A satellite can be in a geostationary orbit around earth at a distance $r$ from the center. If the angular velocity of earth about its axis doubles, a satellite can now be in a geostationary orbit around earth, if its distance from the center is

\frac{r}{2}

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\frac{r}{2 \sqrt{2}}

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\frac{r}{(4)^{1 / 3}}

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\frac{r}{(2)^{1 / 3}}

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Solution

The correct option is C $\frac{r}{(4)^{1 / 3}}$
Given, $ω_{2} = 2 ω$

For case-1:
$T_{1} = \frac{2 π}{ω}$

For case-2:
$T_{2} = \frac{π}{ω}$ (As $ω_{2} = 2 ω$ )

Using Kepler's 3rd law : $T^{2} \propto r^{3}$
or $\frac{T_{1}^{2}}{T_{2}^{2}} = \frac{r^{3}}{r_{2}^{3}}$
So, $\frac{T_{1}^{2}}{T_{2}^{2}} = \frac{4 π^{2}}{ω^{2}} \times \frac{4 ω^{2}}{4 π^{2}} = 4$
here, $\frac{r^{3}}{r_{2}^{3}} = 4$
$\Rightarrow r_{2} = \frac{r}{(4)^{1 / 3}}$

Hence, option C is correct.

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