Answer:
The orbital radius is approximately 42,259 kilometers.
Explanation:
From Newton's Law of Gravitation we find that acceleration experimented by the satellite ([tex]a[/tex]), measured in meters per square second, is defined by:
[tex]a = \frac{G\cdot M}{r^{2}}[/tex] (1)
Where:
[tex]G[/tex] - Gravitational constant, measured in cubic meters per kilogram-square second.
[tex]M[/tex] - Mass of Earth, measured in kilograms.
[tex]r[/tex] - Orbital radius, measured in meters.
By supposing the satellite rotates at constant speed and in a circular path, we find that acceleration is entirely centripetal and can be defined in terms of period, that is:
[tex]\frac{4\pi^{2}\cdot r}{T^{2}} = \frac{G\cdot M}{r^{2}}[/tex]
[tex]4\pi^{2}\cdot r^{3} = G\cdot M\cdot T^{2}[/tex]
[tex]r^{3} = \frac{G\cdot M\cdot T^{2}}{4\pi^{2}}[/tex]
[tex]r = \sqrt[3]{\frac{G\cdot M\cdot T^{2}}{4\pi^{2}} }[/tex]
Where [tex]T[/tex] is period, measured in seconds.
If we know that [tex]G = 6.674\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}}[/tex], [tex]M = 5.98\times 10^{24}\,kg[/tex] and [tex]T = 86400\,s[/tex], then orbital radius of the satellite is:
[tex]r = \sqrt[3]{\frac{\left(6.674\cdot 10^{-11}\,\frac{m^{2}}{kg\cdot s^{2}} \right)\cdot (5.98\times 10^{24}\,kg)\cdot (86400\,s)^{2}}{4\pi^{2}} }[/tex]
[tex]r \approx 42.259\times 10^{6}\,m[/tex]
[tex]r \approx 42.259\times 10^{3}\,km[/tex]
The orbital radius is approximately 42,259 kilometers.
