Calculating the Gravitational Potential of a Spehrical Shell

In summary, Feynman is explaining how a hollow spherical shell acts as though it has all its mass located at the center. He is doing the math for all this, and is using sin theta to find the potential energy. However, he is having difficulty following the mathematics.
  • #1
amolv06
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I'm going through The Feynman Lectures On Physics. I've taken a few classical physics classes before, however Feynman's detailed explanations are much better than anything I've gotten from my teachers. However, I'm having a little difficulty following the mathematics when he tries to show that a hollow spherical shell acts as though it has all its mass located at the center.

I will read the paragraph previous to the math I am having difficulty with, and then read off the math.

"We now demonstrate the correctness of this miracle. In order to do so, however, we shall consider a thin uniform shell instead of the whole earth. Let the total mass of the shell be m, and let us calculate the potential energy of a particle of mass m' a distance R away from the sphere and show that the potential energy is the same as it would be if the mass m were at the center. If we call x the distance of a certain plane section from the center, then all the mass that is in a slice dx is at the same idstance r from P, and the potential due to the ring is -Gm'dm/r. How much mass is in the small slice dx? An amount

[tex]dm=2*pi*y*mu*ds=\frac{2*pi*y*mu*dx}{sin(\vartheta)}=\frac{2*pi*y*mu*dx*a}{y}=2*pi*a*mu*dx[/tex]

Here, a is the radius I believe. I don't know where he pulled y from. Mu is the density of a piece of the spherical shell.

What I can't understand is how he is doing the math for all this. Where does the sin theta come from? I'm slightly lost here.

This is in the Feynman Lectures, by the way.
 
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  • #2
Can you tell me the chapter it's in, I have to see it in the book. I'll try to see if I can figure it out.
 
  • #3
It's in chapter 13-4. Thanks.
 
  • #4
amolv06 said:
Here, a is the radius I believe. I don't know where he pulled y from. Mu is the density of a piece of the spherical shell.

What I can't understand is how he is doing the math for all this. Where does the sin theta come from? I'm slightly lost here.
To understand what he's doing, refer to the diagram next to the equations. The diagram shows the meaning of x and y.

dm refers to the mass of the ring-shaped slice of the sphere. Yes, a is the radius of the sphere. y is the perpendicular (vertical) distance of the ring from the line O-P. Since ds is the length of the side of the ring, the thickness of the ring along the x-axis (which is the line O-P) is [itex]dx = ds \sin\theta[/itex].
This is in the Feynman Lectures, by the way.
Book I, chapter/page 13-9.
 
  • #5
Ahh. That makes perfect sense now. Thanks a lot.
 

1. How is the gravitational potential of a spherical shell calculated?

The gravitational potential of a spherical shell is calculated by using the equation: V = -GM/r, where G is the gravitational constant, M is the mass of the shell, and r is the distance from the center of the shell to the point where the potential is being calculated.

2. What is the difference between the gravitational potential inside and outside of a spherical shell?

The gravitational potential inside a spherical shell is zero, because the gravitational force exerted by the shell cancels out at all points within the shell. Outside of the shell, the potential is calculated using the equation mentioned in question 1.

3. Can the gravitational potential of a spherical shell be negative?

Yes, the gravitational potential of a spherical shell can be negative. This occurs when the point where the potential is being calculated is closer to the center of the shell than the radius of the shell itself.

4. How does the mass of a spherical shell affect its gravitational potential?

The mass of a spherical shell directly affects its gravitational potential. The greater the mass of the shell, the stronger its gravitational pull and the higher its potential at a given distance from the center.

5. Is the gravitational potential of a spherical shell affected by the shape of the shell?

No, the gravitational potential of a spherical shell is not affected by the shape of the shell. This is because the gravitational force exerted by a spherical shell is the same at all points on its surface, regardless of its shape.

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