Question:
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 \(12\ \mathrm{km/s}\). 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 \(11.2\ \mathrm{km/s}\)) ? Give your answer to the nearest integer in kilometer/s _____.
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 \(12\ \mathrm{km/s}\). 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 \(11.2\ \mathrm{km/s}\)) ? Give your answer to the nearest integer in kilometer/s _____.
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Apply conservation of mechanical energy between \(10R\) and \(R\), and use \(GM/R = v_e^2/2\) to express the potential energy change.
Updated On: Nov 5, 2026
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Correct Answer: 16
Solution and Explanation
Escape velocity: \(v_e=\sqrt{2GM/R}\), so \(\dfrac{GM}{R}=\dfrac{v_e^2}{2}\).
Energy conservation per unit mass between \(r=10R\) and \(r=R\):
\(\tfrac12 v_i^2-\dfrac{GM}{10R}=\tfrac12 v_f^2-\dfrac{GM}{R}\)
\(v_f^2=v_i^2+\dfrac{2GM}{R}\left(1-\dfrac{1}{10}\right)=v_i^2+0.9\,v_e^2\)
\(v_f^2=12^2+0.9\times(11.2)^2=144+0.9\times125.44=144+112.896=256.896\)
\(v_f=\sqrt{256.896}\approx 16.03\ \mathrm{km/s}\approx 16\ \mathrm{km/s}\)
Answer: 16
Energy conservation per unit mass between \(r=10R\) and \(r=R\):
\(\tfrac12 v_i^2-\dfrac{GM}{10R}=\tfrac12 v_f^2-\dfrac{GM}{R}\)
\(v_f^2=v_i^2+\dfrac{2GM}{R}\left(1-\dfrac{1}{10}\right)=v_i^2+0.9\,v_e^2\)
\(v_f^2=12^2+0.9\times(11.2)^2=144+0.9\times125.44=144+112.896=256.896\)
\(v_f=\sqrt{256.896}\approx 16.03\ \mathrm{km/s}\approx 16\ \mathrm{km/s}\)
Answer: 16
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