A planet has twice the mass of earth and of identical size. What will be the height above the surface of the planet where its acceleration due to gravity reduces by $36$ of its value on its surface?

Open in App

Solution

Use g= $\frac{G M}{R^{2}}$ and $g_{h}$ =g ${[\frac{R}{R + h}]}^{2}$

Suggest Corrections

Similar questions

Q. At what height the acceleration due to gravity will be reduced by 36 % its value o the surface of the earth?

A planet has twice the density of earth but the acceleration due to gravity on its surface is exactly the same as on the surface of earth. Its radius in terms of radius of earth R will be:

Q. A planet has mass and radius both half of earth. Acceleration due to gravity (g) at its surface should be

Join BYJU'S Learning Program

A planet has twice the mass of earth and of identical size. What will be the height above the surface of the planet where its acceleration due to gravity reduces by 36 of its value on its surface?

Use g=GMR2 and gh=g[RR+h]2

A planet has twice the mass of earth and of identical size. What will be the height above the surface of the planet where its acceleration due to gravity reduces by $36$ of its value on its surface?

Use g= $\frac{G M}{R^{2}}$ and $g_{h}$ =g ${[\frac{R}{R + h}]}^{2}$