Gravitation 3 Question 7
10. 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 hours.
(1987, 2M)
True / False
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Answer:
Correct Answer: 10. $F$
Solution:
$$ \begin{array}{rlrl} & & & \propto r^{3 / 2} \\ & & \frac{T _2}{T _1} & ={\frac{r _2}{r _1}}^{3 / 2} \\ \text { or } & T _2 & ={\frac{r _2}{r _1}}^{3 / 2} \\ & T _1 & =\frac{3.5 R}{7 R}^{3 / 2} \quad(24) h=8.48 h \end{array} $$
( $T _1=24 h$ for geostationary satellite)
