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Question: A geostationary satellite is orbiting the earth at a height of \( 5R \) above the surface of the ear...

A geostationary satellite is orbiting the earth at a height of 5R5R above the surface of the earth. RR being the radius of the earth. The time period of another satellite in hours at a height 2R2R from the surface of the earth is
(A) 626\sqrt 2
(B) 62\dfrac{6}{{\sqrt 2 }}
(C) 55
(D) 1010

Explanation

Solution

The relation between the radius and the time period is needed here. The relation is gained from Kepler’s 3rd3rd law. The law also shows how the time period changes with the change of the height from the earth’s surface in terms of radius for a satellite. Using this relation, the two ratios are to be made regarding the time and height and, thereafter, the ratios are to equate according to the relation (either directly proportional or inversely proportional).
The time period T2∝R3{T^2} \propto {R^3}
RR is the radius of the earth.

Complete answer:
Kepler’s 3rd3rd law gives the relation between the time period and the radius of the earth.
This law can be used in this problem.
The mathematical expression of this law is,
T2∝R3{T^2} \propto {R^3} ,
TT The time period and RR is the radius of the earth.
Now, from this relation, we can write,
T12T22=R13R23\dfrac{{{T_1}^2}}{{{T_2}^2}} = \dfrac{{{R_1}^3}}{{{R_2}^3}}
For the 1st1^{st} satellite, the time period T1=24hr{T_1} = 24hr
And, R1=5R+R=6R{R_1} = 5R + R = 6R
For the 2nd2nd satellite, the time period is T2{T_2}
And, R2=2R+R=3R{R_2} = 2R + R = 3R
⇒(24)2T22=(5R)3(2R)3\Rightarrow \dfrac{{{{(24)}^2}}}{{{T_2}^2}} = \dfrac{{{{(5R)}^3}}}{{{{(2R)}^3}}}
⇒1T22=(6R)3(3R)3×(24)2\Rightarrow \dfrac{1}{{{T_2}^2}} = \dfrac{{{{(6R)}^3}}}{{{{(3R)}^3} \times {{(24)}^2}}}
⇒T22=(3R)3×(24)2(6R)3\Rightarrow {T_2}^2 = \dfrac{{{{(3R)}^3} \times {{(24)}^2}}}{{{{(6R)}^3}}}
⇒T22=(3R6R)3×(24)2\Rightarrow {T_2}^2 = {\left( {\dfrac{{3R}}{{6R}}} \right)^3} \times {(24)^2}
⇒T22=(24)28\Rightarrow {T_2}^2 = \dfrac{{{{(24)}^2}}}{8}
⇒T2=2422\Rightarrow {T_2} = \dfrac{{24}}{{2\sqrt 2 }}
⇒T2=122\Rightarrow {T_2} = \dfrac{{12}}{{\sqrt 2 }}
⇒T2=62\Rightarrow {T_2} = 6\sqrt 2
SO, The TIME PERIOD OF the 2nd2nd Satellite is 62hr6\sqrt 2 hr .
Correct Option is (A).

Note:
Here we take the radius as the sum of the height from the surface of the earth and the radius of the earth. The addition is done to take the length from the center. It seems that the radius of the orbits of the satellites are 6R6R and 3R3R . That’s why we can use Kepler's 3rd3rd law in this problem.
The time period of the 1st1^{st} satellite is taken 24hr24hr since it is given as a geostationary satellite.