Potential energy of a satellite having mass ‘m’ and rotating at a height of 6.4 x10⁶ m from the earth centre is

The correct option is -0.5 mgR_ₑ.

Let's derive the solution step-by-step:
The gravitational potential energy ($U$) of a satellite of mass $m$ at a distance $r$ from the center of the Earth is given by the formula:
$U=-\frac{GMm}{r}$
where $G$ is the universal gravitational constant and $M$ is the mass of the Earth.

We know that the acceleration due to gravity ($g$) at the surface of the Earth is related to $G$ and $M$ by the relation:
$g=\frac{GM}{R_e^2}$
This can be rewritten as:
$GM=g{R}_e^2$
where $R_e$ is the radius of the Earth ($R_e=6.4\times {10}^6\text{m}$).

The satellite is rotating at a height of $6.4\times {10}^6\text{m}$ (which is equal to the Earth's radius, $R_e$) above the Earth's surface. Thus, the total distance ($r$) of the satellite from the center of the Earth is:
$r=R_e+h=R_e+R_e=2R_e$

Substituting the values of $GM$ and $r$ back into the potential energy equation:
$U=-\frac{\left(g{R}_e^2\right)m}{2R_e}$
Simplifying the expression:
$U=-0.5mg{R}_e$
Therefore, the potential energy of the satellite is $-0.5mg{R}_e$.