravity well, thereby making them a believable catastrophe-in-the-making?






gravity orbital-mechanics




gravity orbital-mechanics






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Dec 23 '18 at 6:35







JBH

















asked Dec 23 '18 at 0:32









JBHJBH

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  • 3




    $begingroup$
    "15 degrees above the ecliptic" does not actually mean anything. Please rephrase this. Do you mean that at closest approach to the sun it will appear to be 15 degrees above or below the plane of the ecliptic when observed from earth? If so, calculate the actual distance involved, and then try to reconcile it with your 2 AU number.
    $endgroup$
    – WhatRoughBeast
    Dec 23 '18 at 4:25






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    $begingroup$
    @WhatRoughBeast, when observed from earth? Odd. At a distance of 2 AU from the sun, passing parallel with the ecliptic, 15 degrees above the ecliptic (which, obviously, is referenced from the sun...) I'll put in the minor clarification.
    $endgroup$
    – JBH
    Dec 23 '18 at 6:34






  • 1




    $begingroup$
    "Eons later, ... a battle fought long before Gelato ice cream was invented." Approximately 120,000 years ago. Yep Gelato ice cream wasn't invented in 118 millennium BCE. Certainly a bit later than that.
    $endgroup$
    – a4android
    Dec 23 '18 at 8:37










  • $begingroup$
    In the interests linguistic good practice. "Flotsam" isn't the right word. It's the floating wreckage from wrecks at sea. Jetsam is the stuff that sinks. Floating & sinking wreckage in space doesn't make sense. "Spent munitions" might be a better phrase.
    $endgroup$
    – a4android
    Dec 23 '18 at 8:40






  • 2




    $begingroup$
    If the spent munitions are 2 AU from the Sun, how can there be "a believable catastrophe-in-the-making"? It should miss by 150 million kilometres.
    $endgroup$
    – a4android
    Dec 23 '18 at 8:42














  • 3




    $begingroup$
    "15 degrees above the ecliptic" does not actually mean anything. Please rephrase this. Do you mean that at closest approach to the sun it will appear to be 15 degrees above or below the plane of the ecliptic when observed from earth? If so, calculate the actual distance involved, and then try to reconcile it with your 2 AU number.
    $endgroup$
    – WhatRoughBeast
    Dec 23 '18 at 4:25






  • 1




    $begingroup$
    @WhatRoughBeast, when observed from earth? Odd. At a distance of 2 AU from the sun, passing parallel with the ecliptic, 15 degrees above the ecliptic (which, obviously, is referenced from the sun...) I'll put in the minor clarification.
    $endgroup$
    – JBH
    Dec 23 '18 at 6:34






  • 1




    $begingroup$
    "Eons later, ... a battle fought long before Gelato ice cream was invented." Approximately 120,000 years ago. Yep Gelato ice cream wasn't invented in 118 millennium BCE. Certainly a bit later than that.
    $endgroup$
    – a4android
    Dec 23 '18 at 8:37










  • $begingroup$
    In the interests linguistic good practice. "Flotsam" isn't the right word. It's the floating wreckage from wrecks at sea. Jetsam is the stuff that sinks. Floating & sinking wreckage in space doesn't make sense. "Spent munitions" might be a better phrase.
    $endgroup$
    – a4android
    Dec 23 '18 at 8:40






  • 2




    $begingroup$
    If the spent munitions are 2 AU from the Sun, how can there be "a believable catastrophe-in-the-making"? It should miss by 150 million kilometres.
    $endgroup$
    – a4android
    Dec 23 '18 at 8:42








3




3




$begingroup$
"15 degrees above the ecliptic" does not actually mean anything. Please rephrase this. Do you mean that at closest approach to the sun it will appear to be 15 degrees above or below the plane of the ecliptic when observed from earth? If so, calculate the actual distance involved, and then try to reconcile it with your 2 AU number.
$endgroup$
– WhatRoughBeast
Dec 23 '18 at 4:25




$begingroup$
"15 degrees above the ecliptic" does not actually mean anything. Please rephrase this. Do you mean that at closest approach to the sun it will appear to be 15 degrees above or below the plane of the ecliptic when observed from earth? If so, calculate the actual distance involved, and then try to reconcile it with your 2 AU number.
$endgroup$
– WhatRoughBeast
Dec 23 '18 at 4:25




1




1




$begingroup$
@WhatRoughBeast, when observed from earth? Odd. At a distance of 2 AU from the sun, passing parallel with the ecliptic, 15 degrees above the ecliptic (which, obviously, is referenced from the sun...) I'll put in the minor clarification.
$endgroup$
– JBH
Dec 23 '18 at 6:34




$begingroup$
@WhatRoughBeast, when observed from earth? Odd. At a distance of 2 AU from the sun, passing parallel with the ecliptic, 15 degrees above the ecliptic (which, obviously, is referenced from the sun...) I'll put in the minor clarification.
$endgroup$
– JBH
Dec 23 '18 at 6:34




1




1




$begingroup$
"Eons later, ... a battle fought long before Gelato ice cream was invented." Approximately 120,000 years ago. Yep Gelato ice cream wasn't invented in 118 millennium BCE. Certainly a bit later than that.
$endgroup$
– a4android
Dec 23 '18 at 8:37




$begingroup$
"Eons later, ... a battle fought long before Gelato ice cream was invented." Approximately 120,000 years ago. Yep Gelato ice cream wasn't invented in 118 millennium BCE. Certainly a bit later than that.
$endgroup$
– a4android
Dec 23 '18 at 8:37












$begingroup$
In the interests linguistic good practice. "Flotsam" isn't the right word. It's the floating wreckage from wrecks at sea. Jetsam is the stuff that sinks. Floating & sinking wreckage in space doesn't make sense. "Spent munitions" might be a better phrase.
$endgroup$
– a4android
Dec 23 '18 at 8:40




$begingroup$
In the interests linguistic good practice. "Flotsam" isn't the right word. It's the floating wreckage from wrecks at sea. Jetsam is the stuff that sinks. Floating & sinking wreckage in space doesn't make sense. "Spent munitions" might be a better phrase.
$endgroup$
– a4android
Dec 23 '18 at 8:40




2




2




$begingroup$
If the spent munitions are 2 AU from the Sun, how can there be "a believable catastrophe-in-the-making"? It should miss by 150 million kilometres.
$endgroup$
– a4android
Dec 23 '18 at 8:42




$begingroup$
If the spent munitions are 2 AU from the Sun, how can there be "a believable catastrophe-in-the-making"? It should miss by 150 million kilometres.
$endgroup$
– a4android
Dec 23 '18 at 8:42










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Virtually none of them would be captured and it's unlikely we'd notice anything whizzing past.



Basically, an object that falls into a gravity well from infinity (and a light year off is pretty close to that) will, unless it hits something while passing through, fall right back out again. By definition it picks up escape velocity as it falls in and loses it as it falls back out. So it passes through the solar system on a hyperbolic trajectory. And if the projectiles were moving at high speed before falling into the solar system, it will be a very flat hyperbola -- i.e., they'll hardly be deflected at all by the Sun's gravity.



If the slugs are aimed to pass through at 2AU, there's basically nothing there to hit and all they do is think to themselves, "Here comes the Sun...there goes the Sun...hello interstellar darkness..."



(I'm not quite sure what you mean by "to pass through our system 15 degrees above the ecliptic." It you mean they are aimed to pass well above the plane of the planetary orbits, then there's still less likely to hit anything or even be noticed.)



The planets -- even Jupiter -- would have little effect on the projectiles since most of them would not come near enough to be affected, and the rest would only be deflected a bit. (Remember, they're passing by at greater than escape velocity. The deflection a body can provide depends on the projectile's velocity (the slower it's moving relative, the bigger the defection) and on the impact parameter (the closer the projectile passes the bigger the deflection.)



Hitting the Earth accidentally is hard. The Earth occupies .25(r/R)2 of a 2AU circle, where r is the Earth's radius and R is the Earth's orbital radius (the AU). That fraction is about 5*10-10. So less than one in a billion of the projectiles would hit Earth even if they were aimed to all pass through the ecliptic within 2AU of the Sun.



Finally, if they did hit the Earth we'd probably not notice them. A 100 metric ton slug of metal would be about 10 cubic meters if it's iron, less if it's something denser. That's a sphere less then ten feet in diameter. It would produce a nice bolide, but would come down pretty much intact, making a 50 meter crater with about a 4 kiloton explosion. Nothing you'd want to be next to, of course, but basically unnoticeable more than twenty miles away. Tunguska was close to 100 times bigger (though probably mostly ice) and its explosion was around four thousand times greater, and it was nearly missed. Had it hit water, it would have been missed.






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    +1 for the physics. Also may be worth mentioning re the descriptive text that by the time a 100 ton object (which is pretty small) moving at 2.5 km/s or more becomes visible to the naked eye, any consequences are going to become apparent in less than a second, not hours later.
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    – KerrAvon2055
    Dec 23 '18 at 2:12










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    @KerrAvon2055 Excellent point! (I think your comment says all that needs to be said, though.)
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    – Mark Olson
    Dec 23 '18 at 2:13








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    Is that energy right? It's going to pick up 11 km/sec from hitting Earth and at least 30 km/sec from falling into the solar system. (Note that you can't just add these to 41 km/sec!)
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    – Loren Pechtel
    Dec 23 '18 at 5:04






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    Excellent answer, but you don't address the "don't scatter" part of it. Unless the slugs are tethered, they will drift apart, even if they are all "pointing at" us, the smallest distance between them will place them in different orbits which in turn will spread them.
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    – Diego Sánchez
    Dec 23 '18 at 8:40










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    Don't forget that planets are in motion, and so is the Sun relative to the Galaxy, so anything that passes through a gravity well will see its trajectory and speed modified. See 'gravity assist' or 'slingshot manoeuvre m' at ESA, NASA or Wikipedia. I would link but I'm on my phone. Also, if the Tunguska object would have fallen into the ocean, the resulting tsunami would have been noticed by coastal settlements in the vicinity. And certainly have puzzled scientists since it wouldn't correlate with any known seismic event.
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    – Sava
    Dec 23 '18 at 9:38



















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To be captured by the sun's gravity well means to end up orbiting the sun.



If a projectile is fired from outside the solar system, it may do a flyby around the sun. Its speed at any altitude (relative to the sun) will be greater than the escape velocity at those points.



Scientists believe this is exactly what the Oumuamua object did.



For the flotsam to be captured, it must be decelerated by a planet. Jupiter is the best candidate to do so, but the odds are against it. Even if the flotsam does a flyby by Jupiter, it is more likely to either hit the planet or be accelerated even further, causing it to exit the solar system earlier.



If it does get decelerated by a planet, the flotsam will be locked into an orbit that is unlikely to cross the Earth's own. It will cross Jupiter's orbit during its perihelion, which means it may crash into Jupiter eobs later, and it has an astronomical probability of having an orbit even more inclined relative to the Earth's than Pluto. All this makes an encounter with Earth very unlikely.






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    And note that even if one is captured that doesn't mean another nearby one gets captured.
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    – Loren Pechtel
    Dec 23 '18 at 5:05






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    Oumuamua is a good example. It was too fast to get captured in the gravity well so it went on its way. But that does not mean an object of this sort could not have hit earth.
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    – Willk
    Dec 23 '18 at 17:23





















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Short answer is a space weapon fired at any sort of realistic speed will never be captured by the Sun's gravity well. Consider that mere human technology has created nuclear pumped weapons like project Prometheus, which captured a small fraction of a nuclear explosion to fire a stream of pellets at 100km/sec. Changing some elements allows you to use similar equipment to make shaped charges, Explosively formed projectiles and even streams of star hot plasma moving at speeds measured in percentages of the speed of light. In short, your space armada would be sliced to pieces before they get a shot in using weapons with velocities of 2.5km/sec. In fact, orbital velocity around the Earth is already 7km/sec.



Realistic space weapons will be moving so fast that they will never be captured, while extremely slow moving bodies like you describe will likely not reach Earth at all, but either drift harmlessly past, or be captured by some of the Gas Giant planets in the outer Solar System.






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    Off the top of my head, 100 metric tons, 2 AU from the sun traveling at 2.5 Km/s, (tangent to? and) 15 degrees above the ecliptic, is not a trajectory that will put you into solar orbit (overshot by at least, I'm guessing, 1 Km/s). This question belongs on Math.SE
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    – Mazura
    Dec 23 '18 at 20:33



















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Muzzle velocity seems less important than projectile velocity as seen from Sol.



Give the weapons provided, it seems likely that fleets in motion, starting from bases also in motion, would need to come to (relative) near-rest with each other to have the fight at all...else their weapons would not score at all as the fleets flashed past each other in a fraction of a second and then spent weeks shedding velocity for another attacking run.



That point of relative near-rest-for-the-fight is still moving, from the perspective of everybody else in the galaxy. It orbits the galactic center, and it has a vector from the perspective of all the stars...including Sol.



And if that apparent vector, from Sol's perspective, is greater than about 7km/s at closest approach of 2AU, then Sol will probably not capture it at all. Flotsam just below that limit would likely go into comet-like hyperbolic orbits and may not return to the inner system for millenia.



Flotsam really slowly drifting toward Sol would be captured and become a minor menace to inner planets...but only during the two times each orbit that they intersect the ecliptic...and only until their crazy, unstable orbits dump them into Sol.






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    The fleets could have been passing at well above weapon velocity--they shoot all their weapons as they approach the enemy fleet so they pass through the same place in space. This assumes drives that can build high velocity but only very slowly so they can't simply evade the slower but more maneuverable weapons.
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    – Loren Pechtel
    Dec 23 '18 at 5:08






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    @LorenPechtel, Agreed, or that both fleets are essentially flying through hasty minefields. Both of those possibilities didn't seem to match the romantic sense in the OP's question, so I ignored them.
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    – user535733
    Dec 23 '18 at 13:30



















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They are most likely to end up captured by Jupiter's gravity well first. If they manage to survive that long.



Jupiter is the second biggest object in our solar system after the Sun and possess a vast gravity well. There's a reason why it is used for slingshot maneuvers and accelerations by many of the probes sent throughout the system by the various space agencies. They even sent a solar probe to Jupiter to alter it's trajectory so it could fly by the poles of the sun.



2 AU is a little under the distance of the Asteroid Belt to the Sun, so they would definitely be in the gravitational influence of the Sun.



As for hitting Earth, which is, all things considered, an orbiting mote of dust around the Sun, it seems very improbable. The path of the flotsam would have to intersect Earth's orbit at a point in space and time that is also occupied by Earth at the same moment.



I do not know how to calculate or model the path of your piece of flotsam given the parameters given, but I would rather expect it to either fall into Jupiter or the Sun, or be sent into the depths of interstellar space, than hit anything else in the system.






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    Nearly downvoted. You do realize, I hope, that gravity wells are symmetric? That is, the incoming velocity equals the outgoing velocity? As Mark Olsen discussed in his answer, a body falling toward a large one from far away will end up far away. And as for "fall(ing) into Jupiter or the Sun", that is extremely unlikely, particularly since they are originally pointed at 2 AU out and 15 degrees above the plane of the ecliptic.
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    – WhatRoughBeast
    Dec 23 '18 at 4:23










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    You forget one thing: planets are in motion and so are their gravity wells. Thus, depending upon angle and vector of entry into the gravity well relative to the planet's trajectory around the Sun, any small celestial object will end up accelerating or decelerating and have their orbital path modified. Please look up 'slingshot manoeuvre' or 'gravity assist' at ESA, NASA or even Wikipedia. I would link, but I'm on my phone at the moment.
    $endgroup$
    – Sava
    Dec 23 '18 at 9:30






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    Everything you say is true - but irrelevant. Your opening sentence, "They are most likely to end up captured by Jupiter's gravity well first." is unfounded, unsupported by your statement, and untrue. In the absence of a collision, any object inbound from 1 light-year with an initial velocity of 2.5 km/sec which is out of the plane of the ecliptic is not going to get near enough to an object like Jupiter to be grossly affected.
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    – WhatRoughBeast
    Dec 23 '18 at 16:17










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    Can you prove that my opening sentence is 'unfounded and untrue' and that yours is? Jupiter has a massive gravity well that extends far away in space, it's been calculated that it is responsible from perturbations in Mercury's orbit. And, at 2.5 km/s, the flotsam is moving slowly enough to be easily captured, given that Jupiter captured comets that moved way faster, like Shoemaker-Levy 9, which is believed to have been in solar orbit prior to being 'stolen' by Jupiter.
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    – Sava
    Dec 23 '18 at 22:48










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    After the object has entered the solar system, it will be going way faster than 2.5 km/s—about 20 km/s at Jupiter’s distance from the sun. This is why it has an extremely low probability of being captured by anything.
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    – prl
    Dec 24 '18 at 1:20













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5 Answers
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23












$begingroup$

Virtually none of them would be captured and it's unlikely we'd notice anything whizzing past.



Basically, an object that falls into a gravity well from infinity (and a light year off is pretty close to that) will, unless it hits something while passing through, fall right back out again. By definition it picks up escape velocity as it falls in and loses it as it falls back out. So it passes through the solar system on a hyperbolic trajectory. And if the projectiles were moving at high speed before falling into the solar system, it will be a very flat hyperbola -- i.e., they'll hardly be deflected at all by the Sun's gravity.



If the slugs are aimed to pass through at 2AU, there's basically nothing there to hit and all they do is think to themselves, "Here comes the Sun...there goes the Sun...hello interstellar darkness..."



(I'm not quite sure what you mean by "to pass through our system 15 degrees above the ecliptic." It you mean they are aimed to pass well above the plane of the planetary orbits, then there's still less likely to hit anything or even be noticed.)



The planets -- even Jupiter -- would have little effect on the projectiles since most of them would not come near enough to be affected, and the rest would only be deflected a bit. (Remember, they're passing by at greater than escape velocity. The deflection a body can provide depends on the projectile's velocity (the slower it's moving relative, the bigger the defection) and on the impact parameter (the closer the projectile passes the bigger the deflection.)



Hitting the Earth accidentally is hard. The Earth occupies .25(r/R)2 of a 2AU circle, where r is the Earth's radius and R is the Earth's orbital radius (the AU). That fraction is about 5*10-10. So less than one in a billion of the projectiles would hit Earth even if they were aimed to all pass through the ecliptic within 2AU of the Sun.



Finally, if they did hit the Earth we'd probably not notice them. A 100 metric ton slug of metal would be about 10 cubic meters if it's iron, less if it's something denser. That's a sphere less then ten feet in diameter. It would produce a nice bolide, but would come down pretty much intact, making a 50 meter crater with about a 4 kiloton explosion. Nothing you'd want to be next to, of course, but basically unnoticeable more than twenty miles away. Tunguska was close to 100 times bigger (though probably mostly ice) and its explosion was around four thousand times greater, and it was nearly missed. Had it hit water, it would have been missed.






share|improve this answer









$endgroup$









  • 4




    $begingroup$
    +1 for the physics. Also may be worth mentioning re the descriptive text that by the time a 100 ton object (which is pretty small) moving at 2.5 km/s or more becomes visible to the naked eye, any consequences are going to become apparent in less than a second, not hours later.
    $endgroup$
    – KerrAvon2055
    Dec 23 '18 at 2:12










  • $begingroup$
    @KerrAvon2055 Excellent point! (I think your comment says all that needs to be said, though.)
    $endgroup$
    – Mark Olson
    Dec 23 '18 at 2:13








  • 1




    $begingroup$
    Is that energy right? It's going to pick up 11 km/sec from hitting Earth and at least 30 km/sec from falling into the solar system. (Note that you can't just add these to 41 km/sec!)
    $endgroup$
    – Loren Pechtel
    Dec 23 '18 at 5:04






  • 1




    $begingroup$
    Excellent answer, but you don't address the "don't scatter" part of it. Unless the slugs are tethered, they will drift apart, even if they are all "pointing at" us, the smallest distance between them will place them in different orbits which in turn will spread them.
    $endgroup$
    – Diego Sánchez
    Dec 23 '18 at 8:40










  • $begingroup$
    Don't forget that planets are in motion, and so is the Sun relative to the Galaxy, so anything that passes through a gravity well will see its trajectory and speed modified. See 'gravity assist' or 'slingshot manoeuvre m' at ESA, NASA or Wikipedia. I would link but I'm on my phone. Also, if the Tunguska object would have fallen into the ocean, the resulting tsunami would have been noticed by coastal settlements in the vicinity. And certainly have puzzled scientists since it wouldn't correlate with any known seismic event.
    $endgroup$
    – Sava
    Dec 23 '18 at 9:38
















23












$begingroup$

Virtually none of them would be captured and it's unlikely we'd notice anything whizzing past.



Basically, an object that falls into a gravity well from infinity (and a light year off is pretty close to that) will, unless it hits something while passing through, fall right back out again. By definition it picks up escape velocity as it falls in and loses it as it falls back out. So it passes through the solar system on a hyperbolic trajectory. And if the projectiles were moving at high speed before falling into the solar system, it will be a very flat hyperbola -- i.e., they'll hardly be deflected at all by the Sun's gravity.



If the slugs are aimed to pass through at 2AU, there's basically nothing there to hit and all they do is think to themselves, "Here comes the Sun...there goes the Sun...hello interstellar darkness..."



(I'm not quite sure what you mean by "to pass through our system 15 degrees above the ecliptic." It you mean they are aimed to pass well above the plane of the planetary orbits, then there's still less likely to hit anything or even be noticed.)



The planets -- even Jupiter -- would have little effect on the projectiles since most of them would not come near enough to be affected, and the rest would only be deflected a bit. (Remember, they're passing by at greater than escape velocity. The deflection a body can provide depends on the projectile's velocity (the slower it's moving relative, the bigger the defection) and on the impact parameter (the closer the projectile passes the bigger the deflection.)



Hitting the Earth accidentally is hard. The Earth occupies .25(r/R)2 of a 2AU circle, where r is the Earth's radius and R is the Earth's orbital radius (the AU). That fraction is about 5*10-10. So less than one in a billion of the projectiles would hit Earth even if they were aimed to all pass through the ecliptic within 2AU of the Sun.



Finally, if they did hit the Earth we'd probably not notice them. A 100 metric ton slug of metal would be about 10 cubic meters if it's iron, less if it's something denser. That's a sphere less then ten feet in diameter. It would produce a nice bolide, but would come down pretty much intact, making a 50 meter crater with about a 4 kiloton explosion. Nothing you'd want to be next to, of course, but basically unnoticeable more than twenty miles away. Tunguska was close to 100 times bigger (though probably mostly ice) and its explosion was around four thousand times greater, and it was nearly missed. Had it hit water, it would have been missed.






share|improve this answer









$endgroup$









  • 4




    $begingroup$
    +1 for the physics. Also may be worth mentioning re the descriptive text that by the time a 100 ton object (which is pretty small) moving at 2.5 km/s or more becomes visible to the naked eye, any consequences are going to become apparent in less than a second, not hours later.
    $endgroup$
    – KerrAvon2055
    Dec 23 '18 at 2:12










  • $begingroup$
    @KerrAvon2055 Excellent point! (I think your comment says all that needs to be said, though.)
    $endgroup$
    – Mark Olson
    Dec 23 '18 at 2:13








  • 1




    $begingroup$
    Is that energy right? It's going to pick up 11 km/sec from hitting Earth and at least 30 km/sec from falling into the solar system. (Note that you can't just add these to 41 km/sec!)
    $endgroup$
    – Loren Pechtel
    Dec 23 '18 at 5:04






  • 1




    $begingroup$
    Excellent answer, but you don't address the "don't scatter" part of it. Unless the slugs are tethered, they will drift apart, even if they are all "pointing at" us, the smallest distance between them will place them in different orbits which in turn will spread them.
    $endgroup$
    – Diego Sánchez
    Dec 23 '18 at 8:40










  • $begingroup$
    Don't forget that planets are in motion, and so is the Sun relative to the Galaxy, so anything that passes through a gravity well will see its trajectory and speed modified. See 'gravity assist' or 'slingshot manoeuvre m' at ESA, NASA or Wikipedia. I would link but I'm on my phone. Also, if the Tunguska object would have fallen into the ocean, the resulting tsunami would have been noticed by coastal settlements in the vicinity. And certainly have puzzled scientists since it wouldn't correlate with any known seismic event.
    $endgroup$
    – Sava
    Dec 23 '18 at 9:38














23












23








23





$begingroup$

Virtually none of them would be captured and it's unlikely we'd notice anything whizzing past.



Basically, an object that falls into a gravity well from infinity (and a light year off is pretty close to that) will, unless it hits something while passing through, fall right back out again. By definition it picks up escape velocity as it falls in and loses it as it falls back out. So it passes through the solar system on a hyperbolic trajectory. And if the projectiles were moving at high speed before falling into the solar system, it will be a very flat hyperbola -- i.e., they'll hardly be deflected at all by the Sun's gravity.



If the slugs are aimed to pass through at 2AU, there's basically nothing there to hit and all they do is think to themselves, "Here comes the Sun...there goes the Sun...hello interstellar darkness..."



(I'm not quite sure what you mean by "to pass through our system 15 degrees above the ecliptic." It you mean they are aimed to pass well above the plane of the planetary orbits, then there's still less likely to hit anything or even be noticed.)



The planets -- even Jupiter -- would have little effect on the projectiles since most of them would not come near enough to be affected, and the rest would only be deflected a bit. (Remember, they're passing by at greater than escape velocity. The deflection a body can provide depends on the projectile's velocity (the slower it's moving relative, the bigger the defection) and on the impact parameter (the closer the projectile passes the bigger the deflection.)



Hitting the Earth accidentally is hard. The Earth occupies .25(r/R)2 of a 2AU circle, where r is the Earth's radius and R is the Earth's orbital radius (the AU). That fraction is about 5*10-10. So less than one in a billion of the projectiles would hit Earth even if they were aimed to all pass through the ecliptic within 2AU of the Sun.



Finally, if they did hit the Earth we'd probably not notice them. A 100 metric ton slug of metal would be about 10 cubic meters if it's iron, less if it's something denser. That's a sphere less then ten feet in diameter. It would produce a nice bolide, but would come down pretty much intact, making a 50 meter crater with about a 4 kiloton explosion. Nothing you'd want to be next to, of course, but basically unnoticeable more than twenty miles away. Tunguska was close to 100 times bigger (though probably mostly ice) and its explosion was around four thousand times greater, and it was nearly missed. Had it hit water, it would have been missed.






share|improve this answer









$endgroup$



Virtually none of them would be captured and it's unlikely we'd notice anything whizzing past.



Basically, an object that falls into a gravity well from infinity (and a light year off is pretty close to that) will, unless it hits something while passing through, fall right back out again. By definition it picks up escape velocity as it falls in and loses it as it falls back out. So it passes through the solar system on a hyperbolic trajectory. And if the projectiles were moving at high speed before falling into the solar system, it will be a very flat hyperbola -- i.e., they'll hardly be deflected at all by the Sun's gravity.



If the slugs are aimed to pass through at 2AU, there's basically nothing there to hit and all they do is think to themselves, "Here comes the Sun...there goes the Sun...hello interstellar darkness..."



(I'm not quite sure what you mean by "to pass through our system 15 degrees above the ecliptic." It you mean they are aimed to pass well above the plane of the planetary orbits, then there's still less likely to hit anything or even be noticed.)



The planets -- even Jupiter -- would have little effect on the projectiles since most of them would not come near enough to be affected, and the rest would only be deflected a bit. (Remember, they're passing by at greater than escape velocity. The deflection a body can provide depends on the projectile's velocity (the slower it's moving relative, the bigger the defection) and on the impact parameter (the closer the projectile passes the bigger the deflection.)



Hitting the Earth accidentally is hard. The Earth occupies .25(r/R)2 of a 2AU circle, where r is the Earth's radius and R is the Earth's orbital radius (the AU). That fraction is about 5*10-10. So less than one in a billion of the projectiles would hit Earth even if they were aimed to all pass through the ecliptic within 2AU of the Sun.



Finally, if they did hit the Earth we'd probably not notice them. A 100 metric ton slug of metal would be about 10 cubic meters if it's iron, less if it's something denser. That's a sphere less then ten feet in diameter. It would produce a nice bolide, but would come down pretty much intact, making a 50 meter crater with about a 4 kiloton explosion. Nothing you'd want to be next to, of course, but basically unnoticeable more than twenty miles away. Tunguska was close to 100 times bigger (though probably mostly ice) and its explosion was around four thousand times greater, and it was nearly missed. Had it hit water, it would have been missed.







share|improve this answer












share|improve this answer



share|improve this answer










answered Dec 23 '18 at 2:03









Mark OlsonMark Olson

11.3k12747




11.3k12747








  • 4




    $begingroup$
    +1 for the physics. Also may be worth mentioning re the descriptive text that by the time a 100 ton object (which is pretty small) moving at 2.5 km/s or more becomes visible to the naked eye, any consequences are going to become apparent in less than a second, not hours later.
    $endgroup$
    – KerrAvon2055
    Dec 23 '18 at 2:12










  • $begingroup$
    @KerrAvon2055 Excellent point! (I think your comment says all that needs to be said, though.)
    $endgroup$
    – Mark Olson
    Dec 23 '18 at 2:13








  • 1




    $begingroup$
    Is that energy right? It's going to pick up 11 km/sec from hitting Earth and at least 30 km/sec from falling into the solar system. (Note that you can't just add these to 41 km/sec!)
    $endgroup$
    – Loren Pechtel
    Dec 23 '18 at 5:04






  • 1




    $begingroup$
    Excellent answer, but you don't address the "don't scatter" part of it. Unless the slugs are tethered, they will drift apart, even if they are all "pointing at" us, the smallest distance between them will place them in different orbits which in turn will spread them.
    $endgroup$
    – Diego Sánchez
    Dec 23 '18 at 8:40










  • $begingroup$
    Don't forget that planets are in motion, and so is the Sun relative to the Galaxy, so anything that passes through a gravity well will see its trajectory and speed modified. See 'gravity assist' or 'slingshot manoeuvre m' at ESA, NASA or Wikipedia. I would link but I'm on my phone. Also, if the Tunguska object would have fallen into the ocean, the resulting tsunami would have been noticed by coastal settlements in the vicinity. And certainly have puzzled scientists since it wouldn't correlate with any known seismic event.
    $endgroup$
    – Sava
    Dec 23 '18 at 9:38














  • 4




    $begingroup$
    +1 for the physics. Also may be worth mentioning re the descriptive text that by the time a 100 ton object (which is pretty small) moving at 2.5 km/s or more becomes visible to the naked eye, any consequences are going to become apparent in less than a second, not hours later.
    $endgroup$
    – KerrAvon2055
    Dec 23 '18 at 2:12










  • $begingroup$
    @KerrAvon2055 Excellent point! (I think your comment says all that needs to be said, though.)
    $endgroup$
    – Mark Olson
    Dec 23 '18 at 2:13








  • 1




    $begingroup$
    Is that energy right? It's going to pick up 11 km/sec from hitting Earth and at least 30 km/sec from falling into the solar system. (Note that you can't just add these to 41 km/sec!)
    $endgroup$
    – Loren Pechtel
    Dec 23 '18 at 5:04






  • 1




    $begingroup$
    Excellent answer, but you don't address the "don't scatter" part of it. Unless the slugs are tethered, they will drift apart, even if they are all "pointing at" us, the smallest distance between them will place them in different orbits which in turn will spread them.
    $endgroup$
    – Diego Sánchez
    Dec 23 '18 at 8:40










  • $begingroup$
    Don't forget that planets are in motion, and so is the Sun relative to the Galaxy, so anything that passes through a gravity well will see its trajectory and speed modified. See 'gravity assist' or 'slingshot manoeuvre m' at ESA, NASA or Wikipedia. I would link but I'm on my phone. Also, if the Tunguska object would have fallen into the ocean, the resulting tsunami would have been noticed by coastal settlements in the vicinity. And certainly have puzzled scientists since it wouldn't correlate with any known seismic event.
    $endgroup$
    – Sava
    Dec 23 '18 at 9:38








4




4




$begingroup$
+1 for the physics. Also may be worth mentioning re the descriptive text that by the time a 100 ton object (which is pretty small) moving at 2.5 km/s or more becomes visible to the naked eye, any consequences are going to become apparent in less than a second, not hours later.
$endgroup$
– KerrAvon2055
Dec 23 '18 at 2:12




$begingroup$
+1 for the physics. Also may be worth mentioning re the descriptive text that by the time a 100 ton object (which is pretty small) moving at 2.5 km/s or more becomes visible to the naked eye, any consequences are going to become apparent in less than a second, not hours later.
$endgroup$
– KerrAvon2055
Dec 23 '18 at 2:12












$begingroup$
@KerrAvon2055 Excellent point! (I think your comment says all that needs to be said, though.)
$endgroup$
– Mark Olson
Dec 23 '18 at 2:13






$begingroup$
@KerrAvon2055 Excellent point! (I think your comment says all that needs to be said, though.)
$endgroup$
– Mark Olson
Dec 23 '18 at 2:13






1




1




$begingroup$
Is that energy right? It's going to pick up 11 km/sec from hitting Earth and at least 30 km/sec from falling into the solar system. (Note that you can't just add these to 41 km/sec!)
$endgroup$
– Loren Pechtel
Dec 23 '18 at 5:04




$begingroup$
Is that energy right? It's going to pick up 11 km/sec from hitting Earth and at least 30 km/sec from falling into the solar system. (Note that you can't just add these to 41 km/sec!)
$endgroup$
– Loren Pechtel
Dec 23 '18 at 5:04




1




1




$begingroup$
Excellent answer, but you don't address the "don't scatter" part of it. Unless the slugs are tethered, they will drift apart, even if they are all "pointing at" us, the smallest distance between them will place them in different orbits which in turn will spread them.
$endgroup$
– Diego Sánchez
Dec 23 '18 at 8:40




$begingroup$
Excellent answer, but you don't address the "don't scatter" part of it. Unless the slugs are tethered, they will drift apart, even if they are all "pointing at" us, the smallest distance between them will place them in different orbits which in turn will spread them.
$endgroup$
– Diego Sánchez
Dec 23 '18 at 8:40












$begingroup$
Don't forget that planets are in motion, and so is the Sun relative to the Galaxy, so anything that passes through a gravity well will see its trajectory and speed modified. See 'gravity assist' or 'slingshot manoeuvre m' at ESA, NASA or Wikipedia. I would link but I'm on my phone. Also, if the Tunguska object would have fallen into the ocean, the resulting tsunami would have been noticed by coastal settlements in the vicinity. And certainly have puzzled scientists since it wouldn't correlate with any known seismic event.
$endgroup$
– Sava
Dec 23 '18 at 9:38




$begingroup$
Don't forget that planets are in motion, and so is the Sun relative to the Galaxy, so anything that passes through a gravity well will see its trajectory and speed modified. See 'gravity assist' or 'slingshot manoeuvre m' at ESA, NASA or Wikipedia. I would link but I'm on my phone. Also, if the Tunguska object would have fallen into the ocean, the resulting tsunami would have been noticed by coastal settlements in the vicinity. And certainly have puzzled scientists since it wouldn't correlate with any known seismic event.
$endgroup$
– Sava
Dec 23 '18 at 9:38











10












$begingroup$

To be captured by the sun's gravity well means to end up orbiting the sun.



If a projectile is fired from outside the solar system, it may do a flyby around the sun. Its speed at any altitude (relative to the sun) will be greater than the escape velocity at those points.



Scientists believe this is exactly what the Oumuamua object did.



For the flotsam to be captured, it must be decelerated by a planet. Jupiter is the best candidate to do so, but the odds are against it. Even if the flotsam does a flyby by Jupiter, it is more likely to either hit the planet or be accelerated even further, causing it to exit the solar system earlier.



If it does get decelerated by a planet, the flotsam will be locked into an orbit that is unlikely to cross the Earth's own. It will cross Jupiter's orbit during its perihelion, which means it may crash into Jupiter eobs later, and it has an astronomical probability of having an orbit even more inclined relative to the Earth's than Pluto. All this makes an encounter with Earth very unlikely.






share|improve this answer











$endgroup$









  • 2




    $begingroup$
    And note that even if one is captured that doesn't mean another nearby one gets captured.
    $endgroup$
    – Loren Pechtel
    Dec 23 '18 at 5:05






  • 2




    $begingroup$
    Oumuamua is a good example. It was too fast to get captured in the gravity well so it went on its way. But that does not mean an object of this sort could not have hit earth.
    $endgroup$
    – Willk
    Dec 23 '18 at 17:23


















10












$begingroup$

To be captured by the sun's gravity well means to end up orbiting the sun.



If a projectile is fired from outside the solar system, it may do a flyby around the sun. Its speed at any altitude (relative to the sun) will be greater than the escape velocity at those points.



Scientists believe this is exactly what the Oumuamua object did.



For the flotsam to be captured, it must be decelerated by a planet. Jupiter is the best candidate to do so, but the odds are against it. Even if the flotsam does a flyby by Jupiter, it is more likely to either hit the planet or be accelerated even further, causing it to exit the solar system earlier.



If it does get decelerated by a planet, the flotsam will be locked into an orbit that is unlikely to cross the Earth's own. It will cross Jupiter's orbit during its perihelion, which means it may crash into Jupiter eobs later, and it has an astronomical probability of having an orbit even more inclined relative to the Earth's than Pluto. All this makes an encounter with Earth very unlikely.






share|improve this answer











$endgroup$









  • 2




    $begingroup$
    And note that even if one is captured that doesn't mean another nearby one gets captured.
    $endgroup$
    – Loren Pechtel
    Dec 23 '18 at 5:05






  • 2




    $begingroup$
    Oumuamua is a good example. It was too fast to get captured in the gravity well so it went on its way. But that does not mean an object of this sort could not have hit earth.
    $endgroup$
    – Willk
    Dec 23 '18 at 17:23
















10












10








10





$begingroup$

To be captured by the sun's gravity well means to end up orbiting the sun.



If a projectile is fired from outside the solar system, it may do a flyby around the sun. Its speed at any altitude (relative to the sun) will be greater than the escape velocity at those points.



Scientists believe this is exactly what the Oumuamua object did.



For the flotsam to be captured, it must be decelerated by a planet. Jupiter is the best candidate to do so, but the odds are against it. Even if the flotsam does a flyby by Jupiter, it is more likely to either hit the planet or be accelerated even further, causing it to exit the solar system earlier.



If it does get decelerated by a planet, the flotsam will be locked into an orbit that is unlikely to cross the Earth's own. It will cross Jupiter's orbit during its perihelion, which means it may crash into Jupiter eobs later, and it has an astronomical probability of having an orbit even more inclined relative to the Earth's than Pluto. All this makes an encounter with Earth very unlikely.






share|improve this answer











$endgroup$



To be captured by the sun's gravity well means to end up orbiting the sun.



If a projectile is fired from outside the solar system, it may do a flyby around the sun. Its speed at any altitude (relative to the sun) will be greater than the escape velocity at those points.



Scientists believe this is exactly what the Oumuamua object did.



For the flotsam to be captured, it must be decelerated by a planet. Jupiter is the best candidate to do so, but the odds are against it. Even if the flotsam does a flyby by Jupiter, it is more likely to either hit the planet or be accelerated even further, causing it to exit the solar system earlier.



If it does get decelerated by a planet, the flotsam will be locked into an orbit that is unlikely to cross the Earth's own. It will cross Jupiter's orbit during its perihelion, which means it may crash into Jupiter eobs later, and it has an astronomical probability of having an orbit even more inclined relative to the Earth's than Pluto. All this makes an encounter with Earth very unlikely.







share|improve this answer














share|improve this answer



share|improve this answer








edited Dec 23 '18 at 4:29









L.Dutch♦

83.1k28200411




83.1k28200411










answered Dec 23 '18 at 4:27









RenanRenan

47.8k13111243




47.8k13111243








  • 2




    $begingroup$
    And note that even if one is captured that doesn't mean another nearby one gets captured.
    $endgroup$
    – Loren Pechtel
    Dec 23 '18 at 5:05






  • 2




    $begingroup$
    Oumuamua is a good example. It was too fast to get captured in the gravity well so it went on its way. But that does not mean an object of this sort could not have hit earth.
    $endgroup$
    – Willk
    Dec 23 '18 at 17:23
















  • 2




    $begingroup$
    And note that even if one is captured that doesn't mean another nearby one gets captured.
    $endgroup$
    – Loren Pechtel
    Dec 23 '18 at 5:05






  • 2




    $begingroup$
    Oumuamua is a good example. It was too fast to get captured in the gravity well so it went on its way. But that does not mean an object of this sort could not have hit earth.
    $endgroup$
    – Willk
    Dec 23 '18 at 17:23










2




2




$begingroup$
And note that even if one is captured that doesn't mean another nearby one gets captured.
$endgroup$
– Loren Pechtel
Dec 23 '18 at 5:05




$begingroup$
And note that even if one is captured that doesn't mean another nearby one gets captured.
$endgroup$
– Loren Pechtel
Dec 23 '18 at 5:05




2




2




$begingroup$
Oumuamua is a good example. It was too fast to get captured in the gravity well so it went on its way. But that does not mean an object of this sort could not have hit earth.
$endgroup$
– Willk
Dec 23 '18 at 17:23






$begingroup$
Oumuamua is a good example. It was too fast to get captured in the gravity well so it went on its way. But that does not mean an object of this sort could not have hit earth.
$endgroup$
– Willk
Dec 23 '18 at 17:23













3












$begingroup$

Short answer is a space weapon fired at any sort of realistic speed will never be captured by the Sun's gravity well. Consider that mere human technology has created nuclear pumped weapons like project Prometheus, which captured a small fraction of a nuclear explosion to fire a stream of pellets at 100km/sec. Changing some elements allows you to use similar equipment to make shaped charges, Explosively formed projectiles and even streams of star hot plasma moving at speeds measured in percentages of the speed of light. In short, your space armada would be sliced to pieces before they get a shot in using weapons with velocities of 2.5km/sec. In fact, orbital velocity around the Earth is already 7km/sec.



Realistic space weapons will be moving so fast that they will never be captured, while extremely slow moving bodies like you describe will likely not reach Earth at all, but either drift harmlessly past, or be captured by some of the Gas Giant planets in the outer Solar System.






share|improve this answer









$endgroup$













  • $begingroup$
    Off the top of my head, 100 metric tons, 2 AU from the sun traveling at 2.5 Km/s, (tangent to? and) 15 degrees above the ecliptic, is not a trajectory that will put you into solar orbit (overshot by at least, I'm guessing, 1 Km/s). This question belongs on Math.SE
    $endgroup$
    – Mazura
    Dec 23 '18 at 20:33
















3












$begingroup$

Short answer is a space weapon fired at any sort of realistic speed will never be captured by the Sun's gravity well. Consider that mere human technology has created nuclear pumped weapons like project Prometheus, which captured a small fraction of a nuclear explosion to fire a stream of pellets at 100km/sec. Changing some elements allows you to use similar equipment to make shaped charges, Explosively formed projectiles and even streams of star hot plasma moving at speeds measured in percentages of the speed of light. In short, your space armada would be sliced to pieces before they get a shot in using weapons with velocities of 2.5km/sec. In fact, orbital velocity around the Earth is already 7km/sec.



Realistic space weapons will be moving so fast that they will never be captured, while extremely slow moving bodies like you describe will likely not reach Earth at all, but either drift harmlessly past, or be captured by some of the Gas Giant planets in the outer Solar System.






share|improve this answer









$endgroup$













  • $begingroup$
    Off the top of my head, 100 metric tons, 2 AU from the sun traveling at 2.5 Km/s, (tangent to? and) 15 degrees above the ecliptic, is not a trajectory that will put you into solar orbit (overshot by at least, I'm guessing, 1 Km/s). This question belongs on Math.SE
    $endgroup$
    – Mazura
    Dec 23 '18 at 20:33














3












3








3





$begingroup$

Short answer is a space weapon fired at any sort of realistic speed will never be captured by the Sun's gravity well. Consider that mere human technology has created nuclear pumped weapons like project Prometheus, which captured a small fraction of a nuclear explosion to fire a stream of pellets at 100km/sec. Changing some elements allows you to use similar equipment to make shaped charges, Explosively formed projectiles and even streams of star hot plasma moving at speeds measured in percentages of the speed of light. In short, your space armada would be sliced to pieces before they get a shot in using weapons with velocities of 2.5km/sec. In fact, orbital velocity around the Earth is already 7km/sec.



Realistic space weapons will be moving so fast that they will never be captured, while extremely slow moving bodies like you describe will likely not reach Earth at all, but either drift harmlessly past, or be captured by some of the Gas Giant planets in the outer Solar System.






share|improve this answer









$endgroup$



Short answer is a space weapon fired at any sort of realistic speed will never be captured by the Sun's gravity well. Consider that mere human technology has created nuclear pumped weapons like project Prometheus, which captured a small fraction of a nuclear explosion to fire a stream of pellets at 100km/sec. Changing some elements allows you to use similar equipment to make shaped charges, Explosively formed projectiles and even streams of star hot plasma moving at speeds measured in percentages of the speed of light. In short, your space armada would be sliced to pieces before they get a shot in using weapons with velocities of 2.5km/sec. In fact, orbital velocity around the Earth is already 7km/sec.



Realistic space weapons will be moving so fast that they will never be captured, while extremely slow moving bodies like you describe will likely not reach Earth at all, but either drift harmlessly past, or be captured by some of the Gas Giant planets in the outer Solar System.







share|improve this answer












share|improve this answer



share|improve this answer










answered Dec 23 '18 at 2:13









ThucydidesThucydides

81.8k678243




81.8k678243












  • $begingroup$
    Off the top of my head, 100 metric tons, 2 AU from the sun traveling at 2.5 Km/s, (tangent to? and) 15 degrees above the ecliptic, is not a trajectory that will put you into solar orbit (overshot by at least, I'm guessing, 1 Km/s). This question belongs on Math.SE
    $endgroup$
    – Mazura
    Dec 23 '18 at 20:33


















  • $begingroup$
    Off the top of my head, 100 metric tons, 2 AU from the sun traveling at 2.5 Km/s, (tangent to? and) 15 degrees above the ecliptic, is not a trajectory that will put you into solar orbit (overshot by at least, I'm guessing, 1 Km/s). This question belongs on Math.SE
    $endgroup$
    – Mazura
    Dec 23 '18 at 20:33
















$begingroup$
Off the top of my head, 100 metric tons, 2 AU from the sun traveling at 2.5 Km/s, (tangent to? and) 15 degrees above the ecliptic, is not a trajectory that will put you into solar orbit (overshot by at least, I'm guessing, 1 Km/s). This question belongs on Math.SE
$endgroup$
– Mazura
Dec 23 '18 at 20:33




$begingroup$
Off the top of my head, 100 metric tons, 2 AU from the sun traveling at 2.5 Km/s, (tangent to? and) 15 degrees above the ecliptic, is not a trajectory that will put you into solar orbit (overshot by at least, I'm guessing, 1 Km/s). This question belongs on Math.SE
$endgroup$
– Mazura
Dec 23 '18 at 20:33











1












$begingroup$

Muzzle velocity seems less important than projectile velocity as seen from Sol.



Give the weapons provided, it seems likely that fleets in motion, starting from bases also in motion, would need to come to (relative) near-rest with each other to have the fight at all...else their weapons would not score at all as the fleets flashed past each other in a fraction of a second and then spent weeks shedding velocity for another attacking run.



That point of relative near-rest-for-the-fight is still moving, from the perspective of everybody else in the galaxy. It orbits the galactic center, and it has a vector from the perspective of all the stars...including Sol.



And if that apparent vector, from Sol's perspective, is greater than about 7km/s at closest approach of 2AU, then Sol will probably not capture it at all. Flotsam just below that limit would likely go into comet-like hyperbolic orbits and may not return to the inner system for millenia.



Flotsam really slowly drifting toward Sol would be captured and become a minor menace to inner planets...but only during the two times each orbit that they intersect the ecliptic...and only until their crazy, unstable orbits dump them into Sol.






share|improve this answer









$endgroup$









  • 1




    $begingroup$
    The fleets could have been passing at well above weapon velocity--they shoot all their weapons as they approach the enemy fleet so they pass through the same place in space. This assumes drives that can build high velocity but only very slowly so they can't simply evade the slower but more maneuverable weapons.
    $endgroup$
    – Loren Pechtel
    Dec 23 '18 at 5:08






  • 1




    $begingroup$
    @LorenPechtel, Agreed, or that both fleets are essentially flying through hasty minefields. Both of those possibilities didn't seem to match the romantic sense in the OP's question, so I ignored them.
    $endgroup$
    – user535733
    Dec 23 '18 at 13:30
















1












$begingroup$

Muzzle velocity seems less important than projectile velocity as seen from Sol.



Give the weapons provided, it seems likely that fleets in motion, starting from bases also in motion, would need to come to (relative) near-rest with each other to have the fight at all...else their weapons would not score at all as the fleets flashed past each other in a fraction of a second and then spent weeks shedding velocity for another attacking run.



That point of relative near-rest-for-the-fight is still moving, from the perspective of everybody else in the galaxy. It orbits the galactic center, and it has a vector from the perspective of all the stars...including Sol.



And if that apparent vector, from Sol's perspective, is greater than about 7km/s at closest approach of 2AU, then Sol will probably not capture it at all. Flotsam just below that limit would likely go into comet-like hyperbolic orbits and may not return to the inner system for millenia.



Flotsam really slowly drifting toward Sol would be captured and become a minor menace to inner planets...but only during the two times each orbit that they intersect the ecliptic...and only until their crazy, unstable orbits dump them into Sol.






share|improve this answer









$endgroup$









  • 1




    $begingroup$
    The fleets could have been passing at well above weapon velocity--they shoot all their weapons as they approach the enemy fleet so they pass through the same place in space. This assumes drives that can build high velocity but only very slowly so they can't simply evade the slower but more maneuverable weapons.
    $endgroup$
    – Loren Pechtel
    Dec 23 '18 at 5:08






  • 1




    $begingroup$
    @LorenPechtel, Agreed, or that both fleets are essentially flying through hasty minefields. Both of those possibilities didn't seem to match the romantic sense in the OP's question, so I ignored them.
    $endgroup$
    – user535733
    Dec 23 '18 at 13:30














1












1








1





$begingroup$

Muzzle velocity seems less important than projectile velocity as seen from Sol.



Give the weapons provided, it seems likely that fleets in motion, starting from bases also in motion, would need to come to (relative) near-rest with each other to have the fight at all...else their weapons would not score at all as the fleets flashed past each other in a fraction of a second and then spent weeks shedding velocity for another attacking run.



That point of relative near-rest-for-the-fight is still moving, from the perspective of everybody else in the galaxy. It orbits the galactic center, and it has a vector from the perspective of all the stars...including Sol.



And if that apparent vector, from Sol's perspective, is greater than about 7km/s at closest approach of 2AU, then Sol will probably not capture it at all. Flotsam just below that limit would likely go into comet-like hyperbolic orbits and may not return to the inner system for millenia.



Flotsam really slowly drifting toward Sol would be captured and become a minor menace to inner planets...but only during the two times each orbit that they intersect the ecliptic...and only until their crazy, unstable orbits dump them into Sol.






share|improve this answer









$endgroup$



Muzzle velocity seems less important than projectile velocity as seen from Sol.



Give the weapons provided, it seems likely that fleets in motion, starting from bases also in motion, would need to come to (relative) near-rest with each other to have the fight at all...else their weapons would not score at all as the fleets flashed past each other in a fraction of a second and then spent weeks shedding velocity for another attacking run.



That point of relative near-rest-for-the-fight is still moving, from the perspective of everybody else in the galaxy. It orbits the galactic center, and it has a vector from the perspective of all the stars...including Sol.



And if that apparent vector, from Sol's perspective, is greater than about 7km/s at closest approach of 2AU, then Sol will probably not capture it at all. Flotsam just below that limit would likely go into comet-like hyperbolic orbits and may not return to the inner system for millenia.



Flotsam really slowly drifting toward Sol would be captured and become a minor menace to inner planets...but only during the two times each orbit that they intersect the ecliptic...and only until their crazy, unstable orbits dump them into Sol.







share|improve this answer












share|improve this answer



share|improve this answer










answered Dec 23 '18 at 2:12









user535733user535733

8,67921838




8,67921838








  • 1




    $begingroup$
    The fleets could have been passing at well above weapon velocity--they shoot all their weapons as they approach the enemy fleet so they pass through the same place in space. This assumes drives that can build high velocity but only very slowly so they can't simply evade the slower but more maneuverable weapons.
    $endgroup$
    – Loren Pechtel
    Dec 23 '18 at 5:08






  • 1




    $begingroup$
    @LorenPechtel, Agreed, or that both fleets are essentially flying through hasty minefields. Both of those possibilities didn't seem to match the romantic sense in the OP's question, so I ignored them.
    $endgroup$
    – user535733
    Dec 23 '18 at 13:30














  • 1




    $begingroup$
    The fleets could have been passing at well above weapon velocity--they shoot all their weapons as they approach the enemy fleet so they pass through the same place in space. This assumes drives that can build high velocity but only very slowly so they can't simply evade the slower but more maneuverable weapons.
    $endgroup$
    – Loren Pechtel
    Dec 23 '18 at 5:08






  • 1




    $begingroup$
    @LorenPechtel, Agreed, or that both fleets are essentially flying through hasty minefields. Both of those possibilities didn't seem to match the romantic sense in the OP's question, so I ignored them.
    $endgroup$
    – user535733
    Dec 23 '18 at 13:30








1




1




$begingroup$
The fleets could have been passing at well above weapon velocity--they shoot all their weapons as they approach the enemy fleet so they pass through the same place in space. This assumes drives that can build high velocity but only very slowly so they can't simply evade the slower but more maneuverable weapons.
$endgroup$
– Loren Pechtel
Dec 23 '18 at 5:08




$begingroup$
The fleets could have been passing at well above weapon velocity--they shoot all their weapons as they approach the enemy fleet so they pass through the same place in space. This assumes drives that can build high velocity but only very slowly so they can't simply evade the slower but more maneuverable weapons.
$endgroup$
– Loren Pechtel
Dec 23 '18 at 5:08




1




1




$begingroup$
@LorenPechtel, Agreed, or that both fleets are essentially flying through hasty minefields. Both of those possibilities didn't seem to match the romantic sense in the OP's question, so I ignored them.
$endgroup$
– user535733
Dec 23 '18 at 13:30




$begingroup$
@LorenPechtel, Agreed, or that both fleets are essentially flying through hasty minefields. Both of those possibilities didn't seem to match the romantic sense in the OP's question, so I ignored them.
$endgroup$
– user535733
Dec 23 '18 at 13:30











0












$begingroup$

They are most likely to end up captured by Jupiter's gravity well first. If they manage to survive that long.



Jupiter is the second biggest object in our solar system after the Sun and possess a vast gravity well. There's a reason why it is used for slingshot maneuvers and accelerations by many of the probes sent throughout the system by the various space agencies. They even sent a solar probe to Jupiter to alter it's trajectory so it could fly by the poles of the sun.



2 AU is a little under the distance of the Asteroid Belt to the Sun, so they would definitely be in the gravitational influence of the Sun.



As for hitting Earth, which is, all things considered, an orbiting mote of dust around the Sun, it seems very improbable. The path of the flotsam would have to intersect Earth's orbit at a point in space and time that is also occupied by Earth at the same moment.



I do not know how to calculate or model the path of your piece of flotsam given the parameters given, but I would rather expect it to either fall into Jupiter or the Sun, or be sent into the depths of interstellar space, than hit anything else in the system.






share|improve this answer









$endgroup$









  • 3




    $begingroup$
    Nearly downvoted. You do realize, I hope, that gravity wells are symmetric? That is, the incoming velocity equals the outgoing velocity? As Mark Olsen discussed in his answer, a body falling toward a large one from far away will end up far away. And as for "fall(ing) into Jupiter or the Sun", that is extremely unlikely, particularly since they are originally pointed at 2 AU out and 15 degrees above the plane of the ecliptic.
    $endgroup$
    – WhatRoughBeast
    Dec 23 '18 at 4:23










  • $begingroup$
    You forget one thing: planets are in motion and so are their gravity wells. Thus, depending upon angle and vector of entry into the gravity well relative to the planet's trajectory around the Sun, any small celestial object will end up accelerating or decelerating and have their orbital path modified. Please look up 'slingshot manoeuvre' or 'gravity assist' at ESA, NASA or even Wikipedia. I would link, but I'm on my phone at the moment.
    $endgroup$
    – Sava
    Dec 23 '18 at 9:30






  • 2




    $begingroup$
    Everything you say is true - but irrelevant. Your opening sentence, "They are most likely to end up captured by Jupiter's gravity well first." is unfounded, unsupported by your statement, and untrue. In the absence of a collision, any object inbound from 1 light-year with an initial velocity of 2.5 km/sec which is out of the plane of the ecliptic is not going to get near enough to an object like Jupiter to be grossly affected.
    $endgroup$
    – WhatRoughBeast
    Dec 23 '18 at 16:17










  • $begingroup$
    Can you prove that my opening sentence is 'unfounded and untrue' and that yours is? Jupiter has a massive gravity well that extends far away in space, it's been calculated that it is responsible from perturbations in Mercury's orbit. And, at 2.5 km/s, the flotsam is moving slowly enough to be easily captured, given that Jupiter captured comets that moved way faster, like Shoemaker-Levy 9, which is believed to have been in solar orbit prior to being 'stolen' by Jupiter.
    $endgroup$
    – Sava
    Dec 23 '18 at 22:48










  • $begingroup$
    After the object has entered the solar system, it will be going way faster than 2.5 km/s—about 20 km/s at Jupiter’s distance from the sun. This is why it has an extremely low probability of being captured by anything.
    $endgroup$
    – prl
    Dec 24 '18 at 1:20


















0












$begingroup$

They are most likely to end up captured by Jupiter's gravity well first. If they manage to survive that long.



Jupiter is the second biggest object in our solar system after the Sun and possess a vast gravity well. There's a reason why it is used for slingshot maneuvers and accelerations by many of the probes sent throughout the system by the various space agencies. They even sent a solar probe to Jupiter to alter it's trajectory so it could fly by the poles of the sun.



2 AU is a little under the distance of the Asteroid Belt to the Sun, so they would definitely be in the gravitational influence of the Sun.



As for hitting Earth, which is, all things considered, an orbiting mote of dust around the Sun, it seems very improbable. The path of the flotsam would have to intersect Earth's orbit at a point in space and time that is also occupied by Earth at the same moment.



I do not know how to calculate or model the path of your piece of flotsam given the parameters given, but I would rather expect it to either fall into Jupiter or the Sun, or be sent into the depths of interstellar space, than hit anything else in the system.






share|improve this answer









$endgroup$









  • 3




    $begingroup$
    Nearly downvoted. You do realize, I hope, that gravity wells are symmetric? That is, the incoming velocity equals the outgoing velocity? As Mark Olsen discussed in his answer, a body falling toward a large one from far away will end up far away. And as for "fall(ing) into Jupiter or the Sun", that is extremely unlikely, particularly since they are originally pointed at 2 AU out and 15 degrees above the plane of the ecliptic.
    $endgroup$
    – WhatRoughBeast
    Dec 23 '18 at 4:23










  • $begingroup$
    You forget one thing: planets are in motion and so are their gravity wells. Thus, depending upon angle and vector of entry into the gravity well relative to the planet's trajectory around the Sun, any small celestial object will end up accelerating or decelerating and have their orbital path modified. Please look up 'slingshot manoeuvre' or 'gravity assist' at ESA, NASA or even Wikipedia. I would link, but I'm on my phone at the moment.
    $endgroup$
    – Sava
    Dec 23 '18 at 9:30






  • 2




    $begingroup$
    Everything you say is true - but irrelevant. Your opening sentence, "They are most likely to end up captured by Jupiter's gravity well first." is unfounded, unsupported by your statement, and untrue. In the absence of a collision, any object inbound from 1 light-year with an initial velocity of 2.5 km/sec which is out of the plane of the ecliptic is not going to get near enough to an object like Jupiter to be grossly affected.
    $endgroup$
    – WhatRoughBeast
    Dec 23 '18 at 16:17










  • $begingroup$
    Can you prove that my opening sentence is 'unfounded and untrue' and that yours is? Jupiter has a massive gravity well that extends far away in space, it's been calculated that it is responsible from perturbations in Mercury's orbit. And, at 2.5 km/s, the flotsam is moving slowly enough to be easily captured, given that Jupiter captured comets that moved way faster, like Shoemaker-Levy 9, which is believed to have been in solar orbit prior to being 'stolen' by Jupiter.
    $endgroup$
    – Sava
    Dec 23 '18 at 22:48










  • $begingroup$
    After the object has entered the solar system, it will be going way faster than 2.5 km/s—about 20 km/s at Jupiter’s distance from the sun. This is why it has an extremely low probability of being captured by anything.
    $endgroup$
    – prl
    Dec 24 '18 at 1:20
















0












0








0





$begingroup$

They are most likely to end up captured by Jupiter's gravity well first. If they manage to survive that long.



Jupiter is the second biggest object in our solar system after the Sun and possess a vast gravity well. There's a reason why it is used for slingshot maneuvers and accelerations by many of the probes sent throughout the system by the various space agencies. They even sent a solar probe to Jupiter to alter it's trajectory so it could fly by the poles of the sun.



2 AU is a little under the distance of the Asteroid Belt to the Sun, so they would definitely be in the gravitational influence of the Sun.



As for hitting Earth, which is, all things considered, an orbiting mote of dust around the Sun, it seems very improbable. The path of the flotsam would have to intersect Earth's orbit at a point in space and time that is also occupied by Earth at the same moment.



I do not know how to calculate or model the path of your piece of flotsam given the parameters given, but I would rather expect it to either fall into Jupiter or the Sun, or be sent into the depths of interstellar space, than hit anything else in the system.






share|improve this answer









$endgroup$



They are most likely to end up captured by Jupiter's gravity well first. If they manage to survive that long.



Jupiter is the second biggest object in our solar system after the Sun and possess a vast gravity well. There's a reason why it is used for slingshot maneuvers and accelerations by many of the probes sent throughout the system by the various space agencies. They even sent a solar probe to Jupiter to alter it's trajectory so it could fly by the poles of the sun.



2 AU is a little under the distance of the Asteroid Belt to the Sun, so they would definitely be in the gravitational influence of the Sun.



As for hitting Earth, which is, all things considered, an orbiting mote of dust around the Sun, it seems very improbable. The path of the flotsam would have to intersect Earth's orbit at a point in space and time that is also occupied by Earth at the same moment.



I do not know how to calculate or model the path of your piece of flotsam given the parameters given, but I would rather expect it to either fall into Jupiter or the Sun, or be sent into the depths of interstellar space, than hit anything else in the system.







share|improve this answer












share|improve this answer



share|improve this answer










answered Dec 23 '18 at 1:02









SavaSava

1,5961322




1,5961322








  • 3




    $begingroup$
    Nearly downvoted. You do realize, I hope, that gravity wells are symmetric? That is, the incoming velocity equals the outgoing velocity? As Mark Olsen discussed in his answer, a body falling toward a large one from far away will end up far away. And as for "fall(ing) into Jupiter or the Sun", that is extremely unlikely, particularly since they are originally pointed at 2 AU out and 15 degrees above the plane of the ecliptic.
    $endgroup$
    – WhatRoughBeast
    Dec 23 '18 at 4:23










  • $begingroup$
    You forget one thing: planets are in motion and so are their gravity wells. Thus, depending upon angle and vector of entry into the gravity well relative to the planet's trajectory around the Sun, any small celestial object will end up accelerating or decelerating and have their orbital path modified. Please look up 'slingshot manoeuvre' or 'gravity assist' at ESA, NASA or even Wikipedia. I would link, but I'm on my phone at the moment.
    $endgroup$
    – Sava
    Dec 23 '18 at 9:30






  • 2




    $begingroup$
    Everything you say is true - but irrelevant. Your opening sentence, "They are most likely to end up captured by Jupiter's gravity well first." is unfounded, unsupported by your statement, and untrue. In the absence of a collision, any object inbound from 1 light-year with an initial velocity of 2.5 km/sec which is out of the plane of the ecliptic is not going to get near enough to an object like Jupiter to be grossly affected.
    $endgroup$
    – WhatRoughBeast
    Dec 23 '18 at 16:17










  • $begingroup$
    Can you prove that my opening sentence is 'unfounded and untrue' and that yours is? Jupiter has a massive gravity well that extends far away in space, it's been calculated that it is responsible from perturbations in Mercury's orbit. And, at 2.5 km/s, the flotsam is moving slowly enough to be easily captured, given that Jupiter captured comets that moved way faster, like Shoemaker-Levy 9, which is believed to have been in solar orbit prior to being 'stolen' by Jupiter.
    $endgroup$
    – Sava
    Dec 23 '18 at 22:48










  • $begingroup$
    After the object has entered the solar system, it will be going way faster than 2.5 km/s—about 20 km/s at Jupiter’s distance from the sun. This is why it has an extremely low probability of being captured by anything.
    $endgroup$
    – prl
    Dec 24 '18 at 1:20
















  • 3




    $begingroup$
    Nearly downvoted. You do realize, I hope, that gravity wells are symmetric? That is, the incoming velocity equals the outgoing velocity? As Mark Olsen discussed in his answer, a body falling toward a large one from far away will end up far away. And as for "fall(ing) into Jupiter or the Sun", that is extremely unlikely, particularly since they are originally pointed at 2 AU out and 15 degrees above the plane of the ecliptic.
    $endgroup$
    – WhatRoughBeast
    Dec 23 '18 at 4:23










  • $begingroup$
    You forget one thing: planets are in motion and so are their gravity wells. Thus, depending upon angle and vector of entry into the gravity well relative to the planet's trajectory around the Sun, any small celestial object will end up accelerating or decelerating and have their orbital path modified. Please look up 'slingshot manoeuvre' or 'gravity assist' at ESA, NASA or even Wikipedia. I would link, but I'm on my phone at the moment.
    $endgroup$
    – Sava
    Dec 23 '18 at 9:30






  • 2




    $begingroup$
    Everything you say is true - but irrelevant. Your opening sentence, "They are most likely to end up captured by Jupiter's gravity well first." is unfounded, unsupported by your statement, and untrue. In the absence of a collision, any object inbound from 1 light-year with an initial velocity of 2.5 km/sec which is out of the plane of the ecliptic is not going to get near enough to an object like Jupiter to be grossly affected.
    $endgroup$
    – WhatRoughBeast
    Dec 23 '18 at 16:17










  • $begingroup$
    Can you prove that my opening sentence is 'unfounded and untrue' and that yours is? Jupiter has a massive gravity well that extends far away in space, it's been calculated that it is responsible from perturbations in Mercury's orbit. And, at 2.5 km/s, the flotsam is moving slowly enough to be easily captured, given that Jupiter captured comets that moved way faster, like Shoemaker-Levy 9, which is believed to have been in solar orbit prior to being 'stolen' by Jupiter.
    $endgroup$
    – Sava
    Dec 23 '18 at 22:48










  • $begingroup$
    After the object has entered the solar system, it will be going way faster than 2.5 km/s—about 20 km/s at Jupiter’s distance from the sun. This is why it has an extremely low probability of being captured by anything.
    $endgroup$
    – prl
    Dec 24 '18 at 1:20










3




3




$begingroup$
Nearly downvoted. You do realize, I hope, that gravity wells are symmetric? That is, the incoming velocity equals the outgoing velocity? As Mark Olsen discussed in his answer, a body falling toward a large one from far away will end up far away. And as for "fall(ing) into Jupiter or the Sun", that is extremely unlikely, particularly since they are originally pointed at 2 AU out and 15 degrees above the plane of the ecliptic.
$endgroup$
– WhatRoughBeast
Dec 23 '18 at 4:23




$begingroup$
Nearly downvoted. You do realize, I hope, that gravity wells are symmetric? That is, the incoming velocity equals the outgoing velocity? As Mark Olsen discussed in his answer, a body falling toward a large one from far away will end up far away. And as for "fall(ing) into Jupiter or the Sun", that is extremely unlikely, particularly since they are originally pointed at 2 AU out and 15 degrees above the plane of the ecliptic.
$endgroup$
– WhatRoughBeast
Dec 23 '18 at 4:23












$begingroup$
You forget one thing: planets are in motion and so are their gravity wells. Thus, depending upon angle and vector of entry into the gravity well relative to the planet's trajectory around the Sun, any small celestial object will end up accelerating or decelerating and have their orbital path modified. Please look up 'slingshot manoeuvre' or 'gravity assist' at ESA, NASA or even Wikipedia. I would link, but I'm on my phone at the moment.
$endgroup$
– Sava
Dec 23 '18 at 9:30




$begingroup$
You forget one thing: planets are in motion and so are their gravity wells. Thus, depending upon angle and vector of entry into the gravity well relative to the planet's trajectory around the Sun, any small celestial object will end up accelerating or decelerating and have their orbital path modified. Please look up 'slingshot manoeuvre' or 'gravity assist' at ESA, NASA or even Wikipedia. I would link, but I'm on my phone at the moment.
$endgroup$
– Sava
Dec 23 '18 at 9:30




2




2




$begingroup$
Everything you say is true - but irrelevant. Your opening sentence, "They are most likely to end up captured by Jupiter's gravity well first." is unfounded, unsupported by your statement, and untrue. In the absence of a collision, any object inbound from 1 light-year with an initial velocity of 2.5 km/sec which is out of the plane of the ecliptic is not going to get near enough to an object like Jupiter to be grossly affected.
$endgroup$
– WhatRoughBeast
Dec 23 '18 at 16:17




$begingroup$
Everything you say is true - but irrelevant. Your opening sentence, "They are most likely to end up captured by Jupiter's gravity well first." is unfounded, unsupported by your statement, and untrue. In the absence of a collision, any object inbound from 1 light-year with an initial velocity of 2.5 km/sec which is out of the plane of the ecliptic is not going to get near enough to an object like Jupiter to be grossly affected.
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– WhatRoughBeast
Dec 23 '18 at 16:17












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Can you prove that my opening sentence is 'unfounded and untrue' and that yours is? Jupiter has a massive gravity well that extends far away in space, it's been calculated that it is responsible from perturbations in Mercury's orbit. And, at 2.5 km/s, the flotsam is moving slowly enough to be easily captured, given that Jupiter captured comets that moved way faster, like Shoemaker-Levy 9, which is believed to have been in solar orbit prior to being 'stolen' by Jupiter.
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– Sava
Dec 23 '18 at 22:48




$begingroup$
Can you prove that my opening sentence is 'unfounded and untrue' and that yours is? Jupiter has a massive gravity well that extends far away in space, it's been calculated that it is responsible from perturbations in Mercury's orbit. And, at 2.5 km/s, the flotsam is moving slowly enough to be easily captured, given that Jupiter captured comets that moved way faster, like Shoemaker-Levy 9, which is believed to have been in solar orbit prior to being 'stolen' by Jupiter.
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– Sava
Dec 23 '18 at 22:48












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After the object has entered the solar system, it will be going way faster than 2.5 km/s—about 20 km/s at Jupiter’s distance from the sun. This is why it has an extremely low probability of being captured by anything.
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– prl
Dec 24 '18 at 1:20






$begingroup$
After the object has entered the solar system, it will be going way faster than 2.5 km/s—about 20 km/s at Jupiter’s distance from the sun. This is why it has an extremely low probability of being captured by anything.
$endgroup$
– prl
Dec 24 '18 at 1:20




















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