Question

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

• A. $90\sqrt{9}$ min
• B. 270 min
• C. 720 min
• D. 360 min

Answer 1

Answer: (C)

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