Question #90765
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?
1
Expert's answer
2019-06-13T09:39:31-0400

Newton's second law:

GMm(R+h)2=maa=GM(R+h)2a=v2(R+h)v=2π(R+h)Ta=4π2(R+h)T24π2(R+h)T2=GM(R+h)2\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=4π2(R+h)3GT2==4π2((6371+230)103)36,671011(8960)2=5.971024kgM=\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


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