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I do know the satellite's tangential velocity is always perpendicular to gravity, therefore its speed must remain unchanged, however, I have some confusion.

If I separate both the tangential velocity of the satellite and the gravitational force of the earth, things becomes different.

For example, lets say point O is the Earth and P is a satellite orbiting around the earth in a circular motion, segment LA is a tangential velocity of the satellite at that point, and because segment LA is at an angle, therefore we can separate it into two components, which are LB and LC. In addition to the gravitational force ( PD) , we can also make it into two components, and since EP and FP is in the opposite direction of LC and LB, therefore, the tangential speed of the satellite should be changed.

enter image description here

Can somebody correct me? why couldn't I see it from this aspect?

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    $\begingroup$ The tangential velocity does change, but it only changes direction, not magnitude. One way of seeing this is that since the force is always perpendicualar to the velocity, it does not do any work. And if no work is done, then kinetic energy is unchanged. If $m$ is constant, that means that $v^2 = |\mathbf{v}|^2$ is constant. Hence $|\mathbf{v}|$ is constant. $\endgroup$ May 26, 2020 at 19:54
  • $\begingroup$ yeah, I know that, but I got a different result after using another method which I separate the tangential velocity and the direction of gravity into horizontal and vertical component $\endgroup$
    – Sherri
    May 26, 2020 at 20:14

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The magnitude of the tangential velocity does not change. Let me prove it using simple arguments.

As per your diagram, observe that the vector PD is always perpendicular to the vector LA. As a result, any projection of PD on LA is zero. This means motion along PD is independent of motion along LA and vice-versa. The resolved components, which you have drawn, may add and subtract to give different results but that forms a different coordinate system, rotated with respect to the original frame. Once you go back to the frame in which the two directions are the radial and tangential directions, you will see that the tangential velocity is still the same.

Moreover, this being a Kepler problem, in general, the forces involved are all conservative and the angular momentum is also conserved. So the tangential velocity is also conserved from that view point.

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  • $\begingroup$ Do you mean that even I use a different coordinate system ( the one I have drawn above), the result will always be the same as the original? $\endgroup$
    – Sherri
    May 26, 2020 at 20:22
  • $\begingroup$ @Sherri Yes. It will. $\endgroup$ May 26, 2020 at 20:23

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