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The moon has a period of 27.3 days and a mean distance of 3.90 times 10^5, km from the...

Question:

The moon has a period of {eq}27.3 {/eq} days and a mean distance of {eq}3.90 \times 10^5\, km {/eq} from the center of Earth.

a) Use Kepler's laws to find the period of a satellite in orbit {eq}6.70 \times 10^3\, km {/eq} from the center of Earth.

b) How far above Earth's surface is this satellite?

Johannes Kepler.

He was a German scientist who collected and analyzed in detail the astronomical data recorded by his mentor Tycho Brahe and formulated his three famous laws, which explain the movement of the planets around the sun.

Answer and Explanation:

Variables:

R is the radius of the earth

M is the mass of the earth

G is the universal gravitational constant

r is the orbital radius between the center of the earth and the satellite.

T is the orbital period of the satellite.


{eq}\text{Known data:}\\ r = 6.70\times{10^3}\,km = 6.70\times{10^{6}}\,m\\ M = 5.98\times{10^{24}}\,kg \\ R = 6.38\times{10^{6}}\,m \\ G = 6.674\times{10^{ - 11}}\dfrac{N\,m^2}{kg^2} = 6.674\times{10^{ - 11}}\dfrac{m^3}{kg\,s^2}\\ {/eq}

Part (a)

The satellite period will be determined from Kepler's third law.

{eq}T = \sqrt{\dfrac{4\pi^2(r)^3}{G\,M}} \\ T = \sqrt{\dfrac{4\pi^2(6.70\times{10^{6}}\,m)^3}{\left(6.674\times{10^{ - 11}}\dfrac{m^3}{kg\,s^2}\right)(5.98\times{10^{24}}\,kg)}} = 5454\,s = 90.9\,min \\ \color{blue}{T = 90.9\,min } {/eq}


Part (b)

The orbital height h of the satellite will be determined from the following expression:

{eq}r = R + h \\ h = r - R = 6.70\times{10^{6}}\,m - 6.38\times{10^{6}}\,m = 0.32\times{10^{6}}\,m = 320\times{10^{3}}\,m \\ \color{blue}{h = 320\,km } {/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
47K

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