Question

5) A satellite in a circular orbit of radius R around planet X has an orbital period T. If Planet X had one-fourth as much mass, the orbital period of this satellite in an orbit of the same radius would be:

A) 2T
B) T square root(2)
C) T/4
D) T/2
E) 4

Answer 1

from kepler's third law,
time period T =(2π/√GM)R^3/2

time period T' =(2π/√GM/4)R^3/2 =2T

option A is correct.

Answer 2

You cannot solve this problem without Keplar's law. You can use different energy equation and velocity expression to conserve energy, momentum but it would only be a consquence of the Kepler's law and nothing else. This is because the orbital period that was derived is infact the Kepler's law.

It states,

$$ T^{2} \propto \frac{R^{3}}{M} $$

\(\mathrm{T}\) is the time period, \(\mathrm{R}\) is the orbital radius and \(\mathrm{M}\) is the mass

So, for a same orbital mass

$$ \begin{array}{c} \left(\frac{T_{1}}{T_{2}}\right)^{2}=\left(\frac{M_{2}}{M_{1}}\right) \\ \text { or },\left(\frac{T_{1}}{T_{2}}\right)^{2}=\left(\frac{\frac{1}{4} M_{1}}{M_{1}}\right) \\ \text { or, } T_{2}=2 T_{1}=2 T \end{array} $$