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There are two ways to analyse what happens to the orbit of a satellite when its orbital velocity is suddenly reduced (halved).

By considering the energy, when the velocity is halved, kinetic energy is quartered, so we have \begin{equation} KE_{new}=\dfrac{1}{4}KE_{old}=\dfrac{1}{4}\cdot\dfrac{GMm}{2r}=\dfrac{GMm}{2(4r)}\implies r_{new}=4r_{old} \end{equation} So the radius doubles. However this doesn't fit with intuition, as decreases in velocity results in orbital decay, which decreases the radius of orbit. Obviously from the calculation we have increase in radius. What mistakes did I make two produce these two faulty results?

P.S. After some research, it seems like I have to consider the total energy instead. So the question now becomes, why is it wrong to consider the kinetic energy alone?

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The missing piece is in the "suddenly reduced" bit. How it gets reduced is very important. If you say it remains in a circular orbit, at half of the velocity, the only valid solution is for it to "suddenly" be at a higher orbital radius.

In reality there are more orbits than just circular ones. The actual collection of available orbits are the Keplerian orbits, which are conic sections. If you "suddenly" reduce the velocity but keep the same position, that will typically put you on an elliptical orbit, not a circular one. These orbits should have the natural intuitive feel of decreasing orbital radius. In effect, you did a substantial retrograde burn, and that takes you down towards Earth.

(Of course, if you don't hit the atmosphere, you'll keep following the elliptical orbit and return to the point where you did your retrograde impulse burn)

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  • $\begingroup$ This pretty much answers all of my queries. However while you are here, it would be appreciated for you to answer a few more lol. 1. When the velocity is reduced gradually, the orbit becomes ellptical and the total energy decreases. Does that mean the new orbit is in some sense "smaller" than the previous circular motion (I don't know how to define "smaller", perhaps smaller area? Something to indicate a loss of energy) 2. Is it generally true that in all orbits the total energy of the system stays constant without exterior interference? thank you very much $\endgroup$
    – joshua mason
    Commented Sep 21, 2021 at 10:17
  • $\begingroup$ Another thing is, assuming the velocity does get reduced instantenously and still remains in circular motion, than upon examining change in kinetic energy, we get as shown, a greater radius. However upon examining the total energy of the system, we get a smaller radius. But I thought the only valid solution is to be at a higher orbit. So why is the total energy argument 'wrong' here? $\endgroup$
    – joshua mason
    Commented Sep 21, 2021 at 10:30
  • $\begingroup$ @joshuamason An instantaneous reduction in velocity for an object in circular orbital motion cannot result in circular motion being maintained without teleportation to a higher orbit, or a decrease in the centripetal acceleration such as through rocket thrust, or an instantaneous reduction in mass of the body being orbited. $\endgroup$
    – notovny
    Commented Sep 21, 2021 at 12:33
  • $\begingroup$ @joshuamason 1. it would be best to look at the keplerian orbitals, and find the sense that the orbit is smaller that makes sense to you. The one that makes the most sense to me is that the periapsis (the point of closest approach/lowest altitude) is lower. 2. We generally assume the orbits conserve total energy. Kepler's laws were found via observation, but they can indeed be derived simply by keeping total energy constant. You don't see non conservative effects until you get into more exotic systems where you have more things to deal with (like modeling solar sails) $\endgroup$
    – Cort Ammon
    Commented Sep 21, 2021 at 14:32
  • $\begingroup$ @joshuamason And notovny is right for your third question. You're simply solving a constraint, not actually doing something physical. Change the constraint, change the behavior. I dug up a website talking about specific orbital energy which shows a graph of total energy given an orbit. The graph shows that that function is not simple. Simply constraining one's behavior to that graph yields complicated behaviors. $\endgroup$
    – Cort Ammon
    Commented Sep 21, 2021 at 14:34

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