Question

The amount of work done in lifting a body of mass 'm' from the surface of the earth to a height equal to twice the radius of the earth is​

Answer 1

Explanation:

work done by gravity= -change in gravitational potential energy

W= -(GMem)(1/R -1/3R)

W=-2GMem/3R

Answer 2

The amount of work done in lifting a body the surface of the earth to a height 2R is (2/3)mgR.

Explanation:

Let the mass of the earth is M and radius is R.

The gravitation potential energy of the body,

$U_1=-\frac{GMm}{R}$

Here, G is the gravitational constant and m is the mass of the body.

Now the gravitational potential energy at height equal to twice the radius of the earth is​,

$U_2= -\frac{GMm}{R+2R}$

Use work-energy theorem, work done (W) = change in potential energy.

Therefore,

$W=U_2-U_1= -\frac{GMm}{R+2R}-( -\frac{GMm}{R})$

$W=\frac{2GMm}{3R}$

The acceleration due to gravity is given as

$g=\frac{GM}{R^2}$

hence above formula becomes,

$W=\frac{2}{3}mgR$.

Thus, the amount of work done in lifting a body the surface of the earth to a height 2R is (2/3)mgR.

#Learn More: gravitation potential energy.

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