A spacecraft is in circular Earth orbit at 5400km altitude. By how much will its altitude...

Question:

A spacecraft is in circular Earth orbit at 5400km altitude. By how much will its altitude decrease if it moves to a new circular orbit where its orbital speed is 20% higher?

Orbital Speed:

The orbital speed of an object is the speed by which the object revolves in the given circular orbit. It is defined as the ratio of the distance traveled and the time taken to make one revolution in the given orbit. If an object(e.g. spacecraft, satellite, etc. ) revolves around a source object(e.g. Earth), the orbital speed of the object revolving around the source object in an orbit of radius r is expressed as such:

{eq}v = \sqrt{ \dfrac{ GM }{ r } } {/eq}

Where:

  • {eq}M {/eq} is the mass of the source object.
  • {eq}r {/eq} is the radius of the orbit in which the object revolves around the source object.

Answer and Explanation:


Identify the given information in the problem:

  • A spacecraft is in circular Earth orbit at {eq}h = 5400 \, \rm km {/eq} altitude.
  • Radius of the earth {eq}R_{e} = 6378.1 \, \rm km {/eq}

When the spacecraft revolves around the earth at an altitude of {eq}h {/eq}, then

The orbital speed of the spacecraft can be expressed as follows:

{eq}v = \sqrt{ \dfrac{ GM_{e} }{ (R_{e} + h ) } } {/eq}

Where:

  • {eq}M_{e} {/eq} is the mass of the earth.
  • {eq}R_{e} {/eq} is the radius of the earth.

According to the given problem, the new orbital speed of the spacecraft to revolve, at the new altitude, in the new circular orbit around the earth is:

{eq}v^{'} = v + 20 \% \, \rm of \, v \\ \Rightarrow v^{'} = v + \dfrac{ v }{ 5 } \\ \Rightarrow v^{'} = \dfrac{ 6 }{ 5 } v \,\,.....(i) {/eq}

If we assume that the new altitude be {eq}h^{'} {/eq}, then

The new orbital speed can be expressed as such:

{eq}v^{'} = \sqrt{ \dfrac{ G M_{e}}{ R_{e} + h ^{'} } } {/eq}

Now,

Substituting {eq}v = \sqrt{ \dfrac{ GM_{e} }{ (R_{e} + h ) } } {/eq} and {eq}v^{'} = \sqrt{ \dfrac{ G M_{e}}{ R_{e} + h ^{'} } } {/eq} into relation (i), we have:

{eq}\sqrt{ \dfrac{ G M_{e}}{ R_{e} + h ^{'} } } = \dfrac{ 6 }{ 5 } \, \sqrt{ \dfrac{ GM_{e} }{ (R_{e} + h ) } } {/eq}

On squaring both the sides, we have:

{eq}\dfrac{ G M_{e} }{ R_{e} + h ^{'} } = \dfrac{ 36 }{ 25 } \, \dfrac{ GM_{e} }{ (R_{e} + h ) } {/eq}

After cross multiplication and then rearranging the terms, we have:

{eq}36 h^{'} = - 11 R_{e} + 25 h \\ \Rightarrow h^{'} = \dfrac{ - 11 R_{e} + 25 h }{ 36 } {/eq}

After plugging in the values, we have:

{eq}h^{'} = \dfrac{ - 11 (6378.1 \, \rm km ) + 25 (5400 \, \rm km ) }{ 36 } {/eq}

Simplifying it further, we will get:

{eq}h^{'} = 1801.136 \, \rm km {/eq}

Thus, the decrement in the altitude, if the spacecraft moves to a new circular orbit where its orbital speed is 20% higher, will be:

{eq}\Delta h = h - h^{'} \\ \Delta h = 5400 \, \rm km - 1801.136 \, \rm km \\ \Delta h = 3598.86 \, \rm km {/eq}

Therefore,

{eq}\begin{align} \text{ The percentage decrement in the altitude } &= \dfrac{ \Delta h }{ h } \times 100 \\ & = \dfrac{ 3598.86 \, \rm km }{5400 \, \rm km } \times 100 \\ & = \color{blue}{\boxed{ 66.64 \% }} \end{align} {/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
536

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