A satellite is revolving around a planet of mass M in an elliptical orbit of semi-major axis a. The orbital velocity of the satellite at a distance r from the focus will be

The correct option is: [GM(2/r - 1/a)]⁰.⁵

To find the orbital velocity of the satellite, we apply the law of conservation of mechanical energy. For a satellite of mass m revolving around a planet of mass M in an elliptical orbit of semi-major axis a, the total mechanical energy E remains constant.

The total mechanical energy E of the satellite in an elliptical orbit is given by:
$E=-\frac{GMm}{2a}$
where G is the universal gravitational constant.

At any point in the orbit at a distance r from the focus (where the planet is located), the total mechanical energy is the sum of its kinetic energy (K) and gravitational potential energy (U):
$E=K+U$
where the kinetic energy is:
$K=\frac{1}{2}m{v}^2$
and the gravitational potential energy is:
$U=-\frac{GMm}{r}$
Here, v represents the orbital velocity of the satellite at distance r.

Equating the two expressions for total energy:
$-\frac{GMm}{2a}=\frac{1}{2}m{v}^2-\frac{GMm}{r}$

Dividing both sides of the equation by the mass of the satellite m:
$-\frac{GM}{2a}=\frac{1}{2}{v}^2-\frac{GM}{r}$

Rearranging the terms to solve for v²:
$\frac{1}{2}{v}^2=\frac{GM}{r}-\frac{GM}{2a}$

Multiplying both sides of the equation by 2:
$v^2=GM\left(\frac{2}{r}-\frac{1}{a}\right)$

Taking the square root of both sides gives the orbital velocity of the satellite:
$v={\left[GM\left(\frac{2}{r}-\frac{1}{a}\right)\right]}^{0.5}$