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When moving in an elliptical orbit from the Apoapsis toward the Periapsis could you class yourself as being pulled (falling) towards the Periapsis and then your velocity after is then pushing you towards the apoapsis? If this is the case what could the variance of speed be? I always assumed that you inserted yourself into an orbit at one speed and once you accelerated fast enough you could maintain that.

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    $\begingroup$ The only real force acting on you at any point of the orbit is gravity/weight, towards the center of the planet. You may pick various other frames of reference and include various fictitious forces, like centrifugal, or force of inertia, but they obscure the situation more than help here. You have momentum, and you have the force of gravity, that's all that matters. $\endgroup$ – SF. Mar 5 '17 at 4:08
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In an elliptical orbit, velocity varies over the course of the orbit.

The equation for speed in elliptical orbit is:

$ v=\sqrt{\mu\left({2\over{r}}-{1\over{a}}\right)} $

In this equation, $\mu$ is the gravitational parameter of the body you're orbiting, and $a$ is the semi-major axis of the orbit, both of which are unchanging for a stable orbit. $r$ is the current distance between the orbiting bodies, so it's smallest at periapsis and largest at apoapsis, meaning that speed is fastest at periapsis.

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The model you are looking for is a co-rotating frame of reference. Be careful though! If you thing properly understanding rotating reference frames is easier than properly understanding momentum, you are very likely going to get confused.

That said, the co-rotating model is very useful when working with orbital tethers, solar sail simulation or complex things like the CR3BP.

In this frame of reference, there are three forces acting on a spacecraft. The first one is gravity, and it works in just the same way as in a normal inertial frame:

$$\frac{\mu}{r^2}$$

In addition, you do now have two inertial forces due to momentum being a strange concept in such a frame. The first one is centrifugal force (yes, centrifugal):

$$\frac{v_h^2}{r}$$

Where $v_h$ is horizontal velocity. The second one is Coriolis force (yes, Coriolis):

$$-\frac{2v_vv_h}{r}$$

Where $v_v$ is vertical velocity. This force is normal to the two other forces.

In this frame, a circular orbit is just not moving, with $\frac{\mu}{r^2} - \frac{v_h^2}{r} = 0$. In an elliptical orbit, at periapsis, $\frac{\mu}{r^2} - \frac{v_h^2}{r} < 0$, and at apoapsis $\frac{\mu}{r^2} - \frac{v_h^2}{r} > 0$, just like you said.

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