A satellite is placed between the Earth and the Moon, along a straight line that connects their...

Question:

A satellite is placed between the Earth and the Moon, along a straight line that connects their centers of mass. The satellite has an orbital period around the Earth that is the same as that of the Moon, 27.3 days. How far away from the Earth should this satellite be placed?

Orbital time period:

The orbital time period of a satellite revolving around source mass(e.g earth) is defined as the time taken to revolve one round around the source mass. Mathematically, the orbital time period of a satellite is expressed as follows:

{eq}T = \sqrt{\dfrac{ 4 \pi^{2} \, r^{3} }{ GM } } {/eq}

Where:

  • {eq}M {/eq} is the source mass.
  • {eq}r {/eq} is the orbital radius of the orbit, in which the satellite is revolving around the source mass.
  • {eq}G {/eq} is the gravitational force constant.

Answer and Explanation:


Identify the given information in the problem:

  • Mass of the earth is {eq}M_{earth} = 5.972 \times 10^{24} \, \rm kg {/eq}
  • Mass of the moon is {eq}M_{moon} = 7.348 \times 10^{22} \, \rm kg {/eq}
  • The distance between the center of the earth and the center of the moon is {eq}r = 3.84 \times 10^{8} \, \rm m {/eq}

Let the satellite is placed at a distance of {eq}x {/eq} from the earth. Therefore,

The distance between the moon and the satellite will be equal to {eq}r - x {/eq}

Therefore,

The orbital time period of the satellite around the earth can be expressed as follows:

{eq}T = \sqrt{\dfrac{ 4 \pi^{2} \, x^{3} }{ GM_{earth} } } {/eq}

Similarly,

The orbital time period of the satellite around the moon can be expressed as follows:

{eq}T^{'} = \sqrt{\dfrac{ 4 \pi^{2} \,(r - x)^{3} }{ GM_{moon} } } {/eq}

Now,

According to the problem,

{eq}\begin{align} T^{'} &= T \\ \Rightarrow \sqrt{\dfrac{ 4 \pi^{2} \,(r - x)^{3} }{ GM_{moon} } } &= \sqrt{\dfrac{ 4 \pi^{2} \, x^{3} }{ GM_{earth} } } \\ \Rightarrow (\dfrac{ r - x }{ x })^{3} &= \dfrac{ M_{moon} }{ M_{earth} } \\ \Rightarrow \dfrac{ r - x }{ x } &= (\dfrac{ M_{moon} }{ M_{earth} } )^{1/3} \\ \Rightarrow \dfrac{ r }{ x } - 1 &= (\dfrac{ M_{moon} }{ M_{earth} } )^{1/3} \\ \Rightarrow \dfrac{ r }{ x } &= (\dfrac{ M_{moon} }{ M_{earth} } )^{1/3} + 1 \\ \Rightarrow x &= \dfrac{ r }{ (\dfrac{ M_{moon} }{ M_{earth} } )^{1/3} + 1 } \end{align} {/eq}

After plugging in the values, we have:

{eq}x = \dfrac{ 3.84 \times 10^{8} \, \rm m }{ (\dfrac{ 7.348 \times 10^{22} \, \rm kg }{ 5.972 \times 10^{24} \, \rm kg } )^{1/3} + 1 } {/eq}

Simplifying it further, we will get:

{eq}\color{blue}{\boxed{ x = 3.12 \times 10^{8} \, \rm m }} {/eq}


Learn more about this topic:

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Stable Orbital Motion of a Satellite: Physics Lab

from Physics: High School

Chapter 8 / Lesson 16
565

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