If $ρ$ is the density of the planet, the time period of nearby satellite is given by.

\sqrt{\frac{4 π}{3 G ρ}}

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\sqrt{\frac{4 π}{G ρ}}

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\sqrt{\frac{3 π}{G ρ}}

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\sqrt{\frac{π}{G ρ}}

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Solution

The correct option is C $\sqrt{\frac{3 π}{G ρ}}$
Time period of nearby satellite
$T = 2 π \sqrt{\frac{r^{3}}{G M}}$
$2 π \sqrt{\frac{R^{3}}{G M}}$
$= \frac{2 π (R^{3})^{1 / 2}}{{[G \cdot \frac{4}{3} π R^{3} ρ]}^{1 / 2}}$
$= \sqrt{\frac{3 π}{G ρ}}$

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If ρ is the density of the planet, the time period of nearby satellite is given by.

The correct option is C √3πGρTime period of nearby satelliteT=2π√r3GM2π√R3GM=2π(R3)1/2[G⋅43πR3ρ]1/2=√3πGρ

If $ρ$ is the density of the planet, the time period of nearby satellite is given by.

The correct option is C $\sqrt{\frac{3 π}{G ρ}}$
Time period of nearby satellite
$T = 2 π \sqrt{\frac{r^{3}}{G M}}$
$2 π \sqrt{\frac{R^{3}}{G M}}$
$= \frac{2 π (R^{3})^{1 / 2}}{{[G \cdot \frac{4}{3} π R^{3} ρ]}^{1 / 2}}$
$= \sqrt{\frac{3 π}{G ρ}}$