Question

What is the period of a satellite orbiting around the earth in a radius which is one half that of the distance from the earth to the moon?

Answer 1

The satellite has a period of 204.90 days

Step-by-step explanation:

The period can be determine by means of Kepler's third law:

`T^(2) = r^(3)` (1)

Where T is the period of revolution and r is the radius.

`\sqrt{T^(2)} = \sqrt{r^(3)}`

`T = \sqrt{r^(3)}` (2)

The distance between the moon and the Earth has a value of 384400 km, therefore:

`r = (384400km)*(0.50)`

`r = 192200 km`

Finally, equation 2 can be used:

`T = \sqrt{(192200)^(3)}`

`T = 84261672 km`

However, the period can be expressed in days, to do that it is necessary to make the conversion from kilometers to astronomical units:

An astronomical unit (AU) is the distance between the Earth and the Sun (
`1.50x10^(8) km`)

`T = 84261672 km \cdot (1AU)/(1.50x10^(8) km)` ⇒
`0.561 AU`

But 1 year is equivalent to 1 AU according to Kepler's third law, since 1 year is the orbital period of the Earth.

`T = 0.561 AU \cdot (1year)/(1AU)` ⇒
`0.561 year`

`T = 0.561 year \cdot (365.25 days)/(1year)` ⇒
`204.90 days`

`T = 204.90 days`

Hence, the satellite has a period of 204.90 days.