Topic: Gravity
Concepts: Escape velocity, Gravitational force, Kepler’s laws

Solution:

• (a) Recall the escape velocity is given by
  
  $v_{\text{esc}}=\sqrt{\frac{2GM}{R}} \propto \sqrt{\frac{R^3}{R}}=R$
  
  which would depend on radius.
• (b) Recall the acceleration due to gravity is given by
  
  $g = \frac{GM}{R^2} \propto \frac{R^3}{R^2}=R$
  
  which would depend on radius.
• (c) Recall from Kepler’s 3rd law
  
  $T \propto \sqrt{\frac{r^3}{M}}$
  
  For $r=R$,
  
  $T \propto \sqrt{\frac{R^3}{R^3}}=1$
  
  which is independent of radius.
• (d) Recall from Kepler’s 3rd law
  
  $T \propto \sqrt{\frac{r^3}{M}} \propto \frac{1}{\sqrt{R^3}}$
  
  which would depend on radius.
• (e) Not applicable.

Thus, the answer is C.