A satellite of mass 225 kg is launched from a site on Earth's equator into an orbit at 200 km...

Question:

A satellite of mass 225 kg is launched from a site on Earth's equator into an orbit at 200 km above the surface of Earth. (The mass of the Earth is {eq}5.98\times 10^{24}{/eq} kg, and the radius of the Earth is {eq}6.38\times 10^3{/eq} km.)

(a) Assuming a circular orbit, what is the orbital period of this satellite?

Orbital Period:

The orbital period is the time in which an orbiting body complete its rotation around its orbit. The orbital period is described by Newton's modification of Kepler's third law. Kepler's Third law relates the orbital period of an orbiting object to the semi-major axis of its orbit or the distance from its host body. The modified Kepler's third law is written as

{eq}T^{2}=\displaystyle \frac{4\pi^{2} r^{3}}{GM} {/eq}

Where:

{eq}T {/eq}=period

{eq}r {/eq}=distance

{eq}M {/eq}=mass of the host body

{eq}G {/eq}=gravitational constant({eq}6.67\times10^{-11} \frac{m^{3}}{kg \cdot s^{2}} {/eq})

Answer and Explanation:

The orbital period of the satellite is {eq}T=5310.13 s {/eq} or {eq}1.475 {/eq} hours.

Solution:

Using the modified Kepler's third law

{eq}T^{2}=\displaystyle \frac{4\pi^{2} r^{3}}{GM} {/eq}

finding the period

{eq}T=\displaystyle \sqrt{\frac{4\pi^{2} r^{3}}{GM}} {/eq}

{eq}T=\displaystyle \sqrt{\frac{4\pi^{2} (2.0\times10^{5} m+6.38\times10^{6} m)^{3}}{(6.67\times10^{-11} \frac{m^{3}}{kg \cdot s^{2}})(5.98\times10^{24} kg)}} {/eq}

{eq}=5310.13 s {/eq}

Converting to hours

{eq}5310.13 s \times \frac{1 min}{60 s} \times \frac{1 hr}{60 mins}=1.475 hrs {/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
43K

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