The situation is this: Two space vehicles A and B are doing a race in space. A is in front of B and they are both in orbit around the earth. For simplicity let this orbit be a circle (i.e. neglectable eccentricity). Now, B wants to get past A, that is, B wants to cross the "line" connecting the center of the earth and A in his orbit. That means, he'll catch up a higher angular velocity [tex]\omega[/tex]. B has two choices: he can break or he can accelerate (both in tangential dierction). What should B do?(adsbygoogle = window.adsbygoogle || []).push({});

For simplicity ignore all details: B won't lose any mass, and he'll stay in an (approximately) circular orbit.

One approach would be this: B could brake instantaneously and thereby lower his angular momentum [tex]L=mrv=mr^{2} \omega[/tex], which would get him behind and then make him "fall" to a lower orbit, picking up speed. Looking at [tex]\omega[/tex] as a function of [tex]L[/tex], one gets: [tex]\omega(L)=C\cdot L^{-3}[/tex], [tex]C[/tex] being some constant involving the masses etc. butnotthe radii. This would lead to argue: If [tex]L[/tex] goes down (braking) [tex]\omega[/tex] must go up and B will pass by A.

Now, I'm not so sure about the validity of this argument, as there are several critical points. First of all: If B brakes instantaneously, i.e. his speed [tex]v[/tex] decreases, so does his [tex]L[/tex] (that's ok so far). But this also means, that [tex]\omega[/tex] would decrease. How could I tell that the increase of [tex]\omega[/tex] caused by falling towards the earth is bigger than this decrease?

Second: Are the simplifications made here ignoring important aspects or even not valid at all? This whole low eccentricity thing might spoil our argument.

Thanks for any comments. Best regards...Cliowa

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# Driving past your opponent in a space race

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