work done integration

smileandbehappy

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.


My answer.

I subbed in the values for F getting 6739.24 then integrated with respect to r within the limits of 25000 and 0 and got an answer of 1.68*10^8j! What am i doing wrong???

Cinimod

Yep. You have made a mistake. Work done is calculated using line integrals.
[tex] W = \int F(r).dr = \int \frac{GM_1M_2}{r^2} dr = -\frac{GM_1M_2}{r} + C[/tex]

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.

[tex] W = \frac{GM_1M_2}{r_1} - \frac{GM_1M_2}{r_2} [/tex]

D H

I gather the OP was questioning the magnitude of the answer. Since the gravitational force is approximately constant over the 25 km range in question, the value 1.68e8 joules is roughly correct. The correct value (see post #2) is slightly smaller because the gravitational force is not constant. It takes a lot of energy to lift something off the Moon.