A satellite is revolving round the earth with orbital speed v₀ . If it stops suddenly, the speed with which it will strike the surface of earth would be ( vₑ = escape velocity of a particle on earth’s surface)

The correct option is √(vₑ²-2v₀²).

Let us find the relationship by analyzing the energy of the satellite at different points.

Step 1: Express the orbital speed of the satellite
Let $M$ be the mass of the Earth, $R$ be the radius of the Earth, and $r$ be the orbital radius of the satellite of mass $m$.
The orbital velocity $v_0$ is given by:
$v_0=\sqrt{\frac{GM}{r}}$
Squaring both sides, we get:
$v_0^2=\frac{GM}{r}$ --- (Equation 1)

Step 2: Express the escape velocity on the Earth's surface
The escape velocity $v_e$ from the surface of the Earth is given by:
$v_e=\sqrt{\frac{2GM}{R}}$
Squaring both sides, we get:
$v_e^2=\frac{2GM}{R}$ --- (Equation 2)

Step 3: Apply the Law of Conservation of Energy
When the satellite is suddenly stopped, its kinetic energy becomes zero. Thus, its initial mechanical energy at radius $r$ is solely its gravitational potential energy:
$E_i=-\frac{GMm}{r}$

As the satellite falls and strikes the surface of the Earth (at distance $R$) with speed $v$, its final mechanical energy is the sum of its kinetic and potential energies:
$E_f=\frac{1}{2}m{v}^2-\frac{GMm}{R}$

By conservation of mechanical energy:
$E_i=E_f$
$-\frac{GMm}{r}=\frac{1}{2}m{v}^2-\frac{GMm}{R}$

Divide the entire equation by mass $m$:
$-\frac{GM}{r}=\frac{1}{2}{v}^2-\frac{GM}{R}$

Multiply by 2 and solve for $v^2$:
$v^2=\frac{2GM}{R}-\frac{2GM}{r}$

Substitute the values from Equation 1 ($\frac{GM}{r}=v_0^2$) and Equation 2 ($\frac{2GM}{R}=v_e^2$):
$v^2=v_e^2-2v_0^2$

Taking the square root:
$v=\sqrt{v_e^2-2v_0^2}$