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# Momentum of systems

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If an asteroid bisects a binary star system, does the asteroid lose momentum?

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• 3 weeks later...
On 8/14/2023 at 9:55 PM, DimSon said:

If an asteroid bisects a binary star system, does the asteroid lose momentum?

The Principle of the Conservation of Energy is that Energy can never be created nor destroyed. Energy can only be transferred from one form to another. Therefore Kinetic energy gained = Gravitational potential energy lost to point O, (the point bisecting the stars halfway), Kinetic energy lost = Gravitational potential energy gained on the outbound journey. Therefore the speed of the asteroid at the end is the same as at the beginning of the transit, (3x104 m/s).

Thank you for posting such a thoughtful, interesting question.

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• 2 months later...
On 8/30/2023 at 11:57 AM, spartan45 said:

The Principle of the Conservation of Energy is that Energy can never be created nor destroyed. Energy can only be transferred from one form to another. Therefore Kinetic energy gained = Gravitational potential energy lost to point O, (the point bisecting the stars halfway), Kinetic energy lost = Gravitational potential energy gained on the outbound journey. Therefore the speed of the asteroid at the end is the same as at the beginning of the transit, (3x104 m/s).

Thank you for posting such a thoughtful, interesting question.

I guess perhaps I'm looking for a method on how to approach this problem. It seems like the stars would be attracted to the asteroid with both an x and y component. Their x displacement would even out as the asteroid switches from the left hand to the right hand side of the trip. But the y(and z) displacement would not be restored, so in the end, the two stars would orbit closer together. It seems to me that this reduction in the distance between the two stars would generate a non-symmetrical x-axis component for the asteroid, and the two stars.

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