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Two body problem |
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| Mar27-09, 05:48 AM | #1 |
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Two body problem
1. The problem statement, all variables and given/known data
Two gravitating particles with masses m1 and m2 start from rest a large distance apart. They are allowed to fall freely towards one another. The particles are given equal and opposite impulses I when they are a distance a apart, such that each impulse is perpendicular to the direction of motion. Show that the total angular momentum of the two particles about their centre of mass has magnitude aI /µ, where µ is the reduced mass of the system. 2. Relevant equations Reduced mass=m1m2/(m1+m2) 3. The attempt at a solution Well this is the second last part of quite a long question on the 2-body problem, and I've managed fine until now (showing the position of the centre of mass 'C' - is constant, finding their relative speed etc) but I'm not sure how to go about this part of the question. The two particles are going to be moving in a straight line towards each other before the impulses which should mean 0 angular momentum before, so then the only angular momentum afterwards would be that from the impulses right? But the impulses are perpendicular to the direction of motion so with the r x p cross product we'd just have angular momentum=dist. from C * impulse in each case wouldn't we? The m2 mass particle should have a distance (m1/(m1+m2))a from C and the m1 particle a distance of (m2/(m1+m2))a, but then clearly I've done something wrong because the sum of the impulses will just be (m1+m2/m1+m2)aI=aI. Where am I going wrong? Thanks! |
| Mar27-09, 08:25 AM | #2 |
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That sure seems correct to me. aI/mu doesn't even have the correct units, does it?
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| Mar27-09, 09:06 AM | #3 |
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Actually that's a fair point, whereas aI does - weird! I'll check with someone else to see if the question is mistyped, but it seems odd that the entire "/µ, where µ is the reduced mass of the system." would be a mistake...
I'll let you know if i find out! Do you think just aI is the correct answer then? |
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