Verified Solution • 2025–26 Session moderate

QJEEM20Ph5334: JEE Main 2020 Physics - Class 11 JEEM20Ph5334 Solution: 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 10R (R is the radius of the earth) from the earths centre, it has a speed of 12 km/s. Neglecting the effect of earths atmosphere, what will be the speed of the asteroid when it hits the surface of the earth (escape velocity from the earth is 11.2 km/s) ? Give your answer to the nearest integer in kilometer/s _____.

Step-by-Step Verified Solution

U1 + K1 = U2 + K2

⇒\Rightarrow −GMm10R+12mV12- {{GMm} \over {10R}} + {1 \over 2}mV_1^2 = −GMmR+12mV22- {{GMm} \over R} + {1 \over 2}mV_2^2

⇒\Rightarrow 12V22=12V12+GMR−GM10R{1 \over 2}V_2^2 = {1 \over 2}V_1^2 + {{GM} \over R} - {{GM} \over {10R}}

⇒\Rightarrow V22=V12+95GMRV_2^2 = V_1^2 + {9 \over 5}{{GM} \over R} ....(1)

Given escape velocity Ve = 11.2 km/s

⇒\Rightarrow 2GMR\sqrt {{{2GM} \over R}} = 11.2

⇒\Rightarrow GMR=(11.2)22{{{GM} \over R} = {{{{\left( {11.2} \right)}^2}} \over 2}}

So from (1)

V22=V12+95V_2^2 = V_1^2 + {9 \over 5} ×(11.2)22{ \times {{{{\left( {11.2} \right)}^2}} \over 2}}

= (12)2{\left( {12} \right)^2} + 112.896

⇒\Rightarrow V2 = 16 km/s

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