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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.)

User Rob Hughes
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Answer:

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


User Iwo Kucharski
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