When I see an orbital mechanics problem, I recall the following things from my "toolbox":

  1. Angular momentum is conserved for a body orbiting a planet, since the torque due to gravity is 0.

  2. Kinetic energy and potential gravitational energy sum to a conserved quantity. For an elliptical orbit, this quantity is half the potential at the semi-major axis of an elliptical orbit.

  3. If the velocity is perpendicular to the moment arm of the body with respect to the planet, centripetal acceleration may be helpful.

Now I have the following problem:

An artificial satellite is in a circular orbit around the moon at radius $R = kr$ , where $r$ is the radius of the moon itself. A short burn of the satellite’s motor provides an impulse which halves the satellite’s speed without changing its direction, and this alters the orbit to one that just grazes the moon’s surface. By considering the angular momentum and energy of the satellite at the apoapsis and periapsis of the new orbit, deduce the value of $k$. Try using angular momentum.

The answer to the problem is apparently $k=7$, So let me share my approach on this problem considering each of the tools and perhaps someone could point out what I am doing wrong.

First note that the maneuver sends the satelite into elliptical orbit with apogee $R$ and perigee $r$, thus major axis $R+r$.

By centripetal force considerations, the original speed of the craft before the maneuver is $\sqrt{GM/R}$ so after the manuever it is $0.5\sqrt{GM/R}$.

1.By considering angular momentum at apogee and perigee, $m0.5\sqrt{GM/R}R=mv_1r$.

2.At a perigee, the speed is equal to the tangential speed (the moment arm is perpindicular to the motion of the craft), so we can apply centripetal acceleration to get $v_1=\sqrt{GM/r}$. Combining this with the observation in (1), you get $R=4r$, which is incorrect.

  1. By applying conservation of energy, $0.5v^2-GM/d=-GM/(R+r)$. At apogee, $d=R$ and from (1), $v=0.5\sqrt{GM/R}$. Substituting this yeilds the correct answer $R=7r$. At perigee, $d=r$, and from (1), $v=0.5\sqrt{GM/R}R/r$. Substituting this gives something in terms of $r$ and $R$ but by itself its not useful. (At least it agrees with $R=7r$).

So it seems the observation in (2) is incorrect. But why? Also, angular momentum was not nessecary apparently, although the problem statement suggested to use it. So what is the alternative solution using angular momentum?

  • $\begingroup$ The equation $v = \sqrt{GM/r}$ is only true for a circular orbit because it is derived assuming the gravitational acceleration is equal to the acceleration towards the centre. It is not true at the perigee of an elliptical orbit. $\endgroup$ Commented Jan 23, 2015 at 11:58
  • $\begingroup$ @JohnRennie: it is true even for nonuniform motion that the centripetal acceleration is $v^2/r$ where $v$ is the tangential speed. At the perigee, we "just graze the moon's surface" so the orbit is tangent to the moon's surface, so the tangential speed is equal to the total speed. What's wrong with that reasoning? $\endgroup$
    – math_lover
    Commented Jan 23, 2015 at 18:10
  • $\begingroup$ Yes, that's true. But at the apogee the centripetal acceleration is not equal to the gravitational acceleration $GM/r^2$. The equation $v = \sqrt{GM/r}$ is true only when the centripetal and gravitational accelerations are equal. $\endgroup$ Commented Jan 24, 2015 at 6:40
  • $\begingroup$ What are you talking about. The radial force is always $v^2/r$ where $v$ is tangential speed. The radial force is provided by gravity. Since the tangential speed is equal to total speed at apogee as per above, the expressions are equal. $\endgroup$
    – math_lover
    Commented Jan 24, 2015 at 11:57
  • $\begingroup$ The radial accln is $v^2/r$. The gravitational acceln is $GM/r^2$. If they were equal the net force would be zero and the object would stay at the same distance from the Earth i.e. a circular orbit. At the perigee $v^2/r > GM/r^2$ so the net force is outwards and the body moves away from the Earth. At the apogee $v^2/r < GM/r^2$ so the net force is inwards and the body moves towards the Earth. That's why $r$ changes in an elliptical orbit. The relationship between $v$ and $r$ is given by the vis-viva equation. $\endgroup$ Commented Jan 24, 2015 at 12:12

1 Answer 1


After the burn the orbit will no longer be circular, so the radius will no longer be constant.

The point of the burn will be the apolune (d) of the new orbit. The point of closest approach, the perilune (q), will be on the opposite side of the moon. The semimajor axis (a) is the average of the apolune distance and the perilune distance. $$ d = R $$ $$ q = r_\rm{moon} $$ $$ a = \frac{d+q}{2} = \frac{ R + r_\rm{moon} }{2} $$

Starting with the vis-viva equation: $$ \frac{Gm}{4R} = \frac{2Gm}{R} - \frac{Gm}{a} $$

Solving for apolune distance gives $$ d = R = 7 r_\rm{moon} $$

Also, the semimajor axis (a) after the burn will be: $$ a = \frac{4}{7} R = 4 r_\rm{moon} $$


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