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A satellite of mass 500 kg orbits the Earth with a period of 6000 s. The Earth has a mass of 5.98...

Question:

A satellite of mass {eq}500 \ kg {/eq} orbits the Earth with a period of {eq}6000 \ s {/eq}. The Earth has a mass of {eq}5.98 \times 10^{24} \ kg {/eq}.

(a) Calculate the magnitude of the Earth's gravitational force on the satellite.

(b) Determine the altitude of the satellite above the Earth's surface.

Period of a Satellite:

{eq}\\ {/eq}

A satellite is an object that orbits another celestial object. The period of a satellite is equal to the time required by the satellite to complete one orbit. The centripetal force for the orbital motion of the satellite is provided by the gravitational force between the satellite and the planet.

The time period of a satellite depends on the following factors.

  • The mass of the planet/star/moon.
  • The distance of the satellite from the center of the planet.

Answer and Explanation:

{eq}\\ {/eq}

We are given:

  • The mass of satellite, {eq}m=500\;\rm kg {/eq}
  • The period of the orbital motion of the satellite, {eq}T=6000\;\rm s {/eq}
  • The mass of Earth, {eq}M=5.98\times 10^{24}\;\rm kg {/eq}

a)

The time period of the motion of a satellite around a planet depends only on the mass, {eq}M {/eq}, of the planet and its distance, {eq}r {/eq}, from the center of the planet by the following equation:

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

Here,

  • {eq}G=6.67\times 10^{-11}\;\rm m^3/kg.s^2 {/eq} is the gravitational constant.

After plugging the given values into the above equation, we have:

{eq}\begin{align*} 6000&=2\pi\sqrt {\dfrac{r^3}{6.67\times 10^{-11}\times 5.98\times 10^{24}}}\\ \Rightarrow \dfrac{r^3}{39.89\times 10^{13}}&=\left ( \dfrac{6000}{2\pi} \right )^2\\ \Rightarrow r^3&=39.89\times 10^{13}\times 9.12\times 10^{5}\\ \Rightarrow r&=\left (363.8\times 10^{18} \right )^{1/3}\\ &=7.14\times 10^{6}\;\rm m \end{align*} {/eq}


The radius of the Earth is: {eq}R=6.4\times 10^{6}\;\rm m {/eq}

Therefore, the altitude of the satellite above the surface of the earth is:

{eq}\begin{align*} h&=r-R\\ &=7.14\times 10^{6}-6.4\times 10^{6}\\ &=\boxed{7.4\times 10^{5}\;\rm m} \end{align*} {/eq}


a)

According to Newton's Law of gravitation, the gravitational force between two objects of mass, {eq}m_1 {/eq}, and {eq}m_2 {/eq}, separated by a distance, {eq}r {/eq}, is given by the following equation:

{eq}F=\dfrac{Gm_1m_2}{r^2} {/eq}

Here,

  • {eq}G=6.67\times 10^{-11}\;\rm m^3/kg.s^2 {/eq} is the gravitational constant.

The earth can be approximated as spherical in shape, and a spherical object behaves like a point mass of the same located at the center of the object.

Therefore, the gravitational force on the satellite is:

{eq}\begin{align*} F&=\dfrac{6.67\times 10^{-11}\times 5.98\times 10^{24}\times 500}{\left (7.14\times 10^{6} \right )^2}\\ &=\boxed{3.91\times 10^{3}\;\rm N} \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
565

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