A bullet is fired vertically upwards with velocity v from the surface of a spherical planet. When it reaches its maximum height, its acceleration due to the planet’s gravity is 14th of its value at the surface of the planet. If the escape velocity from the planet is vese=vN , then the value of N is (ignore energy loss due to atmosphere)
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answer is 2.
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Detailed Solution
gh=gR2R+h2⇒g4=gR2R+h2⇒h=RBy law of conservation of energy⇒12mv2+−GMmR=0+−GMmR+h⇒12mv2+−GMmR=−GMm2R⇒12mve2N−12mve2=−12mve22 ∵ve2=2GMR⇒GMR=ve22 & ve=vN⇒12mve2N=12mve212⇒N=2