Question

The period of revolution of an earth's 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)

720 min

Answer 2

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 $\frac{T'}{T}=\left(\frac{4R}{R}\right)^{3/2}$ = $(4)^{3/2}=(2^2)^{3/2}=2^3=8$ = $T' =8T = 8\times90$ = 720 min