Question

A body of mass m is situated at distance $4{{R}_{e}}$ above the earth’s surface, where ${{R}_{e}}$ is the radius of earth. How much minimum energy is given to the body so that it escapes the earth's gravity?

Answer 1

Use the law of conservation of mechanical energy, which says that the sum of the potential energy and the kinetic energy is conserved. In other words, it is constant. Equate the sum of the potential energy at the given position and the minimum energy given to the sum of the potential energy and kinetic energy at infinity.

Formula used:
The potential energy of a body of mass $m$, situated at a distance r from the centre of earth is,
$U=-\dfrac{GMm}{r}$
$G$ is the universal gravitational constant and $M$ is the mass of earth.

Complete step by step answer:
When a body is with the earth’s gravity, it is attracted by the gravitational force exerted by the earth. As a result, the body is bound to the earth and does not leave its gravitational field. Therefore, to make a body to escape the earth’s gravity some energy must be given to the body. Let us calculate the minimum energy to be given to the body such that its escapes the earth’s gravity from its given position. For this we shall use the law of conservation of mechanical energy i.e. the sum of the potential energy and the kinetic energy is conserved (constant). The potential energy of a body of mass m, situated at a distance r from the centre of earth is equal to $U=-\dfrac{GMm}{r}$, G is the universal gravitational constant and M is the mass of earth.

It is given that the body is situated at a distance of $4{{R}_{e}}$ above the surface of earth. This means that it is at a distance of $5{{R}_{e}}$ from the centre of earth.
Therefore, the initial potential energy of the body is ${{U}_{1}}=-\dfrac{GMm}{5{{R}_{e}}}$.
Let us assume that the body is at rest initially and later some energy is given to which increases its kinetic energy to ${{K}_{1}}$.
When the body escapes the earth's gravity, it is no more attracted to earth. Therefore, its potential energy is zero. i.e. ${{U}_{2}}=0$.
Let the kinetic energy of the body at this point be ${{K}_{2}}=0$. (because we want to find the minimum energy to be given).
Now, we know that ${{U}_{1}}+{{K}_{1}}={{U}_{2}}+{{K}_{2}}$
Substitute the known values.
$-\dfrac{GMm}{5{{R}_{e}}}+{{K}_{1}}=0$
$\therefore{{K}_{1}}=\dfrac{GMm}{5{{R}_{e}}}$ …. (i)

This means that the minimum energy to be given to the body such that it escapes the earth’s gravity is equal to $\dfrac{GMm}{5{{R}_{e}}}$.

Note: When the body is at the surface of the earth, it experiences a force whose magnitude is equal to $F=\dfrac{GMm}{R_{e}^{2}}$.
And we know that $F=mg$, where g is the acceleration due to gravity.
Therefore,
$\dfrac{GMm}{R_{e}^{2}}=mg$
$\Rightarrow \dfrac{GM}{R_{e}^{2}}=g$
$\Rightarrow \dfrac{GM}{{{R}_{e}}}=g{{R}_{e}}$.
Substitute this value in (i).
$\therefore {{K}_{1}}=\dfrac{mg{{R}_{e}}}{5}$.
Therefore, the minimum energy can be written in this way too.