Gravitation

JEE Main level, numerical answer, medium

143 of 223

Question 143 of 223

An asteroid is moving directly towards the centre of the earth. When at a distance of 10R10R (RR is the radius of the earth) from the centre of the earth, it has a speed of 12kms−1.12km{s}^{-1}. Neglecting the effect of earth's atmosphere, what will be the speed of the asteroid when it hits the surface of the earth (escape velocity from the earth is 11.2kms−1)11.2km{s}^{-1}) ? Give your answer to the nearest integer in kms−1km{s}^{-1}__________.

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Answer: 16

KEi+PEi=KEf+PEfK{E}_{i}+P{E}_{i}=K{E}_{f}+P{E}_{f}
12mu02+(−GMm10R)=12mv2+(−GMmR)\dfrac{1}{2}m{u}_{0}^{2}+\left(-\dfrac{GMm}{10R}\right)=\dfrac{1}{2}m{v}^{2}+\left(-\dfrac{GMm}{R}\right)
v2=u02+2GMR[1−110]{v}^{2}={u}_{0}^{2}+\dfrac{2GM}{R}\left[1-\dfrac{1}{10}\right]
v=u02+(95)GMRv=\sqrt{{u}_{0}^{2}+\left(\dfrac{9}{5}\right)\dfrac{GM}{R}}
=122+(95)(11.2)22=\sqrt{{12}^{2}+\left(\dfrac{9}{5}\right)\dfrac{{\left(11.2\right)}^{2}}{2}}
=144+0.9(11.2)2=156.896=\sqrt{144+0.9{\left(11.2\right)}^{2}}=\sqrt{156.896}
=16.028kms−1=16.028km{s}^{-1}
≃16kms−1≃16km{s}^{-1}

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