What is the minimum work that has to be performed to bring a spaceship of mass `m=2.0*10^3kg` from the surface of the Earth to the Moon?

Explanation

Between the earth and the moon,the potential energy of the spaceship will have a maximum at the point where the attractions of the earth and the moon balance each other. This maximum P.E. is approximately zero. We can also neglect the contribution of either body to the p.E. of the spaceship sufficiently near the other body. Then the minimum energy that must be imparted to the spaceship to cross the maximum of the P.E. is clearly (using E to denote the earth)
`(gammaM_Em)/(R_E)`
With this energy the spaceship will cross over the hump in the P.E. and coast down the hill of P.E. towards the moon and crashland on it. What the problem seeks is the minimum energy reguired for softlanding. That reguies the use of rockets to loving about the braking of the spaceship and since the kinetic energy of the gases ejected from the rocket will always be positive, the total energy required for softlanding is greater than that required for crashlanding. To calculate this energy we assume that the rockets are used fairly close to the moon when the spaceship has nealy attained its terminal velocity on the moon `sqrt((2gammaM_O)/(R_0))` where `M_0` is the mass of the moon and `R_0` is its radius. In general `dE=vdp` and since the speed of the ejected gases is not less than the speed of the rocket, and momentum transfered to the ejected gases must equal the momentum of the spaceship the energy E of the gass ejected is not less than the kinetic energy of spaceship
`(gammaM_0m)/(R_0)`
Adding the two we get the minimum work done on the ejected gases to bring about the softlanding.
`A_(min)~~gammam((M_E)/(R_E)+(M_0)/(R_0))`
On substitution we get `1*3xx10^8kJ`.