A body of mass m is situated at a distance 4 Rₑ above the earth’s surface, where Rₑ is the radius of earth. How much minimum energy be given to the body so that it may escape

The correct answer is mgRₑ/5.

Let us solve the problem step-by-step:

Step 1: Find the distance of the body from the center of the Earth.
The body of mass $m$ is located at a height $h=4R_e$ above the Earth's surface, where $R_e$ is the radius of the Earth.
The distance $r$ of the body from the center of the Earth is:
$r=R_e+h=R_e+4R_e=5R_e$

Step 2: Calculate the initial total energy of the body.
Assuming the body is initially at rest at this position, its initial kinetic energy is zero. The initial total energy $E_i$ is equal to its gravitational potential energy:
$E_i=-\frac{GMm}{r}$
where $G$ is the universal gravitational constant and $M$ is the mass of the Earth. Substituting $r=5R_e$ gives:
$E_i=-\frac{GMm}{5R_e}$

Step 3: Determine the energy required to escape.
For the body to escape the gravitational field of the Earth, its total final energy $E_f$ at infinity must be at least zero:
$E_f=0$
The minimum energy $E$ that must be supplied to the body is the difference between this final energy and the initial energy:
$E=E_f-E_i=0-\left(-\frac{GMm}{5R_e}\right)=\frac{GMm}{5R_e}$

Step 4: Relate the expression to acceleration due to gravity (g).
We know that the acceleration due to gravity $g$ at the Earth's surface is:
$g=\frac{GM}{R_e^2}$
This can be rewritten to express $GM$ as:
$GM=g{R}_e^2$
Substituting $GM$ back into our minimum energy equation gives:
$E=\frac{\left(g{R}_e^2\right)m}{5R_e}=\frac{mg{R}_e}{5}$

Therefore, the minimum energy that must be given to the body so that it may escape is mgRₑ/5.