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Gravitation

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Question : 17 of 25
Marks: +1, -0
A rocket is fired vertically with a speed of 5 km s−1^{-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= 6.0 \times 10^{24} kg; mean radius of the earth =6.4×106 m  ;G=6.67×10−11 N m2kg−2.= 6.4 \times 10^{6} \, \text{m} \; ; G = 6.67 \times 10^{-11} \, \text{N} \, \text{m}^{2} \text{kg}^{-2}.
Solution:  
Total energy of rocket at the surface of earth
=K.E.+P.E.= \text{K.E.} + \text{P.E.}
=12mv2+(−GMmR)= \frac{1}{2} m v^{2} + \left( \frac{-G M m}{R} \right)
At the highest point, v=0,K.E.=0v = 0, \text{K.E.} = 0 and P.E.=−GMmR+h\text{P.E.} = - \frac{G M m}{R+h}
Total energy =K.E.+P.E.= \text{K.E.} + \text{P.E.}
=0+(−GMmR+h)= 0 + \left( \frac{-G M m}{R+h} \right)
=−GMmR+h= - \frac{G M m}{R+h}
According to law of conservation of energy
12mv2−GMmR=−GMmR+h\frac{1}{2} m v^{2} - \frac{G M m}{R} = - \frac{G M m}{R+h}
or 12v2=GMR−GMR+h\frac{1}{2} v^{2} = \frac{G M}{R} - \frac{G M}{R+h}
=gR2R−gR2R+h= \frac{g R^{2}}{R} - \frac{g R^{2}}{R+h}
=gR(1−RR+h)= g R \left( 1 - \frac{R}{R+h} \right)
=gR(hR+h)= g R \left( \frac{h}{R+h} \right)
or v2(R+h)=2gRhv^{2}(R+h) = 2 g R h
or Rv2=2gRh−v2hR v^{2} = 2 g R h - v^{2} h
=(2gR−v2)h= \left( 2 g R - v^{2} \right) h
or h=Rv22gR−v2h = \frac{R v^{2}}{2 g R - v^{2}}
=(6.4×106)×(5×103)22×9.8×(6.4×106)−(5×103)2= \frac{ (6.4 \times 10^{6}) \times (5 \times 10^{3})^{2} }{2 \times 9.8 \times (6.4 \times 10^{6}) - (5 \times 10^{3})^{2}} =1.6×106 m= 1.6 \times 10^{6} \, \text{m}
Distance of the rocket from the Earth =6.4×106 m+1.6×106 m=8×106 m= 6.4 \times 10^{6} \, \text{m} + 1.6 \times 10^{6} \, \text{m} = 8 \times 10^{6} \, \text{m}
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