Earth's motion around the sun was neglected; that requires using earth's "sidereal" period (about four minutes shorter). notice that earth's mass could be found by substituting the moon's distance and period into this form of kepler's third law. question if the satellite was placed in an orbit three times farther away, about how long would it take to orbit the earth once

T = 5.196 T₀

Explanation:

For this exercise let's use Kepler's readings, the third law is

     T² = (4π² / G M) a³

They tell us that

The distance is three times greater

    a₂ = 3a

As the satellite orbits the earth we use the mass of the earth

     M = 5.98 10²⁴ kg

     T² = 4 pi2 / (66.7 10⁻¹¹ 5.98 10²⁴) a₂³

     T² = 9,898 10¹⁴ a₂³

     T² = 9,898 1014 (3 a)³

     T2 = 9.89 1014 a³ 27

     T₀² = 9.89 10¹⁴ a³

     T² = T₀² 27

     T = T₀ √27

     T = 5.196 T₀

The period increases by 5 times

the answer should be in the picture.