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 _____.

Solution

Physics sir · 4 steps

Energy is conserved all the way in, and the escape velocity hands you 2GM/R without needing G or M.

1. Conserve energy. With no atmosphere, only gravity acts, so kinetic plus potential energy stays constant. Per unit mass, 1/2u² - GM/10R = 1/2v² - GM/R, with u = 12 km s⁻¹ at 10R and v at the surface.
2. Bring in escape velocity. Rearranging gives v² = u² + 2GM/R(1 - 1/10). Since vₑ² = 2GM/R, this becomes the equation below.

   v² = u² + 0.9 vₑ²

3. Put in numbers. vₑ² = (11.2)² = 125.44, so 0.9 vₑ² = 112.896. Then v² = 144 + 112.896 = 256.896 km² s⁻².
4. So. v = √256.896 ≈ 16.03 km s⁻¹, which rounds to 16.

The common slip: 14. Taking vₑ² = GM/R instead of 2GM/R adds only 0.45 vₑ² and gives about 14, not 16.

The idea to keep. For any fall between two distances, write energy conservation per unit mass and replace 2GM/R by vₑ². The potential energy is not zero at the starting point, so keep its term.

Answer from the NTA final answer key. Solution by PrepWiser.