posted by Ugandanmafia on .
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.
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