The accelerate from an elliptical orbit to a circular orbit 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
 A: 
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
A: 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.
