Question

The magnitude of potential energy per unit mass of an object at the surface of earth is $E$, the escape velocity of the object is

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

Answer 1

The correct option is A $\sqrt{2E}$

Let the mass of the earth is $M$, radius is $R$ and mass of the object is $m$.

The gravitational potential energy of an object on surface of earth = $\frac{-GMm}{R}$

Magnitude of potential energy per unit mass$=\left(\frac{GM}{R}\right)=E\dots \left(i\right)$

The escape velocity of an object from the surface of earth is given by
$v_{escape}=\sqrt{\frac{2GM}{R}}$

$\Rightarrow v_{escape}=\sqrt{2E}$ (By using equation $\left(i\right)$)

Hence, option (a) is correct.