Question

Given that the mass of Earth is 5.98x1024 kg, what is the orbital radius of a satellite that has an orbital period of

exactly one day (assume that a day is exactly 24 hours in length)?

Answer 1

The orbital radius is approximately 42,259 kilometers.

Step-by-step explanation:

From Newton's Law of Gravitation we find that acceleration experimented by the satellite (
`a`), measured in meters per square second, is defined by:

`a = (G\cdot M)/(r^(2))` (1)

Where:

`G` - Gravitational constant, measured in cubic meters per kilogram-square second.

`M` - Mass of Earth, measured in kilograms.

`r` - 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:

`(4\pi^(2)\cdot r)/(T^(2)) = (G\cdot M)/(r^(2))`

`4\pi^(2)\cdot r^(3) = G\cdot M\cdot T^(2)`

`r^(3) = (G\cdot M\cdot T^(2))/(4\pi^(2))`

`r = \sqrt[3]{(G\cdot M\cdot T^(2))/(4\pi^(2)) }`

Where
`T` is period, measured in seconds.

If we know that
`G = 6.674* 10^(-11)\,(m^(3))/(kg\cdot s^(2))`,
`M = 5.98* 10^(24)\,kg` and
`T = 86400\,s`, then orbital radius of the satellite is:

`r = \sqrt[3]{(\left(6.674\cdot 10^(-11)\,(m^(2))/(kg\cdot s^(2)) \right)\cdot (5.98* 10^(24)\,kg)\cdot (86400\,s)^(2))/(4\pi^(2)) }`

`r \approx 42.259* 10^(6)\,m`

`r \approx 42.259* 10^(3)\,km`

The orbital radius is approximately 42,259 kilometers.