Finding the gravitational force over a flat infinite sheet

In summary: The equation for circumference instead of area is used because the element has to consist of a region for which d is roughly constant.
  • #1
Phantoful
30
3

Homework Statement


2nno2Tj.png


Homework Equations


F=ma
F=Gm1m2/r2
Gauss' Law?

The Attempt at a Solution


I'm not sure if I should be using Gauss' Law for this question, because I've never heard of it or learned about it. I'm currently taking multi-variable calculus (gradients, vectors, etc.). From what I know, the gravitational force should be found using an integral from R to infinity for the area of the plane... but I'm not sure what to do from there.

G*∫((mσ)/(Z^2))*dA from R to ∞?
 

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  • #2
You can't "dA from R". Decide what your variable of integration is and use that in the range, in the expression for the force and in the expression for the infinitesimal element.
 
  • #3
Okay, I'm starting to visualize it now, but I'm not sure how to put it into mathematics. I want to find the gravitational force for each concentric circle that has mass σπ(r2-R2) because V*σ=Msheet, and just using geometry the distance between the mass and the circles would be (r2+Z2)½, here's where I am:

wI5H0X1.png


However, I know that it's not correct mathematically because I'm not sure how to include dr, which I know is supposed to be in here. How can I make it a part of my integral? Or am I approaching this incorrectly?
 

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  • #4
The dr is included by putting the correct mass of the infinitesimal ring which is σ2πrdr.
 
  • #5
Note also that only the vertical component of the force due to the ring is to be considered.
 
  • #6
grzz said:
The dr is included by putting the correct mass of the infinitesimal ring which is σ2πrdr.
grzz said:
Note also that only the vertical component of the force due to the ring is to be considered.
I'm not sure how to visualize it this way, but what you're saying is that my equation (inside the integral) should be ((Gσ2πr*m)/Z2)*dr? I used the second equation from #2 in my original post, m being the mass of the point and G being the constant. Why is the equation for circumference instead of area used? I made the integral from R to ∞, which I believe is correct, right?
 
  • #7
Phantoful said:
Why is the equation for circumference instead of area used?
Because the element has to consist of a region for which d is roughly constant. That makes it an annulus of internal radius r and external radius r+dr. That has area 2πrdr.
 

Related to Finding the gravitational force over a flat infinite sheet

1. What is the formula for finding the gravitational force over a flat infinite sheet?

The formula for finding the gravitational force over a flat infinite sheet is F = 2Gσ, where F is the force, G is the gravitational constant (6.67 x 10^-11 Nm^2/kg^2), and σ is the surface mass density of the sheet in kg/m^2.

2. How does the distance from the sheet affect the gravitational force?

The distance from the sheet does not affect the gravitational force. Unlike point masses, the gravitational force over a flat infinite sheet is uniform and independent of distance.

3. Can the gravitational force over a flat infinite sheet be negative?

No, the gravitational force over a flat infinite sheet is always positive. This is because the force is always attractive and there is no negative mass.

4. How does the surface mass density of the sheet affect the gravitational force?

The surface mass density of the sheet directly affects the gravitational force. The higher the mass density, the stronger the gravitational force will be.

5. Is the gravitational force over a flat infinite sheet affected by the orientation of the sheet?

No, the gravitational force over a flat infinite sheet is not affected by the orientation of the sheet. This is because the sheet is infinite and has no edges or corners, so there is no preferred direction for the force to act in.

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