Gravitation

JEE Main level, single correct, medium

177 of 223

Question 177 of 223

The time period of a satellite of earth is 2424 hours. If the separation between the earth and the satellite is decreased to one fourth of the previous value, then its new time period will become.

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Answer: (D) 33 hours

From Kepler's law of periods, we know
T2∝R3{T}^{2}\propto {R}^{3}
⇒T12T22=R13R23⇒(T1T2)2=(RR4)3\Rightarrow \dfrac{{T}_{1}^{2}}{{T}_{2}^{2}}=\dfrac{{R}_{1}^{3}}{{R}_{2}^{3}} \Rightarrow {\left(\dfrac{{T}_{1}}{{T}_{2}}\right)}^{2}={\left(\dfrac{R}{\dfrac{R}{4}}\right)}^{3}
⇒T12T22=64\Rightarrow \dfrac{{T}_{1}^{2}}{{T}_{2}^{2}}=64
∴T2=248=3∴{T}_{2}=\dfrac{24}{8}=3 hours

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