Physics Problem

unitedshoes86

I have been having difficulty with the following problem, if anyone can point me in the right direction it would be much appreciated.

I am trying to find the Gravity that Mass M1 exerts on Mass M2, I am having trouble once I had the basic equation of F=G(M1*M2)/(r^2), because to integrate I need eliminate variables and am having problems there

I have a solid sphere: -Radius= R
-Sphere with a Mass = M1
- A second Mass of M2
-With M2 a distance of X away

Here is an image I made:

JebusUD

I guess I don't understand the question.

Find for X or something?

Phisti

Handy diagram, but the problem part of the OP seems to be missing. Is this an angular momentum question? Is it a find X question? is it a ... there a bunch of possibilities so if you could fill us in we'd be much more helpful.

Six

Maybe the problem is that he's not sure if it's a good diagram.

It's a pretty good diagram. Hope that helps.

unitedshoes86

sorry about that fixed it now

JebusUD

http://en.wikipedia.org/wiki/Newton's_law_of_universal_gravitation

Phisti

Jebus pretty much hit the nail right on the head. Newton is the place to go and you have your answer. Note that R in the case of Newton's formula is not the radius of either object, but the distance between the centres of mass.

unitedshoes86

yah I understand, the basic equation and setting it up, I just am having problems integrating it, because I cannot get rid of variables. So more what I am asking is to see how any of you would set up your equation, right before you would integrate. and yes I understand what r is, but the radius is given because it is needed for the problem because we are working with a sphere.

physi_marc

I don't understand why you're trying to integrate. If the mass is uniformly distributed in the sphere (which is a reasonable assumption to make in this case), you can treat the large sphere as a point mass of mass M1, located at the centre of mass of the sphere (which is just the centre of the sphere in this case).

Jebus314

physi_marc wrote: »

I don't understand why you're trying to integrate. If the mass is uniformly distributed in the sphere (which is a reasonable assumption to make in this case), you can treat the large sphere as a point mass of mass M1, located at the centre of mass of the sphere (which is just the centre of the sphere in this case).

Pretty much what I was thinking. If you're looking for the force exerted on mass 2 by mass 1, then it is just a simple plug and chug problem. You have the equation, your looking for F, M1 and M2 are given, G is a constant, and r is the distance between the two centers of mass. In other words r != x ( != means does NOT equal).

nuclearalchemist

If you are calculating the force of one sphere on the other, and the spheres are external to each other, then you can just replace M1 and M2 by the total massed of the spheres and then the distance between them. This comes from the analogous form of Gauss's law for gravitation; once you are external to the mass density (charge density in EM), then you can just look at it like a point charge!

unitedshoes86

ok awesome, that explains it, I always tend to over complicate problems. Thanks a lot! So then a similar problem I have, if I was looking at the force exerted by solid circular disk on a point of mass, what method should I use to approach this. will it be necessary to evaluate a dm of the disk and integrate, or am I allowed to treat it uniformly, because the (X,Y) cancel out and can then treat it as a point? Anyway here is an image I drew up real fast.

Phisti

From the sounds of the complexity of the problems you're allowed to treat it uniformly (I may be wrong here, not a physicist, I just took uni physics as an elective).

Use the centre point of the disc as the centre of mass. Otherwise you'd get into fairly complex gravitational field calculations (also on the wiki page Jebus linked) since the distance from the disc to the other mass changes the further toward the edge of the disc you travel and the distance from the centre of the disc's mass itself changes.

From the sounds of things you aren't there yet, so stick with the single point method.

ProPatriaMori

unitedshoes86 wrote: »

ok awesome, that explains it, I always tend to over complicate problems. Thanks a lot! So then a similar problem I have, if I was looking at the force exerted by solid circular disk on a point of mass, what method should I use to approach this. will it be necessary to evaluate a dm of the disk and integrate, or am I allowed to treat it uniformly, because the (X,Y) cancel out and can then treat it as a point? Anyway here is an image I drew up real fast.

You can exploit the symmetry of the disc. The net force is going to be directed down, since all the lateral forces will cancel. However, I don't think you can avoid integration in that problem by acting like all the mass of the disc is concentrated in the center, because the edges of the disc are further away from the point mass and are thus exerting less force.

The simplest way to do the second one might be to integrate out in rings. Each ring will express no lateral forces but will have some sum downforce that varies with the radius of the ring (and thus the distance of its component mass from the point mass).

EDIT: I think the people who are telling you that you aren't at that level of complexity are totally wrong. They're giving you all the information you need to correctly address the problem, i.e. to account for the shape of the objects. I'd imagine there's a reason that they're doing so. That should be clear from class participation, though.

Dunadan019

these questions all have the same theme.

in this case, what you do is notice that the tiny piece will be an arc segment with width dr and a length of dtheta*r where r is the position of the arc from 0 to R.

you find the area density by taking the mass and dividing it by the total area (PiR^2).

now the area of your piece is r*dr*dtheta and since you assume the density is constant you can find the mass dm of this small piece with relation to the entire solid. in order to have something to substitute in your double integral.

however what you have to do first is take the force component for each small piece that is vertical (since the horizontal components will cancel out) that force is GdmM2/x^2*cos(phi) where phi is the angle between vertical and the line connecting M2 to your little piece.

the relationship between Phi and r can be had by noticing that tan(phi) = r/x.

then you just combine all of that together and integrate. I did this stuff about 7 years ago so I may have gotten some of it wrong but thats the ghist.

unitedshoes86

alright so I figured it all out, ProPatriaMori was right and integration is necessary in each. The sphere is the integral of [GMm4(pi)r^(2)/x^2]. and for a disk the thickness(t) included (unless its a lamina), would be the angles for the width of the disk where the final solution is F= 2(pi)GmMt(cos?) evaluated at the boundaries.

ProPatriaMori

Wokka wokka wokka! Glad to help.

Doctorstrongbad

Do you need anymore help with Physics?