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

Answer:

The satellite has a period of 204.90 days

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 ($1.50x10^{8} km$)

$T = 84261672 km \cdot \frac{1AU}{1.50x10^{8} km}$ ⇒ $0.561 AU$

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

$T = 0.561 AU \cdot \frac{1year}{1AU}$ ⇒ $0.561 year$

$T = 0.561 year \cdot \frac{365.25 days}{1year}$ ⇒ $204.90 days$

                                                   

$T = 204.90 days$

Hence, the satellite has a period of 204.90 days.