Question

A satellite that goes around the earth once every 24 hours iscalled a geosynchronous satellite. If a geosynchronoussatellite is in an equatorial orbit, its position appearsstationary with respect to a ground station, and it is known as ageostationary satellite
Find the radius Rof the orbit of a geosynchronous satellite that circles the earth.(Note that Ris measured from the center of the earth, not the surface.)

Answer 1

Answer:

R=4.22*10⁴km

Explanation:

The tangential speed $v$ of the geosynchronous satellite is given by:

$v=\frac{2\pi R}{T}$

Because $2\pi R$ is the circumference length (the distance traveled) and T is the period (the interval of time).

Now, we know that the centripetal force of an object undergoing uniform circular motion is given by:

$F_c=\frac{mv^{2} }{R}$

If we substitute the expression for $v$ in this formula, we get:

$F_c=\frac{m(\frac{2\pi R}{T})^{2}}{R}=\frac{4m\pi ^{2}R}{T^{2}}$

Since the centripetal force is the gravitational force $F_g$ between the satellite and the Earth, we know that:

$F_g=\frac{GMm}{R^{2}}\\\\\implies \frac{GMm}{R^{2}}=\frac{4m\pi ^{2}R}{T^{2}}\\\\R^{3}=\frac{GMT^{2}}{4\pi^{2}} \\\\R=\sqrt[3]{\frac{GMT^{2}}{4\pi^{2}} }$

Where G is the gravitational constant ($G=6.67*10^{-11} Nm^{2}/kg^{2}$) and M is the mass of the Earth ($M=5.97*10^{24}kg$). Since the period of the geosynchronous satellite is 24 hours (equivalent to 86400 seconds), we finally can compute the radius of the satellite:

$R=\sqrt[3]{\frac{(6.67*10^{-11}Nm^{2}/kg^{2})(5.97*10^{24}kg)(86400s)^{2}}{4\pi^{2}}}\\\\R=4.22*10^{7}m=4.22*10^{4}km$

This means that the radius of the orbit of a geosynchronous satellite that circles the earth is 4.22*10⁴km.