Question

The $\text{K.E}$. of a satellite in an orbit close to the surface of the earth is $E$. Its maximum $\text{K.E.}$ so as to escape from the gravitational field of the earth is.

• A. $2E$
• B. $4E$
• C. $2\sqrt{2}E$
  
• D. $\sqrt{2}E$

Answer 1

The correct option is A $2E$

Orbital velocity of the satellite near the surface of the earth will be, $$v=\sqrt{\frac{GM}{R}}$$So, the $\text{K.E.}$ of the satellite will be

$E=\frac{1}{2}m{v}^2=\frac{GMm}{2R}...\left(1\right)$

Also the escape velocity of satellite is given by

$v_e=\sqrt{\frac{2GM}{R}}$

So, the maximum kinetic energy to escape from the orbit,

$E^{\prime }=\frac{1}{2}m{v}_e^2$

Substituting the value of $v_e$,

$\Rightarrow E^{\prime }=\frac{GMm}{R}$

From equation $\left(1\right)$,

$\Rightarrow E^{\prime }=2E$

Hence ,option (a) is the correct answer.