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Question: If \['g'\] is the acceleration due to gravity on earth then the increase in P.E of a body of mass \[...

If ′g′'g' is the acceleration due to gravity on earth then the increase in P.E of a body of mass ′m′'m' up to a distance equal to the radius of earth from the earth surface will be?

Explanation

Solution

The potential energy associated with gravitational force, such as rising items against the Earth's gravity, is known as gravitational energy. The gravitational potential energy of a system of respective masses at a certain distance, using gravitational constant GG is U=−GMmrU = - \dfrac{{GMm}}{r} .

Complete answer:
Let us consider;
Radius of the earth is given by RR
Mass of the earth is given by MM
Mass of the given body will be mm and
Acceleration due to gravity on earth is given by gg
When a body is on the surface of the earth, we can write the force of attraction of the earth on it.
U1=−GMmR......(1){U_1} = - \dfrac{{GMm}}{R}......\left( 1 \right)
Where, G=G = Gravitational constant.
When the body is at a height of hh from the earth's surface, the PE of the system is given by
U2=−GmMR+h{U_2} = - \dfrac{{GmM}}{{R + h}}
The potential energy of the object at height h=Rh = R from the surface of the earth.
U2=−GmMR+R=−GmM2R......(2){U_2} = - \dfrac{{GmM}}{{R + R}} = - \dfrac{{GmM}}{{2R}}......\left( 2 \right)
As a result of the displacement of the body of mass m from the surface to a height equal to the radius (R)\left( R \right) of the earth, the rise in PE is given by-
ΔU=U2−U1 ΔU=−GMmR+R+GMmR ΔU=−GMm2R+GMmR ΔU=12GMmR  \Delta U = {U_2} - {U_1} \\\ \Delta U = - \dfrac{{GMm}}{{R + R}} + \dfrac{{GMm}}{R} \\\ \Delta U = - \dfrac{{GMm}}{{2R}} + \dfrac{{GMm}}{R} \\\ \Delta U = \dfrac{1}{2}\dfrac{{GMm}}{R} \\\
But we know that GM=gR2GM = g{R^2}
Hence, ΔU=12gR2mR\Delta U = \dfrac{1}{2}\dfrac{{g{R^2}m}}{R}
Now, one RR will be cancelled out so the equation will become

ΔU=12gRm ∴ΔU=12mgR  \Delta U = \dfrac{1}{2}gRm \\\ \therefore \Delta U = \dfrac{1}{2}mgR \\\

Therefore the increase in P.E of a body of mass ′m′'m' up to a distance equal to the radius of earth from the earth surface will be 12mgR\dfrac{1}{2}mgR.

Note: It should be noted that the formula for gravitational acceleration is g=GMr2g = \dfrac{{GM}}{{{r^2}}} . It is determined by the earth's mass and radius. This enables us to comprehend the following:-
1. Gravity accelerates all bodies at the same rate, regardless of their mass.
2. It's worth on Earth is determined by the mass of the planet, not the mass of the object.