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I have a satellite in a stable trajectory around Earth, of a Mass $m$. I know that its velocity is:

$$ v = \sqrt{mG/r}\, $$

But now it begins accelerating directly against the gravity vector (i.e. away from the center of the earth) by an acceleration, $a$. How do I describe the velocity of the satellite under a constant acceleration away from Earth?

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Since the force is radial, you are not changing the angular momentum, but you are adding potential energy: this tells you what must happen to the tangential velocity (decreases) and radial velocity (increases). I will leave it up to you to figure out by how much.

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  • $\begingroup$ how tangential velocity goes down if the radial goes up? You speak of direction? Or you speak of magnitude? $\endgroup$ – Sofia Mar 2 '15 at 22:38
  • $\begingroup$ @Sofia - conservation of angular momentum says that if you end up in a wider orbit without changing your angular momentum, your tangential velocity must be lowered so $v_{\theta} r $ is constant. $\endgroup$ – Floris Mar 2 '15 at 22:41
  • $\begingroup$ yes, it's clear but in the beginning I took you "goes down and goes up", literally. $\endgroup$ – Sofia Mar 2 '15 at 22:43
  • $\begingroup$ I see how that might be confusing. I changed the wording to "decreases" and "increases" - is that better? $\endgroup$ – Floris Mar 2 '15 at 22:57
  • $\begingroup$ Yeah, I've got the general idea that the tangential velocity would decrease. I was hoping to get something more quantifiable, like a formula. $\endgroup$ – Quarkly Mar 2 '15 at 23:20
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I have an answer, but I was hoping to see confirmation from another source. I used the centripetal force formula to come up with:

$$ v = \sqrt{Gm/r-ar} $$ Where v is the tangential velocity of the orbit and a is the radial acceleration away from Earth. Is this right?

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  • $\begingroup$ What is the direction of this velocity? Is it the total velocity, or the radial component? Is it sufficient for you to have the answer as a function of $r$ (rather than say $t$)? You might want to show your entire derivation if you would like us to comment $\endgroup$ – Floris Mar 2 '15 at 23:22
  • $\begingroup$ Sorry, I thought it was understood. This is the tangential velocity of the orbit. $\endgroup$ – Quarkly Mar 2 '15 at 23:24
  • $\begingroup$ Why does the tangential velocity depend on $a$. Why is it not simply $v = \sqrt{GM/R_0}\frac{R_0}{r}$ (conservation of angular momentum)? $\endgroup$ – Floris Mar 2 '15 at 23:25

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