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JEE Main 20208 JanuaryEvening Shift
Exam
JEE Main
Year
2020
Date
8 January
Shift
Evening Shift
Subject
Physics
Chapter
Gravitation

JEE Main 2020 · 8 January · Evening Shift · Physics · Gravitation

An asteroid is moving directly towards the centre of the earth....

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}__________.

Correct answer

0

Step-by-step solution

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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