A geostationary satellite orbit having a radius of 156 E6 m is established around a planet whose...

Question:

A geostationary satellite orbit having a radius of 156 E6 m is established around a planet whose mass is 9.2 E26 kg. Determine the period of the orbit in Earth-hours.

Orbital Period:


The formula for the orbital time period obtained from the third law of Kepler is shown below:

{eq}\displaystyle T=\frac{2\pi}{\sqrt{GM}}r^{\frac{3}{2}}{/eq}, where:

  • {eq}G=\rm 6.67\times 10^{-11}\ N\cdot m^2/kg^2{/eq} is known as the universal gravitational constant.
  • r is orbital radius.
  • M is mass of the planet.

Answer and Explanation:


We are given the following data:

  • The radius of the orbit of a geostationary satellite is {eq}r=\rm 156\times 10^{6}\ m{/eq}.
  • The mass of the geostationary satellite is {eq}M=\rm 9.2 \times 10^{26}\ kg{/eq}.


By the general formula of orbital period {eq}\displaystyle T=\frac{2\pi}{\sqrt{GM}}r^{\frac{3}{2}}{/eq} and the above values, we have:

{eq}\begin{align*} \displaystyle T&=\frac{2\pi}{\sqrt{(\rm 6.67\times 10^{-11}\ N\cdot m^2/kg^2)(\rm 9.2 \times 10^{26}\ kg)}}(\rm 156\times 10^{6}\ m)^{\frac{3}{2}}\\ &=\displaystyle \rm \frac{2\pi}{\sqrt{61.364\times 10^{-11+26}}}(\sqrt{156\times 10^{6}})^3\ s\\ &=\displaystyle \rm \frac{2\pi}{\sqrt{61.364\times 10^{15}}}(12.49\times 10^{3})^3\ s\\ &=\displaystyle \rm \frac{12242.417\times 10^{9}}{24.772\times 10^{7}}\ s\\ &\approx \boxed{\displaystyle \rm 494.20\times 10^{2}\ s}\\ \end{align*}{/eq}


Converting the unit of obtained value, we get:

{eq}\begin{align*} \displaystyle T&=\displaystyle \rm 494.20\times 10^{2}\ s\times \frac{1\ h}{3600\ s}\\ &=\boxed{\displaystyle \rm 13.73\ hours}\\ \end{align*}{/eq}


Thus, the period of the orbit is 13.73 hours.


Learn more about this topic:

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Kepler's Three Laws of Planetary Motion

from Basics of Astronomy

Chapter 22 / Lesson 12
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