How does one do this problem? The answer is wrong but the steps taken don't produce any reasonable answer. I left the last comment on that page with the answer that I got which essentially uses the same steps shown by one of the online tutors that attempted the problem.

http://www.chegg.com/homework-help/questions-and-answers/newton-s-law-universal-gravitation-strictly-applies-perfectly-spherical-bodies-many-celest-q6055533

Question

How does one do this problem? The answer is wrong but the steps taken don't produce any reasonable answer. I left the last comment on that page with the answer that I got which essentially uses the same steps shown by one of the online tutors that attempted the problem.

http://www.chegg.com/homework-help/questions-and-answers/newton-s-law-universal-gravitation-strictly-applies-perfectly-spherical-bodies-many-celest-q6055533

irishboy123 · Answer

off top of head, i'd integrate F and thus make it a measure of potential energy and compare energy levels at the aph and peri.....

i would expect however for the difference to be accounted for by the change in orbital speed of the comet so i do not see how that can be categorised as "work" done.

matrixation · Answer

Thanks for the quick reply. What you said sounds similar to the solution I got but I am not sure if it's right. I'm sending you my solution for the same problem but with different values for the constants in the integrand. The solution is just the essentials...basically, the bare-bones of the solution. What I wrote should be pretty striaght-foward.

matrixation · Answer

Here is the original problem as an attachment.

irishboy123 · Answer

$$F = GMm (\frac{1}{r^{2}} + \frac{3JR^{2}}{2} \frac {1}{r^4} )$$
$$U = -GMm ( -\frac{1}{r} + \frac{3JR^{2}}{2} (-\frac{1}{3})\frac{1}{r^{3}} )$$
$$U(r) = -GMm ( \frac{1}{r} + \frac{JR^{2}}{2} \frac{1}{r^{3}} )$$

NB: normally, one would expect U to have a negative sign to reflect fact that potential energy increases as bodies move further apart so i have switched signs.

with that in mind, at this point i would compare U, and its sub-components at the aph and peri.  

i get  7,506,411.56 J as the difference for the correction term alone, and 1.6 10^18 for difference in main term.

in each case a reduction in potential energy......

matrixation · Answer

I see what you did. Yes, that essentially is along the same path I took. I got 7506.08 Joules and that's multiplying the corrective term by GMm. I'm uploading the hint they give which seems to be in the same direction we're in.

matrixation · Answer

It seems like someone else said that their answer to the same question, but with different numbers, was around 1225J and I did the same problem with different numbers and got 980J. So, my answer doesn't seem so unreasonable but still, I don't know how to verify it. It seems like the orbital time is not taken into account.

matrixation · Answer

Also, note the limits of integration. By the way, thanks for trying to help me.

irishboy123 · Answer

i get 7506J too.

i had all the distances in km.......

irishboy123 · Answer

numbers in here for what its worth.....

matrixation · Answer

I just checked my work with the professor and he said I was correct all along. I confirmed my solution with the online assignment and I was correct. The answer is 7506J. I'll give you best answer anyway since you really put time into this. Peace brother.