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Calculate the escape velocity on surface of Earth of radius of Moon is 1740km and mass is 7.4...

Question:

Calculate the escape velocity on surface of Earth of radius of Moon is 1740km and mass is 7.4 *10^{22}

Escape Velocity:

The escape velocity of an object is defined as the minimum velocity required to project it into space so that it never returns. The escape velocity of an object from the earth's surface is {eq}11.186 \, \rm km/s {/eq}. Mathematically, the escape velocity of the object from a planet's surface can be expressed as follows:

{eq}v_{e} = \sqrt{ \dfrac{ 2 GM }{ R } } {/eq}

Where:

  • {eq}M {/eq} is the mass of the planet.
  • {eq}R {/eq} is the distance of the planet' surface from its center.
  • {eq}G {/eq} is the universal gravitational constant.

Answer and Explanation:


Identify the given information in the problem:

  • Mass of the moon is {eq}M = 7.4 \times 10^{22} \, \rm kg {/eq}
  • Radius of the moon is {eq}R = 1740 \, \rm km = 1.74 \times 10^{6} \, \rm m {/eq}

The escape velocity of an object from the moon surface is:

{eq}v_{e} = \sqrt{ \dfrac{ 2 GM }{ R } } {/eq}

After plugging in the values, we have:

{eq}v_{e} = \sqrt{ \dfrac{ 2 (6.67 \times 10^{-11} N.m^{2} /kg^{2} )(7.4 \times 10^{22} \, \rm kg) }{ 1.74 \times 10^{6} \, \rm m } } {/eq}

Simplifying it further, we will get:

{eq}\color{blue}{\boxed{ v_{e} = 2.4 \times 10^{3} \, \rm m/s }} {/eq}

Or,

{eq}\color{blue}{\boxed{ v_{e} = 2.4 \, \rm km/s }} {/eq}


Learn more about this topic:

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Circular Velocity & Escape Velocity

from Basics of Astronomy

Chapter 25 / Lesson 5
7.3K

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