KEi+PEi=KEf+PEf
21mu02+(−10RGMm)=21mv2+(−RGMm)
v2=u02+R2GM[1−101]
v=u02+(59)RGM
=122+(59)2(11.2)2
=144+0.9(11.2)2=156.896
=16.028kms−1
≃16kms−1
An asteroid is moving directly towards the centre of the earth. When at a distance of 10R (R is the radius of the earth) from the centre of the earth, it has a speed of 12kms−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) ? Give your answer to the nearest integer in kms−1__________.
Held on 8 Jan 2020 · Verified 6 Jul 2026.
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Work through every JEE Main Mechanics PYQ, year by year.