# Newton's Law of Universal Gravitation (Differential Equation Question)

Bazzinga
I'm having trouble with part a) of this question...

[PLAIN]http://img69.imageshack.us/img69/5815/98157006.png [Broken]

So I started off by solving the DE above a), and I've gotten it down to:

$$\frac{1}{2} m v^{2} = \frac{mgR^{2}}{(x + R)} + C$$

I can tell I'm getting close, but I'm a little confused where h is coming from, and how to get rid of the $$R^{2}$$. Also, what do I do with C??

I know it says x = x(t) which is the height, so does that mean that x(t) = h? I've never taken physics (and this was in my Calculus homework) so this is all new to me!

Last edited by a moderator:

Homework Helper
Good so far, to find C set v=0 when x=h, then to find v_0 set x=0.

Homework Helper
Hi, you should use a definite integration

$$\int_{v_0}^{0} mv \ dv = \int_{0}^{h} -\frac{mgR^2}{(R+x)^2} dx$$

That way you'll get the formula in the book.

As for the limit, I think it's trivial, right ?

Homework Helper
or, he could have done what I told him to to find the constant C, both methods are equivalent.

Bazzinga
So C = (-mgR^2) / (h+R) ?

Homework Helper
Correct. From there you can easily derive the required answer.

Bazzinga
Thanks a lot guys! So what about b)? will V0 approach infinity because the top approaches infinity faster than the bottom?

Homework Helper
divide the top and bottom of the fraction and you will have on the demoninator a R/h which is the only place h appears, take the limit as h tends to infinity and that yerm dissapears.