Question

The planet shown has a mass of M, and the satellite is in a circular orbit of radius r.

a) In terms of M, r, and the universal gravitational constant G, what is the period Tof the satellite? Derive the formula.

b) By what factor would the period change if the mass of the planet doubled?

c) By what factor would the period change if the radius of the orbit doubled?

Answer 1

According to the equation of motion   
centripetal force = gravitational attraction force
  
  $\Rightarrow v^2=\frac{GM}{r}$
  $\Rightarrow v=\sqrt{\frac{GM}{r}}$
So, the time period is

$T=\frac{2\pi r}{v}=\frac{2\pi r}{\sqrt{\frac{GM}{r}}}$
   $\Rightarrow T=\frac{2\pi}{\sqrt{GM}}r^{3/2}$
b)
As the time period depends upon the mass M as
$\Rightarrow T\sim \frac{1}{\sqrt{M}}$
if the mass M gets doubled, the time period decreases and decreases by a factor of square root 2.
   $\Rightarrow \frac{T'}{T}=\frac{\sqrt{M}}{\sqrt{2M}}=\frac{1}{\sqrt{2}}$
  $\Rightarrow {T'}=\frac{T}{\sqrt{2}}$
c)
  The time period depends upon the mass radius r as
$\Rightarrow T\sim r^{3/2}$   
if the radius r gets doubled, the time period increases and increases by a factor of 2 times square root 2.
   $\Rightarrow \frac{T'}{T}=\frac{r'^{3/2}}{r^{3/2}}=\frac{(2r)^{3/2}}{r^{3/2}}$
  $\Rightarrow T'=2\sqrt{2}T$