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# A bullet is fired vertically upwards with velocity υ from the surface of a spherical planet. When it reaches its maximum heights, its acceleration due to the planet's gravity is 1/4th of its value at the surface of the planet. If the escape velocity from the planet is υesc=υ√N, then the value of N is (ignore energy loss due to atmosphere)

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#### We know that at the surface of the earth, value of g=GMR2At height h above the Earth's surface, the value of acceleration due to gravity g′=GM(R+h)2So it is given that g′=g4When the bullet reaches maximum height, acceleration due to gravity is 14th of that at planet's surface. That implies GM4R2=GM(R+h)2 →h=R By conservation of mechanical energy, −GMmR+12mv2=−GMmh+R+0 since velocity is zero at max height. ⇒12mv2=GMm2R v=√GMR=√2GM2R=1√2√2GMR=1√2vesc ⇒vesc=√2v ⇒N=2

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