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)

Question

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)

Infinity Learn · Accepted Answer

$$gh=gR2R+h2$$​⇒$$g4=gR2R+h2$$​⇒$$h=R$$​By law of conservation of energy​⇒$$12mv2+$$−$$GMmR=0+$$−$$GMmR+h$$⇒$$12mv2+$$−$$GMmR=$$−$$GMm2R$$​⇒$$12mve2N$$−$$12mve2=$$−$$12mve22$$ ∵$$ve2=2GMR$$⇒$$GMR=ve22$$ & $$ve=vN$$⇒$$12mve2N=12mve212$$​⇒$$N=2$$

Motion of planets and satellites

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Motion of planets and satellites

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gh=gR2R+h2​⇒g4=gR2R+h2​⇒h=R​By 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

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