Estimate the decrease in the period of the satellite to the nearest one-hundredth hour.... According to Kepler's Third Law, the period T (in days) of a satellite moving in a circular orbit x mi above the surface of the earth is given by $T=.0588(1+\frac{x}{3959})^{\frac{3}{2}}$ Suppose that a communications satellite is moving in a circular orbit 28,000 mi above the earth's surface. Because of friction, the satellite drops down to a new orbit 27,500 mi above the earth's surface. Estimate the decrease in the period of the satellite to the nearest one-hundredth hour.

So this is what I think I have, not sure whether I'm right or wrong. $ΔT=T'dx$ So $dx=-500$ and $T'=.0588(\frac{3}{2}(1+\frac{x}{3959})^{.5}*\frac{1}{3959}$ So $T'(28,000)=.000063298$ and then I multiply $.000063298(-500)=-.03(rounded)$ but it's wrong and I suppose I'm approaching the problem wrong. A precise step by step solution would be greatly appreciated, thanks.
Could you just sub in T(28000) - T(27500)?
Edit: After a quick read of linear approximation, it looks like what you're doing is right. Did you check your units though?
Could you just sub in T(28000) - T(27500)?
Edit: After a quick read of linear approximation, it looks like what you're doing is right. Did you check your units though?