A projectile is thrown straight upward from the earth's surface with an initial speed v = α v E , where α is a constant and v E is the escape speed. The projectile travels upto a height 800 km from earth's surface, before it comes to rest. The value of the constant α is, (Radius of the earth = 6400 km )

Step-by-step solution

Escape velocity is given by: v E = 2 G M R At Earth's surface: Kinetic energy of the particle K E 1 = 1 2 m v 2 = 1 2 m α 2 v E 2 = 1 2 m α 2 × 2 G M R = G M m α 2 R Potential energy of the particle: P E 1 = - G M m R At certain height the particle stops for a moment: K E 2 = 0 P E 2 = - G M m R + h Applying conservation of mechanical energy, K E 1 + P E 1 = K E 2 + P E 2 ⇒ G M m α 2 R + - G M m R = 0 + - G M m R + h ⇒ 1 R α 2 - 1 = - 1 R + h ⇒ 1 6400 α 2 - 1