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7.Gravitation
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The time period of a geostationary satellite is $24\; \mathrm{h}$, at a helght $6 \mathrm{R}_{\mathrm{E}}( \mathrm{R}_{\mathrm{E}}$ is radius of earth) from surface of earth. The time period of another satellite whose helght is $2.5 \mathrm{R}_{\mathrm{E}}$ from surface will be
A
$6 \sqrt{2} \mathrm{h}$
B
$12 \sqrt{2} \mathrm{h}$
C
$\frac{24}{2.5} \mathrm{h}$
D
$\frac{12}{25} \mathrm{h}$
(NEET-2019) (AIIMS-2011)
Solution
Kepler's Third Law :
$T\propto r^{3 / 2}$
$\frac{\mathrm{T}_{2}}{\mathrm{T}_{1}}=\left(\frac{\mathrm{r}_{2}}{\mathrm{r}_{1}}\right)^{3 / 2}=\left(\frac{\mathrm{R}+2.5 \mathrm{R}}{\mathrm{R}+6 \mathrm{R}}\right)^{3 / 2}=\frac{1}{2 \sqrt{2}} $
$\Rightarrow \mathrm{T}_{2}=\frac{24}{2 \sqrt{2}}=6 \sqrt{2} \mathrm{hours}$
Standard 11
Physics
