A) \[{{v}^{2}}{{R}^{2}}/(2GR-Mv)\]
B) \[{{v}^{2}}{{R}^{2}}/(2GM+{{v}^{2}}R)\]
C) \[{{v}^{2}}{{R}^{2}}/(2GM-{{v}^{2}}R)\]
D) \[{{v}^{2}}{{R}^{2}}/(2Gv+RM)\]
Correct Answer: C
Solution :
If m is the mass of rocket, M that of the earth and R is the radius of earth, then gravitational potential energy of rocket near the surface of earth. \[{{U}_{1}}=-\frac{GMm}{R}\] Gravitational potential energy of rocket at a height h from earth's surface. \[{{U}_{2}}=-\frac{GMm}{(R+h)}\] Increase in gravitational potential energy of rocket \[\Delta U={{U}_{2}}-{{U}_{1}}=-\frac{GMm}{(R+h)}\,+\frac{GMm}{R}\] Or \[\Delta U\,=\frac{GMm}{(R+h)R}\] \[\Rightarrow \,m{{v}^{2}}{{R}^{2}}+m{{v}^{2}}Rh\,=2GMmh\] \[\Rightarrow \,{{v}^{2}}{{R}^{2}}=(2GM\,-{{v}^{2}}R)h\Rightarrow \,h=\frac{{{v}^{2}}{{R}^{2}}}{2GM-{{v}^{2}}R}\]
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