Question

A ball is launched from the top of Mt. Everest which is at elevation of 9000 m. The ball moves in circular orbit around earth. Acceleration due to gravity near the earth’s surface is g. The magnitude of the ball’s acceleration while in orbit is

• A. Close to g/2
• B. Zero
• C. Much greater than g.
• D. Nearly equal to g.

Answer 1

The correct option is D

Nearly equal to g.

$\frac{m{v}^2}{r}$=m$g^{{}^{\prime }}$(Where $g^{{}^{\prime }}$ is nearly equal to g)