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A rocket is launched from the surface of Earth with an initial speed of 10 kms-1. The maximum distance of rocket from the surface of Earth is n×104 km. The value of n is __________.(Radius of Earth = 6400 km and escape velocity from Earth's surface = 11.2 km/s) Neglect atmospheric resistance.

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answer is 2.58.

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Detailed Solution

−GMmR+12mv2=-GMm(R+h)⇒−GMR+12v2=−GMR+h⇒12v2=−GM(R+h)+GMR⇒12v2=GM-1R+h+1R ⇒v2=2GMR1-11+hR here, 2GMR=ve2=11.22 ⇒1011.22=hR+h ⇒h=2.58×104km

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