Question #293817

A satellite orbits Earth at an altitude of 356 kilometers above the planet's surface. Compute for its (a) radius (b) speed (c) orbital period. (ME=5.97x10^24 kg)


1
Expert's answer
2022-02-04T07:40:51-0500

Solution

Altitude h=356Km

Radius R=6400km

Mass of earth ME=5.97x1024 kg

a) radius is given

r=R+h=356Km+6400km

R=6956km

b)velocity

v=GMr=6.67×1011×5.97×10246956000=5.72×107m/secv=\sqrt{\frac{GM}{r}}\\=\sqrt{\frac{6.67\times10^{-11}\times5.97\times10^{24}}{6956000}}\\=5.72\times10^{-7}m/sec

c) time period

T=2πRv=2×3.14×69560005.72×107=7.6×1013sec.=\frac{2\pi R }{v}\\=\frac{2\times3.14\times6956000}{5.72\times10^{-7}}=7.6\times10^{13}sec.


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