a satellite to be projected with a speed of 4500 m/s in a direction 45 degree above the surface of the earth. calculate the maximum height and the time taken to reach the maximum height
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Mass of the Earth, M=6.0×1024 kg
m=200 kg
Re=6.4×106 m
G=6.67×10−11 Nm2kg−2
Height of the satellite,h=400 km=4×105 m
Total energy of the satellite at height h=(1/2)mv2+(−G(Re+h)Mem)
Orbital velocity of the satellite, v=Re+hGMe
Total energy at height h =21Re+hGMem−Re+hGMem
Total Energy=−21Re+hGM
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