Question

Our moon is 240,000 miles away and takes about 28 days to orbit the earth. How high up (in miles) would a satellite have to be placed if it were to orbit the earth in 24 hours. (This satellite orbits over the same spot on the earth in an equatorial orbit.)

Answer 1

The moon takes 28 days to orbit the earth at a distance of 240,000 miles. The satellite will take 24 hours to orbit the earth at an unknown distance. 

We want to convert hours into days so that we have a constant unit of time for both the moon and satellite.

There are 24 hours in a day. Divide the amount of hours by 24:

$24 \div 24 = 1$

The satellite will take 1 day to orbit the Earth.

This question involves Kepler's 3rd law of planetary motion. This law can use a simplified formula:

$\frac{T^2}{d^3} = k$

T is the orbital period, and d is the distance from the earth for the moon and satellite.

We can set k to equal the same formula:

$\frac{T^2}{d^3} = \frac{T^2}{d^3}$

Plug in your values for the moon and satellite:

$\text{Moon:} \ t = 28, d = 240,000$
$\text{Satellite:} \ t = 1, d = x$

$\frac{28^2}{240000^3} = \frac{1^2}{x^3}$

We can rearrange this proportion so that we can solve for x:

$x^3 = 240000^3 \times (1^2 / 28^2)$

$\sqrt[3]{x^3} = \sqrt[3]{240000^3 \times \frac{1}{784}}$

$x = 240000 \times \sqrt[3]{\frac{1}{784}}$

$x = 26,027.9054$

The altitude of this satellite will be 26,027.9 miles. Keeping significant figures in mind, the altitude will be 26,000 miles.