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Suppose we want a satellite to revolve around the Earth 5 times a day in a circular orbit. What...

Question:

Suppose we want a satellite to revolve around the Earth 5 times a day in a circular orbit. What should the radius of its circular orbit be?

(Hint: A geosynchronous orbit is at 42,164 km.)

The time period of motion:

The square of the time period of the satellite is directly proportional to the cube of the radius of orbit as per Kepler's third law. The mathematical expression is as follows:

{eq}\rm T^{2} = \dfrac{4 \pi^{2}R^{3}}{GM} {/eq}

where

  • R is the radius of the orbit

Answer and Explanation:

Given Data:

  • Time period {eq}\rm T = \dfrac{24 \ hr}{5} \\ T = \dfrac{24 \times 3600}{5} \\ T = 17280 \ s {/eq}

Now, the time period is given by

{eq}\begin{align} \rm T^{2} &= \rm \dfrac{4 \pi^{2}R^{3}}{GM} \\ (17280)^{2} &= \rm \dfrac{4\pi^{2} \times R^{3}}{(6.67 \times 10^{-11})(5.98 \times 10^{24} \ kg)} \\ \rm R &= \rm 1.44 \times 10^{7} \ m \\ \end{align} {/eq}


Learn more about this topic:

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Kepler's Three Laws of Planetary Motion

from Basics of Astronomy

Chapter 22 / Lesson 12
46K

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