Question

Escape velocity on the surface of the Earth is 11.2 km/s. If any object is thrown with double the escape velocity, what would be its speed at maximum distance? Neglect the presence of Sun and other celestial bodies.

Answer 1

Escape velocity from the Earth’s surface,
v_e = 11.2 km-s^-1
Projection velocity from the Earth,
v₁ = 2v_e
∴ Kinetic energy of projection

Now applying the principle” of conservation of energy,
Total energy on Earth’s surface = Total energy at infinity