A space ship of mass 2 × 10⁴ kg is launched into a circular orbit close to the earth surface. The additional velocity to be imparted to the space ship in the orbit to overcome the gravitational pull will be (if g=10 m / s² and radius of earth =6400 km ):

Solution

Physics sir · 4 steps

You already orbit at √gR. Escape needs √2 times that, so add the difference.

1. Speed in orbit. Close to the surface, gravity supplies the centripetal force: mg = mv₀²/R. So the orbital speed is v₀ = √gR.
2. Put in numbers. v₀ = √(10 × 6.4 × 10⁶) = √(6.4 × 10⁷) = 8000 m s⁻¹ = 8 km s⁻¹.
3. Escape from there. To escape from the same height, the total energy must reach zero: 1/2mvₑ² = GMm/R, which gives vₑ = √2gR = √2 v₀ = 8√2 km s⁻¹. The mass of the ship cancels, so 2 × 10⁴ kg is not needed.

   Δ v = vₑ - v₀ = 8√2 - 8 = 8(√2-1) km s⁻¹

4. So. The extra speed to give the ship is 8(√2-1) km s⁻¹, which is option (d).

The idea to keep. Near the surface, escape speed is always √2 times orbital speed. The extra kick from orbit is (√2-1)√gR, and the mass never matters.

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