A projectile with mass of 139 kg is launched straight up from the Earth's surface with an initial speed vi. What magnitude of vi enables the projectile to just reach a maximum height of 5.8RE, measured from the center of the Earth? Ignore air friction as the projectile goes through the Earth's atmosphere.

Answer 1

The initial kinetic energy must equal the potential energy change going from Re to 5.8Re. The potential energy in this situation is
-G*Me*m/r, where Me is the mass of the Earth and Me*G/Re^2 = g , the value of the acceleration of gravity at the Earth's surface. G is the un iversal gravity constant. I find it easier to remember and work with g (9.8 m/s^2), as I do below. I used Re = 6380 km

Vi^2/2 = G*Me/Re - G*Me/(5*Re)
= Re*g(1 - 1/5.8) = 0.83*Re*g
Vi^2 = 1.68*Re*g

Vi = 10,200 m/s