Question

If the period of revolution of a nearest satellite around the planet of radius R is T then its period of revolution around other planet, having radius $3R$ and same density will be
A. $T$
B. $3T$
C. $3\sqrt 3 T$
D. $9T$

Answer 1

Hint: Use the relation between density of the planet and time period of a satellite around that planet.

Complete step-by-step answer:
We know that time period is the distance travelled around the planet divided by the velocity of the object.
$T = \dfrac{{2\pi R}}{v}$
Where,
$R = $radius of planet.
$v = $velocity of satellite around that planet.
Also,
In a stable orbit, gravitational force=centripetal force
$
  G\dfrac{{Mm}}{{{R^2}}} = \dfrac{{m{v^2}}}{R} \\
  v = \sqrt {\dfrac{{GM}}{R}} \\
 $
Putting the value of v, we get,
$T = 2\pi \sqrt {\dfrac{{{R^3}}}{{GM}}} $
We know that,
$M = \rho \dfrac{4}{3}\pi {R^3}$
Putting the value of M in time period,
$T \propto \sqrt {\dfrac{1}{\rho }} $

From this equation, we can say that time period only depends upon the density of the planet. It is given in the question that the density of the planet is the same in both cases. So, the time period will remain the same as before.

Option A, is correct.

Additional information:
A geostationary satellite is an earth-orbiting satellite, placed at an altitude of approximately 35,800 kilometres directly over the equator, that revolves in the same direction the earth rotates (west to east). At this altitude, one orbit takes 24 hours, the same length of time as the earth requires to rotate once on its axis. The term geostationary comes from the fact that such a satellite appears nearly stationary in the sky as seen by a ground-based observer.

Note: The alternate approach to this question is by solving the result of time period for both the cases. This approach is relatively more time consuming.