Question

A ball is launched from the top of Mr.Everest which is at elevation of $9000m$. 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 $g$)

As we can write $g^{\prime }=g(1-\frac{h}{R})$ where $R$ is the radius of the earth. Now in comparison with the radius of earth the elevation of Everest is very less so we can finally write $g^{\prime }=g$.

Hence D is correct answer.