Respuesta :
Answer:
a) The altitude of the satellite is approximately 2129 kilometers.
b) The gravitational acceleration at the location of the satellite is 5.517 meters per square second.
Explanation:
a) At first we assume that Earth is a sphere with a uniform distributed mass and the satellite rotates on a circular orbit at constant speed. From Newton's Law of Gravitation and definition of uniform circular motion, we get the following identity:
[tex]G\cdot \frac{m\cdot M}{(R_{E}+h)^{2}} = m\cdot \omega^{2}\cdot (R_{E}+h)[/tex] (Eq. 1)
Where:
[tex]G[/tex] - Gravitational constant, measured in newton-square meters per square kilograms.
[tex]m[/tex] - Mass of the satellite, measured in kilograms.
[tex]M[/tex] - Mass of the Earth, measured in kilograms.
[tex]\omega[/tex] - Angular speed of the satellite, measured in radians per second.
[tex]R_{E}[/tex] - Radius of the Earth, measured in meters.
[tex]h[/tex] - Height of the satellite above surface, measured in meters.
Then, we simplify the formula and clear the height above the surface:
[tex]G\cdot M = \omega^{2}\cdot (R_{E}+h)^{3}[/tex]
[tex](R_{E}+h)^{3} =\frac{G\cdot M}{\omega^{2}}[/tex]
[tex]R_{E}+h = \sqrt[3]{\frac{G\cdot M}{\omega^{2}} }[/tex]
[tex]h = \sqrt[3]{\frac{G\cdot M}{\omega^{2}} }-R_{E}[/tex]
From Rotation physics, we know that angular speed is equal to:
[tex]\omega = \frac{2\pi}{T}[/tex] (Eq. 2)
And (Eq. 1) is now expanded:
[tex]h = \sqrt[3]{\frac{G\cdot M\cdot T^{2}}{4\pi^{2}} }-R_{E}[/tex] (Eq. 3)
If we know that [tex]G = 6.674\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}}[/tex], [tex]M = 5.972\times 10^{24}\,kg[/tex], [tex]T = 7800\,s[/tex] and [tex]R_{E} = 6.371\times 10^{6}\,m[/tex], then the altitude of the satellite is:
[tex]h = \sqrt[3]{\frac{\left(6.674\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}} \right)\cdot (5.972\times 10^{24}\,kg)\cdot (7800\,s)^{2}}{4\pi^{2}} }-6.371\times 10^{6}\,m[/tex]
[tex]h\approx 2.129\times 10^{6}\,m[/tex]
[tex]h \approx 2.129\times 10^{3}\,km[/tex]
The altitude of the satellite is approximately 2129 kilometers.
b) The value for the gravitational acceleration of the satellite ([tex]g[/tex]), measured in meters per square second, is derived from the Newton's Law of Gravitation, that is:
[tex]g = G\cdot \frac{M}{(R_{E}+h)^{2}}[/tex] (Eq. 4)
If we know that [tex]G = 6.674\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}}[/tex], [tex]M = 5.972\times 10^{24}\,kg[/tex],[tex]R_{E} = 6.371\times 10^{6}\,m[/tex] and [tex]h\approx 2.129\times 10^{6}\,m[/tex], then the value of the gravitational acceleration at the location of the satellite is:
[tex]g = \frac{\left(6.674\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}} \right)\cdot (5.972\times 10^{24}\,kg)}{(6.371\times 10^{6}\,m+2.129\times 10^{6}\,m)^{2}}[/tex]
[tex]g = 5.517\,\frac{m}{s^{2}}[/tex]
The gravitational acceleration at the location of the satellite is 5.517 meters per square second.
