The orbital period of the artificial satellite is 1.5 hours.
Given the data in the question;
Acceleration due to gravity; \(a = 9.00 m/s^2\)
Orbital period of the satellite; \(T =\ ?\)
To determine the orbital period of the satellite, we make use of Kepler's third law:
\(T^2= [\frac{4\pi ^2}{GM_E}]r^3\)
First we find the orbital radius of the artificial satellite
Using Newton's law of universal gravitation
\(F = \frac{Gm_Em}{r^2}\)
Where m is mass of satellite, \(m_E\) is mass of earth ( \(5.972 * 10^{24} kg\) ), G is the universal gravitational constant( \(6.67430*10^{-11} N.m^2/kg^2\) ) and r is the orbital radius.
We know that, Force; \(F = mass *gravity = mg\)
So
\(mg = \frac{Gm_Em}{r^2}\\\\g = \frac{Gm_E}{r^2}\\\\r^2 = \frac{Gm_E}{g}\\\\r = \sqrt{\frac{Gm_E}{g}}\)
We substitute in the values
\(r = \sqrt{\frac{(6.67430*10^{-11}N.m^2/kg^2)(5.972*10^{24}kg}{9.00m/s^2}}\\\\r = 6.65*10^6 m\)
Now, using the of Kepler's third law:
\(T^2= [\frac{4\pi ^2}{GM_E}]r^3\)
We substitute our values into the equation
\(T^2 = [\frac{4\pi ^2}{6.67430*10^{-11}N.m^2/kg^2)(5.972*10^{24}kg} ](6.65*10^6m)^3\\\\T^2 = 2.9127*10^{7}s^2\\\\T = \sqrt{2.9127*10^{7}s^2}\\\\T = 5396.94s\\\\T = 1.5hrs\)
Therefore, the orbital period of the artificial satellite is 1.5 hours.
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