# Planet Z-34 has a mass equal to four times that of Earth and a radius equal to one-fourth that of...

## Question:

Planet Z-34 has a mass equal to four times that of Earth and a radius equal to one-fourth that of Earth. If ve is the escape velocity for Earth, the escape velocity for Z-34 is

(A) ve/4.

(B) 2 ve.

(C) ve/9.

(D) 4 ve.

## Escape velocity

Gravitational force is an attractive force that acts between two objects. Escape velocity is the minimum velocity required for an object to escape the gravitational influence of a massive object like planetary gravitational field. The escape velocity can be derived from the conservation of mechanical energy.

Escape velocity is the minimum velocity required for an object to escape the gravitational field of a massive body. It is given by the equation

{eq}\displaystyle v = \sqrt{ \frac{GM}{r}} {/eq}

where G is the universal gravitational constant ({eq}G = 6.67 \times 10^{-11} m^3 \cdot kg^{-1} \cdot s^{-2} {/eq}), M is the mass of the planet, r is the radius of the planet.

Suppose the escape velocity in Earth is given by

{eq}\displaystyle v_{e} = \sqrt{ \frac{GM_{Earth}}{r_{Earth}}} {/eq}.

If the mass of planet Z-34 is four times that of Earth, {eq}M_{Z-34} = 4M_{Earth} {/eq} and the radius is {eq}r_{Z-34} = \frac{1}{4}r_{Earth} {/eq}, we can now calculate the escape velocity in planet Z-34 as:

{eq}\displaystyle \begin{align*} v_{Z-34} &= \sqrt{\frac{GM_{Z-34}}{r_{Z-34}}}\\ &= \sqrt{\frac{G(4M_{Earth})}{\frac{1}{4}r_{Earth}}}\\ &= \sqrt{16\frac{G(M_{Earth})}{r_{Earth}}}\\ &= 4 \sqrt{\frac{G(M_{Earth})}{r_{Earth}}}\\ v_{Z-34} &= 4 v_{e}\\ \end{align*} {/eq}

If the escape velocity of Earth is {eq}v_e {/eq}, therefore, the escape velocity for Z-34 is {eq}\boxed{\text{ 4 times the escape velocity in Earth or (D) 4} v_e} {/eq} 