Gravitational force from a square plate

In summary, the student was trying to calculate the gravitational force from a square plate. He was having trouble with the expression for ##dm## and the integral. He thinks breaking the plate up into parallel line segments might be easier to work out.
  • #1
Emspak
243
1
I ordinarily would put this up in a homework/ coursework forum, but this isn't either one, its just something I was curious about given that I am in a Mechanics class.

So, I have done calculations for finding the gravitational force from a sphere and from a ring, and a flat circular plate. All these are symmetrical, tho. I was thinking about what happens with a square plate.

Now, here's the thing. Let's put our particle at a distance z from the surface of the plate. We wil put it in the center, to make the forces as symmetrical as possible.

Kind of intuitively, I would say that the way to do that calculation is to pretend you are dealing with a wire, (kind of like what you do calculating the electrical fields with a square wire). But this is a plate, so it's different (I think).

We have the mass of the plate. At any "radius" b from the center, if the plate is density ρ, with a side of "a" units, the mass is going to be [itex]2b^2ρ[/itex]. So I thought we could have a dM = 4bρ.

That wold say to me that th eintegral would look like this: [tex] \int dF_z = \int_0^a \frac{mG4d\rho dM}{s^2}[/tex] where s^2 is the distance to that dM, and that will be [itex]\sqrt{b^2+z^2}[/itex] so the integral would end up as: [tex] \int dF_z = mG\rho \int_0^a \frac{4b db}{\sqrt{b^2+z^2}}[/tex]

but anyway, I'd be curios to know if I am approaching this right. I was thinking of the old science fiction tropes with these huge space stations, and wondering what wold happen if one was constructed as a plane square -- also Stephen Baxter -- a scientist himself -- wrote a novel called Raft which takes place in a universe where gravitational forces are a billion times stronger, and a rat of metal plates actually has enough "pull" to live on. (there's more to it than that, but y'all get the idea).
 
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  • #2
I'm going to go out on a limb on this one and put out what I got after quickly doing it out on paper. I didn't actually check the integral, so I'm not sure it's 100% right.

You have the wrong expression for ##dm##, it should be ##dm=\rho dA = \rho dx dy##.

I think this problem is easier to work out by not treating it as a bunch of lines.

I think the integral should be $$Gm\rho \int ^{b/2}_{-b/2} \int ^{b/2}_{-b/2} \frac{dxdy}{(x^2+y^2+z^2)} cos\theta= Gm\rho z \int ^{b/2}_{-b/2} \int ^{b/2}_{-b/2} \frac{dxdy}{(x^2+y^2+z^2)^{3/2}} $$

I think this makes sense because $$dF = \frac{Gm\rho dA}{r^2}cos\theta$$

Don't take my word for it though, you should evaluate it and check to see if it makes sense.
 
  • #3
There are fairly simple expressions for the field and potential of a charged rod, in its midplane. See http://www.lightandmatter.com/area1sn.html , section 10.3.1. I would try breaking up the square into parallel line segments, taking the expression for the potential and integrating it. The integral would probably be ugly, but you could put it into software such as maxima (which is free and open source).
 

1. What is gravitational force from a square plate?

Gravitational force from a square plate refers to the force of attraction between two objects due to their mass and distance from each other, where one object is a square plate and the other is another object.

2. How is the gravitational force from a square plate calculated?

The gravitational force from a square plate can be calculated using the equation F = G * (m1 * m2) / d^2, where F is the force, G is the gravitational constant, m1 and m2 are the masses of the two objects, and d is the distance between them.

3. Does the size of the square plate affect the gravitational force?

Yes, the gravitational force from a square plate is directly proportional to the size of the plate. A larger plate will have a stronger gravitational force than a smaller plate, assuming all other factors remain the same.

4. How does the distance between the square plate and another object affect the gravitational force?

The gravitational force between two objects is inversely proportional to the square of the distance between them. This means that as the distance increases, the force decreases. Therefore, the closer the square plate is to another object, the stronger the gravitational force will be.

5. Is the gravitational force from a square plate the same on all sides?

No, the gravitational force may vary slightly on different sides of the square plate, depending on the distribution of mass within the plate. However, the overall force will be relatively uniform as long as the distance from the other object remains constant.

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