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7.Gravitation
hard
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?
A
$24\ h$
B
$365\ days$
C
$10\ min$
D
$1\ hr\ 24\ min$
Solution
By Kepler's law
$\mathrm{T}^{2}=\frac{4 \pi^{2}}{\mathrm{GM}} \mathrm{R}^{3}=\frac{4 \pi^{2}}{\mathrm{G} \rho} \times \frac{4 \pi}{3}$
$\mathrm{R}=\mathrm{R}$ asteroid
$\Rightarrow$ so, time period depends only on density
of planet
$\mathrm{T}=\mathrm{TP}$ of near earth orbit satellite $=$
$84 \min =1$ hr $24\, min$
Standard 11
Physics
