Gravitation

JEE Main level, single correct, medium

129 of 223

Question 129 of 223

A body is projected vertically upwards from the surface of earth with a velocity equal to one third of escape velocity. The maximum height attained by the body will be
(Take radius of earth =6400km=6400km and g=10ms−2g=10{ms}^{-2})

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Answer: (A) 800km800km

Total mechanical energy of the body at the surface of earth
Mi=−GMmR+12m[13(2GMR)]2=−GMmR+GMm9R=−8GMm9R{M}_{i}=-\dfrac{GMm}{R}+\dfrac{1}{2}m{\left[\dfrac{1}{3}\left(\sqrt{\dfrac{2GM}{R}}\right)\right]}^{2} =-\dfrac{GMm}{R}+\dfrac{GMm}{9R} =-\dfrac{8GMm}{9R}
Total mechanical energy of the body at the highest point
Mf=−GMmR+h{M}_{f}=-\dfrac{GMm}{R+h}
Using conservation of mechanical energy,
Mi=Mf⇒−GMmR+h=−8GMm9R{M}_{i}={M}_{f} \Rightarrow -\dfrac{GMm}{R+h}=-\dfrac{8GMm}{9R}
⇒R+h=9R8\Rightarrow R+h=\dfrac{9R}{8}
⇒h=R8=64008=800km\Rightarrow h=\dfrac{R}{8}=\dfrac{6400}{8}=800km

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