According to Kepler's Third Law, the periodT(in days) of a satellite moving in a circular orbitxmi above the surface of the earth is given by

T= 0.0588

1 +

x

3959

	3/2

.

Suppose that a communications satellite is moving in a circular orbit29,000mi above the earth's surface. Because of friction, the satellite drops down to a new orbit28,500mi above the earth's surface. Estimate the decrease in the period of the satellite to the nearest one-hundredth hour.