A spherical asteroid having the same density | vedclass

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(D) The time period $T$ of a satellite revolving near the surface of a spherical body is given by $T = 2\pi \sqrt{\frac{R^3}{GM}}$.Since the mass $M$

Vedclass · Accepted Answer

$1 \ 	ext{hr} \ 24 \ 	ext{min}$

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(D) The time period $T$ of a satellite revolving near the surface of a spherical body is given by $T = 2\pi \sqrt{\frac{R^3}{GM}}$.Since the mass $M$ of the asteroid can be expressed in terms of its density $ho$ and radius $R$ as $M = ho 	imes \frac{4}{3} \pi R^3$,we substitute this into the formula:$T = 2\pi \sqrt{\frac{R^3}{G (ho \cdot \frac{4}{3} \pi R^3)}} = 2\pi \sqrt{\frac{3}{4 \pi G ho}} = \sqrt{\frac{3\pi}{Gho}}$.This shows that the time period depends only on the density $ho$ of the body.Since the asteroid has the same density as Earth,the time period of the pebble revolving near its surface is the same as the time period of a satellite revolving near the surface of the Earth.This value is approximately $84 \ 	ext{min}$,which is equal to $1 \ 	ext{hr} \ 24 \ 	ext{min}$.

$A$ 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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