The value of g at depth h is two third the value that on the earth's surface. The value of h in terms of radius of earth R is

Write down the average value of g on the earth's surface.

The value of 'g' on earth surface depends :-

The depth 'd' at which the value of acceleration due to gravity becomes (1)/(n) times the value at the earth's surface is (R = radius of earth)

At what height above the earth's surface, the value of g is same as that at a depth of 100 km ?

At what altitude, the acceleration due to gravity reduces to one fourth of its value as that on the surface of the earth ? Take radius of earth as 6.4 xx 10^(6) m , g on the surface of the earth as 9.8 ms^(-2) .

At what altitude, the acceleration due to gravity reduces to half of its value as that on the surface of the earth ? Take radius of earth as 6.410^(6) m, g on the surface of the earth as 9.8 m//s^(2) .

The value of g at a certain height h above the free surface of the earth is x//4 where x is the value of g at the surface of the earth. The height h is

A what height above the earth's surface the value of g becomes 25% of its value on the earth if radius of the earth is 6400km.

The mass of a spaceship is 1000kg. It is to be launched from the earth's surface out into free space. The value of g and R (radius of earth) are 10m/s^2 and 6400 km respectively. The required energy for this work will be:

At what height above the earth's surface does the value of g becomes 36% of the value at the surface of earth ?