Question

The force of attraction on an object below earth's surface is directly proportional to it's distance from earth's center.Find the work done in moving a weight of w lb located A miles below earth's surface up to the surface itself. Assume earth's radius is a constant r mi.

Answer 1

I would like someone to check this over. The work I have done so far:
F = k*x

Now the question is a little confusing, how to find k.
If we assume that it is w lb at the surface of earth then
w = k* r
k = w/r

So then we integrate

\[\int\limits_{r-a}^{r}\frac{w}{r}xdx= \frac{w(2ar-a^2)}{2r} \]

Answer 2

the book answer was
\[\frac{w(2ar-a^2)}{2}\]

Answer 3

@Michele_Laino maybe you could look at it

Answer 4

w is GMm/r^2 so the extra r you are putting up there in integration is absurd

Answer 5

please note that force F is given by the subsequent fromula:
F = k*r
so the differential work dW done by gravity acting inside the earth, is :
dW = Fdr
dW = kr*dr
where dr is the differential of r, so we have:
\[W = \int\limits_{r - a}^a {k\,r\,dr} = \frac{k}{2}\left. {{r^2}} \right|_{r - a}^a = \frac{k}{2}\left( {2ar - {a^2}} \right)\]
now, the gravity g(r) inside the earth is giveen by the subsequent formula:
\[g\left( r \right) = {g_0}\frac{r}{a}\]
where g_0 is the gravity onsurface of earth, so we can write:
\[k \cdot r = m{g_0}\frac{r}{a} = w\frac{r}{a}\]
then we get:
\[k = \frac{w}{a}\]
and substituting into the formula for the work done W, we can write:
\[W = \frac{k}{2}\left( {2ar - {a^2}} \right) = \frac{k}{2}a\left( {2r - a} \right) = \frac{w}{2}\left( {2r - a} \right)\]
Note that W<0 since W is the work done against the earth gravity.

Answer 6

please I have made an error, since W>0, not W<0

Answer 7

@Michele_Laino i think there is a mistake , the integral should be from r-a to r
and you should get k = w/r
unless i am doing something wrong.
r is the total radius of earth

Answer 8

also there it is posted here
https://www.physicsforums.com/threads/center-of-earth-work-problem-some-calculus.799459/