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:
[tex]24 \div 24 = 1[/tex]
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:
[tex] \frac{T^2}{d^3} = k [/tex]
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:
[tex]\frac{T^2}{d^3} = \frac{T^2}{d^3}[/tex]
Plug in your values for the moon and satellite:
[tex]\text{Moon:} \ t = 28, d = 240,000[/tex]
[tex]\text{Satellite:} \ t = 1, d = x[/tex]
[tex] \frac{28^2}{240000^3} = \frac{1^2}{x^3} [/tex]
We can rearrange this proportion so that we can solve for x:
[tex]x^3 = 240000^3 \times (1^2 / 28^2)[/tex]
[tex] \sqrt[3]{x^3} = \sqrt[3]{240000^3 \times \frac{1}{784}}[/tex]
[tex]x = 240000 \times \sqrt[3]{\frac{1}{784}}[/tex]
[tex]x = 26,027.9054[/tex]
The altitude of this satellite will be 26,027.9 miles. Keeping significant figures in mind, the altitude will be 26,000 miles.
