The acceleration due to gravity on a planet is the same as that on Earth,and its radius is four times that of Earth. What will be the value of the escape velocity on that planet if it is $v_e$ on Earth?

What is the escape energy of an object of mass $m$ at a height of $4R_e$ from the surface of the Earth? ($R_e$ = radius of the Earth)

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

$A$ body is projected vertically upwards from the Earth's surface. If the velocity of projection is $\left(\frac{1}{3}\right)$ of the escape velocity,then the height up to which the body rises is $(R = \text{radius of Earth})$

The masses and radii of the earth and moon are $(M_1, R_1)$ and $(M_2, R_2)$ respectively. Their centers are at a distance $r$ apart. Find the minimum escape velocity for a particle of mass $m$ to be projected from the midpoint between these two masses.

$A$ body is projected vertically upwards from the surface of the Earth with a velocity equal to one-third of the escape velocity. The maximum height attained by the body will be $...... \ km$. (Take radius of Earth $R = 6400 \ km$ and $g = 10 \ m/s^2$)