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 _____.
Answer (integer)
16
Solution
U<sub>1</sub> + K<sub>1</sub> = U<sub>2</sub> + K<sub>2</sub>
<br><br>$\Rightarrow$ $- {{GMm} \over {10R}} + {1 \over 2}mV_1^2$ = $- {{GMm} \over R} + {1 \over 2}mV_2^2$
<br><br>$\Rightarrow$ ${1 \over 2}V_2^2 = {1 \over 2}V_1^2 + {{GM} \over R} - {{GM} \over {10R}}$
<br><br>$\Rightarrow$ $V_2^2 = V_1^2 + {9 \over 5}{{GM} \over R}$ ....(1)
<br><br>Given escape velocity V<sub>e</sub> = 11.2 km/s
<br><br>$\Rightarrow$ $\sqrt {{{2GM} \over R}}$ = 11.2
<br><br>$\Rightarrow$ ${{{GM} \over R} = {{{{\left( {11.2} \right)}^2}} \over 2}}$
<br><br>So from (1)
<br><br>$V_2^2 = V_1^2 + {9 \over 5}$${ \times {{{{\left( {11.2} \right)}^2}} \over 2}}$
<br><br>= ${\left( {12} \right)^2}$ + 112.896
<br><br>$\Rightarrow$ V<sub>2</sub> = 16 km/s
About this question
Subject: Physics · Chapter: Gravitation · Topic: Kepler's Laws
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