Gravitation Ques 30
- A geostationary satellite is orbiting the earth at a height of $6 R$ above the surface of the earth where $R$ is the radius of earth. The time period of another satellite at a height of $3.5 R$ from the surface of the earth is 1.9 hours.
(1987, 2M)
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Answer:
Correct Answer: 30.8.48
Solution:
Formula:
$ \begin{array}{rlrl} T \propto r^{3 / 2} \\ \frac{T_{2}}{T_{1}} ={\frac{r_{2}}{r_{1}}}^{3 / 2} \\ \text { or } T_{2} ={\frac{r_{2}}{r_{1}}}^{2 / 3} \ T_{1} =\frac{3.5 R}{7 R}^{3 / 2} \quad(24) \mathrm{h}=8.48 \mathrm{~h} \end{array} $
($T_{1}=24 \mathrm{~h}$ for geostationary orbit)
BETA
