Question

A satellite is put in an orbit just above the earths atmosphere with a velocity тИЪ(1.5) times the velocity for a circular orbit at that height. The initial velocity imported is horizontal. What would be the maximum distance of the satellite from the surface of the earth when it is the orbit?

Answer 1

Given velocity of satellite(vi) when it is projected is 1.5−−−√voWhere vi is the initial velocity and vo is the velocity of setelite in the circular orbit of radius (R+h)

Here we have to find h (maximum distance from the earth surface)

According to energy conservation

EA=EQ⇒−GMmR+12mv2i=−GMm(R+h)+12mv2o⇒−GMmR+12m×1.5v2o=−GMm(R+h)+12mv2o⇒12m×1.5v2o−12mv2o=−GMm(R+h)+GMmR=GMm(−R+R+h)R(R+h)⇒12m×0.5v2o=GMm(h)R(R+h)⇒12×0.5v2o=GM(h)R(R+h)..............⎛⎝1⎞⎠And At point QGMm(R+h)2=mv2o(R+h)⇒GM(R+h)=vUnknown node type: subsupputting this valu in equation ⎛⎝1⎞⎠we have 12×0.5GM(R+h)=GM(h)R(R+h)520=(h)Ror, h=R4

Hence maximum distance from the earth surface when it is in orbit is R/4.

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