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A satellite is geostationary in a particular orbit. It is allowed to go to another orbit having an orbital radius 2 times that of the earlier orbit from the centre of the earth. The time period in the second orbit is
A. 48 hours
B. 24 hours
C. \[48\sqrt 2 \,{\text{hours}}\]
D. \[24\sqrt 2 \,{\text{hours}}\]

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Last updated date: 20th Jun 2024
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Answer
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Hint: Use Kepler’s third law to determine the orbital period of the satellite in the second orbit. For a geostationary satellite, the orbital period of the satellite in orbit is the same as the rotation period of the earth.

Formula used:
\[{T^2} = \dfrac{{4{\pi ^2}}}{{GM}}{a^3}\]
Here, G is the universal gravitational constant, M is the mass of earth and a is the semimajor axis of the orbit of satellite.

Complete step by step answer:
According to Kepler’s third law, the period of satellite moving in a orbit of semimajor axis a is given as,
\[{T^2} = \dfrac{{4{\pi ^2}}}{{GM}}{a^3}\]
Here, G is the universal gravitational constant, M is the mass of earth and a is the semimajor axis of the orbit of satellite.
For the first orbit, the period of the satellite in its orbit will be,
\[T_1^2 = \dfrac{{4{\pi ^2}}}{{GM}}a_1^3\] …… (1)
For the second orbit, the period of the satellite in that orbit will be,
\[T_2^2 = \dfrac{{4{\pi ^2}}}{{GM}}a_2^3\] …… (2)
Divide equation (2) by equation (1).
\[\dfrac{{T_2^2}}{{T_1^2}} = \dfrac{{\dfrac{{4{\pi ^2}}}{{GM}}a_2^3}}{{\dfrac{{4{\pi ^2}}}{{GM}}a_1^3}}\]
\[ \Rightarrow \dfrac{{T_2^2}}{{T_1^2}} = {\left( {\dfrac{{{a_2}}}{{{a_1}}}} \right)^3}\]
\[\therefore {T_2} = {\left( {\dfrac{{{a_2}}}{{{a_1}}}} \right)^{\dfrac{3}{2}}}{T_1}\]
For a geostationary orbit, the orbital period is 24 hours.
We have given, the radius of the second orbit is twice of the first orbit. Therefore, substitute \[2{a_1}\] for \[{a_2}\] and 24 hours for \[{T_1}\] in the above equation.
\[{T_2} = {\left( {\dfrac{{2{a_1}}}{{{a_1}}}} \right)^{\dfrac{3}{2}}}\left( {24\,{\text{hours}}} \right)\]
\[ \Rightarrow {T_2} = {\left( 2 \right)^{\dfrac{3}{2}}}\left( {24\,{\text{hours}}} \right)\]
\[ \Rightarrow {T_2} = 2\sqrt 2 \left( {24\,{\text{hours}}} \right)\]
\[\therefore {T_2} = 48\sqrt 2 {\text{ hours}}\]

So, the correct answer is “Option C”.

Note:
As the radius of the orbit increases, the period of the satellite also increases.
To check whether your answer is right or wrong refer to the above conclusion.