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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
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. – nijankowski Feb 20 '13 at 17:58

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


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

therefore the angular momentum is


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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The comments seem to provide the answer.

$$ L = p \times r $$

$$ L = m ( v \times r ) $$

Thus if r (the distance to the focus point of the orbit) changes, the velocity can change without the angular momentum changing. This is not possible for circular orbits where v is always perpendicular to r and the magnitude of r is constant. However, in elliptical and hyperbolic orbits, the angular momentum is conserved while the magnitude of velocity varies as the magnitude of r varies.

A note about JKL's answer: The gravitational force is not balanced by anything. It's a net force that accelerates the satellite towards the focal point of the orbit. If the orbit is circular, than the acceleration can be described by $\frac{v^2}r$ and thus the gravitational force that causes this acceleration must be equal to $\frac{m\:v^2}r$. However, this is not a force balance, and it is only true for circular orbits.

For a general way to describe the velocity during an orbit, an energy balance is appropriate:

$$ E = KE + GPE $$ $$ E = \frac12mv^2 - \frac{mGM}{|r|}$$ $$ |v| = \sqrt{\frac{2GM}{|r|}+C}$$ Where the constant $C=\frac{2E}m$, (For circular orbits $C= -\frac{GM}{|r|}$ and for parabolic orbits $C=0$)

Additionally, angular momentum is the cross product between momentum and the position vector. Therefore, $L\leq m\:|v|\:|r|$ where equality only holds for circular orbits (or temporarily whenever the velocity is perpendicular to the position vector).

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+1. Sometimes a late answer is just late. But since the only other answer to date is not that good, better late than never. – David Hammen Oct 22 '14 at 20:38

By increasing its radius.

Angular momentum is the cross product of radius and velocity:

$$\mathbf{L} = \mathbf{r} \times \mathbf{v}$$

It's a trade-off between radius and speed if the angular momentum is constant (which is for a satellite)

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protected by Qmechanic Jul 17 at 8:33

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