107 139
Assignments Done
100%
Successfully Done
In January 2023

# Answer to Question #112859 in Mechanics | Relativity for Clara

Question #112859
The Hubble Space Telescope is in a "Low-Earth Orbit" with an orbital period of 95min. Calculate the altitude of the HST's orbit above the earth's surface. Express your answer either in meters.

Earth's mass: 5.97x1024kg
1
2020-04-30T10:41:23-0400

Solution.

"T = 95 min = 5700s;"

"R_{\\bigoplus} = 6.371\\sdot10^6m;"

"M_{\\bigoplus} = 5.97\\sdot10^{24}kg;"

"H - ?;"

The linear velocity of a satellite orbiting the Earth is sought by the formula:

"\\upsilon = \\sqrt{G\\sdot \\dfrac{M_{\\bigoplus}}{R_{\\bigoplus}+H}};"

In addition, the linear velocity of the body in a circle is sought by the following formula:

"\\upsilon = \\dfrac{2\\pi R}{T};"

For our case we have:

"\\dfrac{2\\pi( R_{\\bigoplus} +H)}{T} = \\sqrt{G\\sdot \\dfrac{M_{\\bigoplus}}{R_{\\bigoplus}+H}};"

"\\dfrac{4\\pi^2( R_{\\bigoplus} +H)^2}{T^2} = {G\\sdot \\dfrac{M_{\\bigoplus}}{R_{\\bigoplus}+H}};"

"\\dfrac{4\\pi^2( R_{\\bigoplus} +H)}{T^2} = {G\\sdot M_{\\bigoplus}};"

"R_{\\bigoplus} +H =\\dfrac{ G\\sdot M_ {\\bigoplus}T^2}{4\\pi^2};"

"H = \\dfrac{ G\\sdot M_ {\\bigoplus}T^2}{4\\pi^2} - R_{\\bigoplus};"

"H = \\dfrac{6.67\\sdot 10^{-11}\\sdot5.97\\sdot10^{24}\\sdot5700^2}{4\\sdot3.14^2} -6.371\\sdot10^6="

"=3.28\\sdot10^{20}m;"

Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!