Answer to Question #90765 in Classical Mechanics for rashid ali

Question

an earth satellite,in circular orbit at an altitude h of 230 km above the earth's surface,has a period T of 89 minutes.What mass of the earth follows from these data?

Answer 1

Newton's second law:

$$\frac{GMm}{(R+h)^2} = ma\\ a= \frac{GM}{(R+h)^2}\\ a= \frac{v^2}{(R+h)}\\ v=\frac{2\pi (R+h)}{T}\\ a=\frac{4 \pi^2 (R+h)}{T^2}\\ \frac{4 \pi^2 (R+h)}{T^2}=\frac{GM}{(R+h)^2}\\$$

and then the mass of Earth is

$$M=\frac{4 \pi^2 (R+h)^3}{GT^2}=\\ =\frac{4\pi^2 ((6371+230)\cdot 10^3)^3}{6,67\cdot 10^{-11}\cdot (89\cdot 60)^2} = 5.97\cdot 10^{24} kg$$