Topic: Gravity
Concepts: Circular orbits, Gravitational force, Kepler’s laws, Reduced mass

Solution:

To simplify calculating the angular momentum and period, we use the concept of reduced mass to turn the two-body problem of masses $m$ and $M$ interacting via gravity into the equivalent one-body problem of a particle orbiting a fixed source. Recall the reduced mass of $m$ and $M$ is given by

In order for the single particle of mass $\mu$ to experience the same force $F=GMm/R^2$ as the gravity between $m$ and $M$, the fixed source must have mass $m+M$.

Now, we go through the possible choices:

• A) Not correct since the gravitational force is
  
  $F=\frac{GMm}{R^2}$
  
  If $m \rightarrow m-\delta m$ and $M \rightarrow M+\delta m$ then
  
  $mM \rightarrow (m-\delta m)(M+\delta m)=mM-(M-m)\delta m-(\delta m)^2<mM$
  
  so the gravitational force decreases.
• B) Not correct since the gravitational force decreases.
• C) Not correct since the angular momentum is
  
  $L=\mu vR=\frac{mM}{m+M}\sqrt{\frac{G(m+M)}{R}}R=mM\sqrt{\frac{GR}{m+M}}$
  
  where we used the velocity of a circular orbit $v_{\text{circ}}=\sqrt{GM_{\text{source}}/R}$. Since $mM$ decreases while $m+M$ stays constant, the angular momentum decreases.
• D) Not correct since the angular momentum decreases.
• E) Correct since the period is
  
  $T^2=\frac{4\pi^2}{G(m+M)}R^3$
  
  where we used Kepler’s 3rd law with $M_{\text{source}}=m+M$. Since $m+M$ stays constant, the period remains constant.

Thus, the answer is E.