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Suppose we have a satellite orbiting the Earth in an elliptical-closed orbit. When the satellite reaches the perigee, its velocity is increased in the cross-radial direction. According to all the sources, the perigee remains the same, however, the distance to the apogee increases, as the eccentricity increases. Let us assume a situation, where the eccentricity increases but doesn't become more than or equal to one i.e. the orbit remains elliptical.

I suppose this happens because the total energy increases as one increases the kinetic energy, the total energy of the orbit increases. Since the distance to the perigee remains the same, the orbit must compensate by increasing the distance to the apogee i.e. raising the apogee.

My question is, why is the perigee distance constant? Generalizing the question, why does the orbit remain closed even after increasing the velocity? The distance to the perigee remains constant, while the distance to the apogee increases. The new orbit is closed from the beginning it seems. Should it not be more like, the apogee and the perigee remain the same, while the total velocity at both of these points increases? Or something like the perigee and the apogee both increase?

What guarantees that after the velocity is increased at the perigee, the satellite would return to the same point with the same increased velocity after each orbit? What guarantees that the orbit remains closed immediately after increasing velocity ? Can someone explain or show me the proof of this?

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There are two conservation laws at work.

By applying a tangential boost to the velocity you have applied a torque to the object and increased its angular momentum to $v_p r_p$ where $v_p$ is the new tangential velocity at perigee. This angular momentum must then be conserved such that at apogee $v_a r_a = v_p r_p$.

Conservation of energy then means that when it returns to $r_p$ it must have the same speed $v_p$, and conservation of momentum means that this speed must be directed tangentially and so this still defines the perigee.

The orbit remains closed because it is still a Keplerian orbit. There is no sense in which the object retains a "memory" of what its orbit was prior to the boost. You are setting up a new orbit which has a new tangential velocity component $v_p$ at $r_p$. Such an orbit is a closed ellipse (if the kinetic energy is low enough) and it returns to its starting point like every other Keplerian orbit.

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  • $\begingroup$ Is this closed orbit guaranteed by Bertrand's theorem ? $\endgroup$ Commented Nov 19, 2021 at 10:49
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After accelerating, your spaceship has some velocity vector at some location (and it doesn't matter whether that is perigee, apogee, or any other point in the original orbit). By the very assumption of your problem, the spaceship is (still) on a closed, elliptical orbit. After one period, it therefore will be back at this location with that velocity. There are no spiral orbits in the Kepler problem.

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  • $\begingroup$ Is this closed orbit guaranteed by Bertrand's theorem? Is that the reason, the orbit would be closed and the distance to that point where speed is increased, remains constant ? In case we had some other force law, the orbit might have been a spiral where both the point and the one opposite to it, increases, as the satellite sort of spirals out ? $\endgroup$ Commented Nov 19, 2021 at 11:04
  • $\begingroup$ @NakshatraGangopadhay yes, exactly. in general, the orbit ellipse precesses, so over time the orbit eventually fills in the entire region between the circle at periapsis and the circle at apoapsis. $\endgroup$
    – Ryan C
    Commented Sep 13 at 19:50

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