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If there is a satellite and it is in an elliptical orbit. At the aphelion, it accelerates from its elliptical orbit to a circular orbit.

The thing that confuses me is that when the satellite is at aphelion, there's no radius velocity. Thus, it's proper to use the formula for a circular motion ($G\frac{Mm}{R^2}=\frac{mv^2}{R}$) to calculate the velocity of the satellite at aphelion. Then if it is accelerated to a circular orbit at aphelion. In order to calculate the velocity when the satellite is in a circular orbit, we still use the same formula with the same radius, same mass, and so on to calculate the velocity of the satellite.

Then what about the change in angular momentum. When the satellite is in an elliptical orbit, in order to calculate the angular momentum, we can calculate the velocity when the satellite is at aphelion, the radius of aphelion, and the mass of the satellite. However, when the satellite is after acceleration and in a circular orbit, the velocity, radius, and mass do not change. As a result, the angular momentum is not changed, but this is impossible since we need the impulse to let the satellite leave its original elliptical orbit. Where I go wrong? Thanks

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  • $\begingroup$ " the formula for a circular motion" What formula would that be? That's crucial to understanding your question. And please use MathJax to format the formula. A tutorial for MathJax is linked in the sidebar of the Question form where you posted your question. $\endgroup$
    – Bill N
    Nov 24, 2021 at 2:53
  • $\begingroup$ I add the formula and I hope it will be helpful $\endgroup$
    – M_k
    Nov 24, 2021 at 2:59
  • $\begingroup$ You can use the vis-viva equation to find the speed at any point on an orbit. See physics.stackexchange.com/a/676872/123208 $\endgroup$
    – PM 2Ring
    Nov 24, 2021 at 5:30

2 Answers 2

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The formula for relating the gravitational force to mass times centripetal acceleration, $$\frac{GMm}{R^2}=\frac{mv^2}{R},$$ is not true for an elliptical orbit because the instantaneous center of curvature at the aphelion (call it $r_x$) is not equal to the distance from the satellite to the focus ($R$ in your formula). The curvature of an elliptical orbit is different from a circular orbit.

It is true that $$\frac{GMm}{R^2}=\frac{mv^2}{r_x},$$ but one should use a little calculus to find the center of curvature of the ellipse at the apogee.

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The thing that confuses me is that when the satellite is at aphelion, there's no radius velocity. Thus, it's proper to use the formula for a circular motion to calculate the velocity of the satellite at aphelion

This is not correct, or at least, not without being very careful about the meaning of $R$.

In an arbitrary path, at any point where the acceleration is perpendicular to the velocity, it is possible to find a circle which is tangent to the path at that point and such that the perpendicular acceleration matches the centripetal acceleration for the path’s tangential velocity. This is a sort of local “best fit” circle which matches the actual path to second order.

The radius of this circle is the $R$ that you would use in that equation. It is not the distance to the center of the earth. At apogee that circle will be smaller than the circular orbit, and at perigee it will be larger.

However, when the satellite is after acceleration and in a circular orbit, the velocity, radius, and mass do not change.

The velocity does change, as does the angular momentum

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  • $\begingroup$ I think I kind of understand what you mean. You mean that for the apogee(also including every point in the elliptical), there is a "best-fit" circle for it and at that point, the satellite is orbiting that best-fit circle. Then it seems that all I need is the radius of that best-fit circle for apogee, then I can use the radius to calculate the correct angular momentum for the satellite when it is in its elliptical orbit, but how can I find the radius? $\endgroup$
    – M_k
    Nov 24, 2021 at 3:29
  • $\begingroup$ @Yink you would use the formula BillN posted and solve for $r_x$. However, I don’t see how knowing $r_x$ will help you with the angular momentum. Why would you be interested in the angular momentum about that circle’s axis. It is not conserved as the circle changes. $\endgroup$
    – Dale
    Nov 24, 2021 at 3:58
  • $\begingroup$ I want to find a way to calculate the impulse by the difference in angular momentum. My idea is first to calculate the angular momentum when the satellite is on an elliptical orbit, then calculate the angular momentum when it is on a circular orbit. The circular one is ok for me, just use $G\frac{Mm}{R^2}=\frac{mv^2}{R}$ to calculate the velocity and $L=mRv$. My idea for the elliptical one is also to use the formula $L=mrv$. Then I find that I have difficulty finding $r$ and $v$ in the elliptical one. $\endgroup$
    – M_k
    Nov 24, 2021 at 4:26
  • $\begingroup$ Thus, is there any method that works well for finding the impulse of turning the elliptical orbit into a circular one? $\endgroup$
    – M_k
    Nov 24, 2021 at 4:28
  • $\begingroup$ @Yink It would be a good idea to ask that directly as a separate question. The approach you are attempting here will not work. You need to already know $v$ in order to calculate $r_x$ $\endgroup$
    – Dale
    Nov 24, 2021 at 13:28

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