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 12 hours. (This satellite orbits over the same spot on the earth in an equatorial orbit.)

Answer 1

16,400 miles

Step-by-step explanation:

Kepler's third law states that the ratio between the cube of the distance of a satellite from its planet and the square of its orbital period is constant for all the satellite orbiting around that planet:

`(d^3)/(T^2)=const.`

where d is the distance of the satellite from the planet and T is the orbital period.

By applying this law to the Moon and the other satellite of this problem, we can write

`(d_M^3)/(T_M^2)=(d_S^3)/(T_S^2)`

where
`d_M=240,000 miles` is the distance of the Moon from the Earth,
`T_M=28 d` is its orbital period,
`T_S=12 h=0.5 d` is the orbital period of the satellite. Re-arranging the equation and replacing the numbers, we can find dS, the distance of the satellite from the Earth:

`d_S=\sqrt[3]{(d_M^3 T_S^2)/(T_M^2)}= \sqrt[3]{((240,000)^3 (0.5)^2)/((28)^2)}=16,400 miles`