A satellite which is geostationary in a particular orbit is taken to another orbit. Its distance from the center of the earth in a new orbit is 2 times than that of the earlier orbit. The time period in the 2^nd orbit is given by-

(a) 4.8 hours
(b) \[48\sqrt 2 \] hours
(c) 24 hours
(d) \[24\sqrt 2 \] hours

Hint:
1. Kepler’s 3rd law of planetary motion gives direct proportionality relation between radius of the orbit and approximate time taken by the planet to complete 1 revolution.
2. Time period of revolution of the geostationary satellite is T= 24hrs Here, in question, there is no mention of Mass and other parameters so we should not think along gravitational force direction.

Formula Used:
According to Kepler’s 3^rd law, for a given orbit around its Sun: $\dfrac{{{T^2}}}{{{R^3}}} = $ constant …… (1)
Where,
T is time period of revolution
R is radius of the circular orbit or path followed

Complete step by step answer:
Given:
1.Radius of the geostationary satellite: R where, R is radius of earth
Let, the time period of geostationary satellite: T
Let, New radius for the satellite: ${R_{new}} = 2R$ …… (a)
Let, new time period of the satellite in shifted orbit be: ${T_{new}}$
To find: New time period of revolution ${T_{new}}$

Step 1:
From equation (1) we know initially: $\dfrac{{{T^2}}}{{{R^3}}} = $ k …… (2)
where, k is constant

Step 2:
Similarly, Kepler’s 3^rd law is valid for new orbit as well.
So, using equation (1) for new orbit we can again say: ${({T_{new}})^2} = k.{({R_{new}})^3}$ …… (3)

Step 3:
Dividing equation (2) by equation (3), we get-
$\dfrac{{{{(T)}^2}}}{{{{({T_{new}})}^2}}} = \dfrac{{k.{{(R)}^3}}}{{k.{{({R_{new}})}^3}}}$ …… (4)

Step 4:

From the hint we know, the time period of revolution is T=24 hours for geostationary satellites.
Using above fact, and substituting value from equation (a) in equation (4) we get-

$
   \Rightarrow \dfrac{{{{(T)}^2}}}{{{{({T_{new}})}^2}}} = \dfrac{{k.{{(R)}^3}}}{{k.{{(2R)}^3}}} \\
   \Rightarrow \dfrac{{{{(24)}^2}}}{{{{({T_{new}})}^2}}} = \dfrac{1}{8} \Rightarrow {(24)^2} \times 8 = {({T_{new}})^2} \\
   \Rightarrow {T_{new}} = 24 \times 2\sqrt 2 \\
 $

Final Answer
(b) \[48\sqrt 2 \] hours


Note: New time period is different than 24 hours. Hence the satellite is no more geo-stationary, for a satellite to be geo-stationary its time period should be the same as time taken for earth to rotate about its axis i.e 24hrs.