First slide

Keplers 3 Laws

Question

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,

Moderate

A

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

$6 \sqrt{2} h$
C

$12 \sqrt{2} h$
D

$\frac{24}{2.5} h$

Solution

$T^{2} \propto r^{3} T^{2} \propto {(R_{E} + h)}^{3} \frac{T_{1}^{2}}{T_{2}^{2}} = \frac{{(R_{E} + 6 R_{E})}^{3}}{{(R_{E} + 2.5 R_{E})}^{3}} \frac{T_{1}^{2}}{T_{2}^{2}} = \frac{7^{3}}{{(\frac{7}{2})}^{3}} \frac{T_{1}^{2}}{T_{2}^{2}} = 8 T_{2} = \frac{T_{1}}{2 \sqrt{2}} T_{2} = \frac{24}{2 \sqrt{2}} T_{2} = 6 \sqrt{2} h$

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