1. Tardigrade
  2. Question
  3. Physics
  4. The period of revolution of an earths satellite close to surface of earth is 90 min. The time period of another satellite in an orbit at a distance of four times the radius of earth from its surface will be

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Q. The period of revolution of an earths satellite close to surface of earth is $90 \min$. The time period of another satellite in an orbit at a distance of four times the radius of earth from its surface will be

AIIMSAIIMS 2009

Solution:

From Kepler's law
$T^{2} \propto R^{3}$
or $ T \propto R^{3 / 2}$
$\frac{T'}{T}=\left(\frac{R'}{R}\right)^{3 / 2}$
or $ \frac{T'}{T}=\left(\frac{4 R}{R}\right)^{3 / 2}$
$=(4)^{3 / 2}=\left(2^{2}\right)^{3 / 2}=2^{3}=8$
$\therefore T'=8 T=8 \times 90$
$=720 \min$