Question 38: Gravitation

Question

A satellite can be in a geostationary orbit around a planet if it is at a distance R from the centre of the planet. If the planet starts rotating about its axis with double the angular velocity, then to make the satellite geostationary, its orbital
radius should be

Accepted Answer

We know that, angular velocity,
$$\omega  = {{2\pi } \over T}$$
$$ \Rightarrow T = {{2\pi } \over \omega }$$
$$ \Rightarrow {{{T_2}} \over {{T_1}}} = {{{\omega _1}} \over {{\omega _2}}}$$
$$ = {{{\omega _1}} \over {2{\omega _1}}}$$ [$$\because$$ $${\omega _2} = 2{\omega _1}$$]
$$ \Rightarrow {{{T_2}} \over {{T_1}}} = {1 \over 2}$$
$$ \Rightarrow {T_2} = {{{T_1}} \over 2}$$ .... (i)
According to Kepler's law,
$$\begin{aligned}
T^2 & \propto R^3 \
\Rightarrow \quad & R\propto T^{2 / 3} \
\Rightarrow \quad \frac{R_2}{R_1} & =\left(\frac{T_2}{T_1}ight)^{2 / 3} \
& =\left(\frac{T_1 / 2}{T}ight)^{2 / 3} \
& =\left(\frac{1}{2}ight)^{2 / 3} \
R_2 & =\frac{R_1}{2^{2 / 3}} \
& =\frac{R}{4^{1 / 3}}
\end{aligned}
$$

Answer

$$2R$$

Answer

$$\frac{R}{2}$$

Answer

$$\frac{R}{2^{1/3}}$$

Answer

$$\frac{R}{4^{1/3}}$$

Gravitation

A satellite can be in a geostationary orbit around a planet if it is at a distance R from the centre of the planet. If the planet starts rotating about its axis with double the angular velocity, then to make the satellite geostationary, its orbital radius should be

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