An unmanned spacecraft of mass 521 kg, is in a circular orbit around the earth,

Question

An unmanned spacecraft of mass 521 kg, is in a circular orbit around the earth, observing the earth's surface from an altitude of 47.7 km.
1) What is the orbital speed of the spacecraft?
2) What is the period of the spacecraft's orbit?
3) What is the escape velocity of the spacecraft?

To the dismay of scientists, and electrical fault causes an on-board thruster to fire, decreasing the speed of the spacecraft by 23.7m/s. If nothing is done to correct its orbit, with what speed will the spacecraft crash into the earth's surface?

Explanation / Answer

centripetal force is equal to gravitational force. mv^2/r=GMm/r^2 so v^2=GM/r. so v=sqrt(GM/r)=sqrt(6e24*G/6447.7e3)=7880(m/s). ------ 2) T=2piR/v=2pi*6447.7e3/7880=5141(s). 3) its v=sqrt(2GM/r)=11140(m/s). --------------- v'=7880-23.7=7856(m/s). conservation of energy. mv0^2/2-GMm/6400e3=m*v'^2/2-GMm/6447.7e3. so we have v0=7915(m/s)

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