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7.Gravitation
easy
The time period of a satellite of earth is $24$ 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 $.......\,hours$
A
$4$
B
$6$
C
$12$
D
$3$
(JEE MAIN-2023)
Solution
$T ^2 \propto R ^3$
$\frac{ T _1^2}{ T _2^2}=\frac{ R _1^3}{ R _2^3} \Rightarrow\left(\frac{ T _1}{ T _2}\right)^2=\left(\frac{ R }{\frac{ R }{4}}\right)^3$
$\frac{ T _1^2}{ T _2^2}=64$
$T _2^2=\frac{ T _1^2}{64}$
$T _2=\frac{24}{8}=3$
Standard 11
Physics
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