MCQ
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
    $10\ min$
  • $1\ hr\ 24\ min$

Answer

Correct option: D.
$1\ hr\ 24\ min$
d
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 $24\, min$

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