Answer to Question #112859 in Mechanics | Relativity for Clara

Question

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 radius: 6.371x106 m

Earth's mass: 5.97x1024kg

Answer 1

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:

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

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

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

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

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

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

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

Answer: $H = 3.28\sdot10^{20}m.$