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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.

Answer and Explanation:


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}


Learn more about this topic:

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Circular Velocity & Escape Velocity

from Basics of Astronomy

Chapter 25 / Lesson 5
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