## work done integration

Ok I have an answer for this but my method is so simple is must be wrong.

Question

The work done against a firce F(r) in moving an object from r=r1 to r=r2 is integralF(r) dr limits R2 at top and r1 at bottom. The gravitational attraction between two masses m1 and m2 at distance d is given be F=GMM/d^2, G=6.67*10^-11. Find the work done in lifting a 4000kg payload from the surface on the moon to a height of 25000m above its surface. The mass of the moon can be taken as 7.3*10^22 and its radius at 1.7*10^6.

 Yep. You have made a mistake. Work done is calculated using line integrals. $$W = \int F(r).dr = \int \frac{GM_1M_2}{r^2} dr = -\frac{GM_1M_2}{r} + C$$ Since the integral is definite, then the constant of integration can be ignored, and the answer can be found simply by putting the values given into the equation. $$W = \frac{GM_1M_2}{r_1} - \frac{GM_1M_2}{r_2}$$