# Define gravitational potential energy. Deduce an expression for itfor a mass in the gravitational field of the Earth.

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Potential energy=m x g x h

Where:- m→Mass of the body

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Gravitational force between an object of mass m and the Earth is given by: F = m g.

We assume that the force is constant over distance.

Gravitational potential energy is the work done against the Earth's gravitational field when the object is displaced by a unit length ie. , 1 meter.

Suppose the object is displaced by h meters. (from a height 0 to height h,

Work done = Force * displacement = m g * h

So this is the potential energy of the mass at height h.

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We assume that the force (weight) on the object is not a constant over distance.

R = radius of Earth, h = altitude of object h above Earth's surface.

Newtons laws: F = G M m / x²

work done =

On the surface of Earth, h = 0, d = R,

GM/R^2 = g and PE = - m g R

at an altitude h: (h << R)

Potential energy = - m g R^2 / (R + h) ≈ - m g R (1 - h / R)

≈ - m g R + m g h

so change in potential energy when the object moves up by a height h

≈ m g h

We assume that the force is constant over distance.

Gravitational potential energy is the work done against the Earth's gravitational field when the object is displaced by a unit length ie. , 1 meter.

Suppose the object is displaced by h meters. (from a height 0 to height h,

Work done = Force * displacement = m g * h

So this is the potential energy of the mass at height h.

===========================

We assume that the force (weight) on the object is not a constant over distance.

R = radius of Earth, h = altitude of object h above Earth's surface.

Newtons laws: F = G M m / x²

work done =

On the surface of Earth, h = 0, d = R,

GM/R^2 = g and PE = - m g R

at an altitude h: (h << R)

Potential energy = - m g R^2 / (R + h) ≈ - m g R (1 - h / R)

≈ - m g R + m g h

so change in potential energy when the object moves up by a height h

≈ m g h