A spherical asteroid having the same density | vedclass

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Question

By Kepler's law

$\mathrm{T}^{2}=\frac{4 \pi^{2}}{\mathrm{GM}} \mathrm{R}^{3}=\frac{4 \pi^{2}}{\mathrm{G} ho} 	imes \frac{4 \pi}{3}$

$\mat

Vedclass · Accepted Answer

$1\ hr\ 24\ min$

Vedclass · Answer

By Kepler's law

$\mathrm{T}^{2}=\frac{4 \pi^{2}}{\mathrm{GM}} \mathrm{R}^{3}=\frac{4 \pi^{2}}{\mathrm{G} ho} 	imes \frac{4 \pi}{3}$

$\mathrm{R}=\mathrm{R}$ asteroid

$\Rightarrow$ so, time period depends only on density

of planet

$\mathrm{T}=\mathrm{TP}$ of near earth orbit satellite $=$

$84 \min =1$ hr $24\, min$

A spherical asteroid having the same density as that of earth is floating in free space. A small pebble is revolving around the asteroid under the influence of gravity near the surface of the asteroid. What is the approximate time period of the pebble?

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