Question

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

• A. $90\sqrt{80}$min
• B. 360 min
• C. 720 min
• D. 270 min

Answer 1

The correct option is C

720 min

Distance of second satellite from the centre of earth has become four times.

Now $T^2\propto r^3$ or $T\propto r^{\frac{3}{2}}$

$\therefore$ Time period of second satellite will become $(4{)}^{\frac{3}{2}}$ or 8 times.

Hence T'= 8T = (8) (90) = 720 min