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I have no idea what the answer is. I'm supposed to answer it within 3-4 sentences.

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This sounds like a homework and/or test question. Can you provide any sort of reasoning you may have? Otherwise, this question might be closed by the moderators. –  joshphysics Feb 20 '13 at 17:39
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Angular momentum $L = v \times r$... $L$ is angular momentum, $v$ is velocity, $r$ is distance from orbital center. How could you change $v$ to a smaller value and keep $L$ the same? –  KDN Feb 20 '13 at 17:39
    
Based on the formula for the angular momentum $L=r\times p$ , if you move to a higher orbit (and increase in r) you can keep the same angular momentum, at the cost of decreasing the speed of the satellite. –  Leonida Feb 20 '13 at 17:58
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1 Answer 1

Let us analyse the problem to see how this can happen.

The satellite is kept in orbit "being balanced" by two forces in the equation

$\frac{mv^2}{r}=\frac{GMm}{r^2}$

From this we get $v=\sqrt\frac{GM}{r}$

therefore the angular momentum is

$L=pr=mvr=m(\sqrt{GM})r^{1/2}$.

This equation shows that, as long as the mass of the satellite does not change, putting it on a higher orbit will not necessarely keep the angular momentum fixed. The angular momentum can remain the same by taking into account the fact that the satellite mass is reduced by fuel consumption in order to be placed to a higher orbit.

Therefore, if $r\rightarrow \alpha r$ and $m\rightarrow \frac m{\alpha^{1/2}}$ where $\alpha\gt1$ then the satellite will keep the same angular momentum.

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