A rocket is fired vertically with a speed of 5 km s-1 from the earth’s surface. How far from the earth does the rocket go before returning to the earth? Mass of the earth = 6.0 × 1024 kg; mean radius of the earth = 6.4 × 106 m; G = 6.67 × 10–11 N m2 kg–2
Velocity of the rocket, v=5×103 m/s Mass of the Earth, Me=6×1024 kg Radius of the Earth, Re=6.4×106 Height reached by rocket mass m=h At the surface of the Earth, Total energy of the rocket = Kinetic energy + Potential energy =21mv2+(−ReGMem) At the highest point h, v=0⟹Kinetic energy=0 Potential energy=−Re+hGMem Total energy of the rocket =−Re+hGMem By law of conservation of energy, total energy at surface = total energy at height h 21mv2−ReGMem=−Re+hGMem v2/2=Re(Re+h)GMeh Substituting g=GMe/Re2 and rearranging the terms, h=2gRe−v2Rev2 =2×9.8×6.4×106−(5×103)26.4×106×(5×103)2 =1.6×106 m Height achieved by the rocket with respect to the earth is Re+h=6.4×106+1.6×106=8.0106m
