The value of gravity (g) at surface of earth is proportional to `(1)/(R^(2))`, R is radius of earth, what is the value of gravity at height h from the surface.

The acceleration due to gravity (g) on the surface of the the earth is 9.8 m/s^2 . At what height 'h' will the value of 'g' be half of that on the surface of the earth?

If g is acceleration due to gravity on the surface of the Earth and R is radius of the Earth, then the gravitational potential on the surface of the Earth is

Fill in the blanks R- Radius of earth, g-Acceleration due to gravity, h-Height from surface of earth, d- Depth from surface of earth The value of g at a height is (a) g_(h) =_______. The value of g decreases with height.

What will be the acceleration due to gravity at height h if h gt gt R . Where R is radius of earth and g is acceleration due to gravity on the surface of earth

The value of acceleration due to gravity on the surface of earth is x . At an altitude of h from the surface of the earth, its value is y . If R is the radius of earth, then the value of h is

Acceleration due to gravity ' g ' for a body of mass ' m ' on earth's surface is proportional to (Radius of earth= R , mass of earth= M

Given, acceleration due to gravity on surface of earth is g. if g' is acceleration due to gravity at a height h above the surface of earth, then

A particle of mass 'm' is raised to a height h = R from the surface of earth. Find increase in potential energy. R = radius of earth. g = acceleration due to gravity on the surface of earth.

A particle of mass 'm' is raised to a height h = R from the surface of earth. Find increase in potential energy. R = radius of earth. g = acceleration due to gravity on the surface of earth.