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

R=4.22*10⁴km

Step-by-step explanation:

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

`v=(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=(mv^(2) )/(R)`

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

`F_c=(m((2\pi R)/(T))^(2))/(R)=(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=(GMm)/(R^(2))\\\\\implies (GMm)/(R^(2))=(4m\pi ^(2)R)/(T^(2))\\\\R^(3)=(GMT^(2))/(4\pi^(2)) \\\\R=\sqrt[3]{(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]{((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.