The time period of a geostationary satellite is 24 h, at a height ( is radius of earth) from surface of earth. The time period of another satellite whose height is 2.5 from surface will be, [2019]

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Solution

Q.
The time period of a geostationary satellite is 24 h, at a height $6 R_{E}$ ( $R_{E}$ is radius of earth) from surface of earth. The time period of another satellite whose height is 2.5 $R_{E}$ from surface will be, [2019]

1 $6 \sqrt{2}$ $h$

2 $12 \sqrt{2}$ $h$

3 $\frac{24}{2.5}$ $h$

4 $\frac{12}{2.5}$ $h$

Ans.
(1)

Time period of Geostationary satellite is,

$T = 2 π \sqrt{\frac{a^{3}}{G M}} \Rightarrow T^{2} \propto a^{3}$

$∴$ $\frac{T_{1}^{2}}{T_{2}^{2}} = \frac{a_{1}^{3}}{a_{2}^{3}} \Rightarrow \frac{{(24)}^{2}}{T_{2}^{2}} = \frac{{(7 R_{E})}^{3}}{{(3.5 R_{E})}^{3}}$

$\Rightarrow T_{2}^{2} = \frac{{(24)}^{2} \times {(3.5)}^{3}}{{(7)}^{3}} \Rightarrow T_{2} = \frac{\sqrt{{(24)}^{2}}}{\sqrt{8}} = 6 \sqrt{2}$ $h .$

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