• Support PF! Buy your school textbooks, materials and every day products Here!

Integration constants, gravitational potential of sphere

  • Thread starter Vir
  • Start date
  • #1
Vir
18
0

Homework Statement


So I'm calculating the gravitational potential of a sphere at at point P. R = radius of sphere, r = distance from center of sphere to point P. I'm looking at two scenarios; r > R (1) and r < R (2). So I have the following integral:

\begin{equation} V(r) = \int \frac{-GM}{Rr} ds
\end{equation}

The Attempt at a Solution


Now, for (1) I integrate from r-R to r+R. However for (2) I'm thinking I have to integrate from -(R-r) to R+r, but this gives me the same constants as in (1). To my understanding the potential is supposed to be constant for all r < R, but then my constants are giving me the wrong answer.

If I remove the - and integrate from R-r to R+r I get a correct answer, but this makes no sense to me. That would mean I was integrating from the center of the sphere out to the edge, effectively ignoring half of the sphere? What am I missing?
 

Answers and Replies

  • #2
PeroK
Science Advisor
Homework Helper
Insights Author
Gold Member
12,377
5,157

Homework Statement


So I'm calculating the gravitational potential of a sphere at at point P. R = radius of sphere, r = distance from center of sphere to point P. I'm looking at two scenarios; r > R (1) and r < R (2). So I have the following integral:

\begin{equation} V(r) = \int \frac{-GM}{Rr} ds
\end{equation}

The Attempt at a Solution


Now, for (1) I integrate from r-R to r+R. However for (2) I'm thinking I have to integrate from -(R-r) to R+r, but this gives me the same constants as in (1). To my understanding the potential is supposed to be constant for all r < R, but then my constants are giving me the wrong answer.

If I remove the - and integrate from R-r to R+r I get a correct answer, but this makes no sense to me. That would mean I was integrating from the center of the sphere out to the edge, effectively ignoring half of the sphere? What am I missing?
I'm confused about what you are trying to do. ##\frac{-GM}{r}## is the gravitational potential for a solid sphere at a point outside the sphere, a distance r from the centre.

What are you trying to calculate?
 
  • #3
Vir
18
0
What are you trying to calculate?
I'm trying to arrive at the gravitational potential at a distance r < radius of sphere(R). The integral is

\begin{equation}
V(r<R) = \int \frac{-GM}{2rR}ds,
\end{equation}

but I'm not sure how to select my integration limits.
 
  • #4
PeroK
Science Advisor
Homework Helper
Insights Author
Gold Member
12,377
5,157
Is this a hollow sphere? And are you trying to integrate over a surface?
 
  • #5
Vir
18
0
Yes, a hollow sphere. I'm integrating small circles of the surface of the sphere. I guess this image describes it well.
500px-Shell-diag-1.png
 
  • #6
PeroK
Science Advisor
Homework Helper
Insights Author
Gold Member
12,377
5,157
That's a point r outside the sphere. In that diagram, you first need to work out the gravitational potential of the ring marked in blue. So you're going to need to take θ into account and calculate dM as well.
 
  • #7
Vir
18
0
Ok. So
\begin{equation}
dV(r<R) = \frac{-GdM}{S}
\end{equation}

\begin{equation}
dM = \frac{M}{4\pi R^2}d \theta R Sin \theta R 2 \pi
\end{equation}

\begin{equation}
dM = \frac{M}{2}Sin \theta d \theta
\end{equation}

\begin{equation}
S = \sqrt{R^2+r^2-2Rr Cos \theta}
\end{equation}

\begin{equation}
\frac{dS}{d \theta} = \frac{Rr Sin \theta}{S}
\end{equation}

Inserting in dV(r<R):

\begin{equation}
dV(r<R) = \frac{-GM Sin \theta S ds}{2SRr Sin \theta}
\end{equation}

\begin{equation}
dV(r<R) = \frac{-GM}{2Rr} ds
\end{equation}

\begin{equation}
V(r<R) = \int \frac{-GM}{2Rr} ds
\end{equation}

Is that right?
 
  • #8
PeroK
Science Advisor
Homework Helper
Insights Author
Gold Member
12,377
5,157
Not quite. If you think in terms of the force exerted by the ring, what is the direction of the force?
 
  • #9
Vir
18
0
When the ring is to the right of P the force points left, and vice versa.
 
  • #10
PeroK
Science Advisor
Homework Helper
Insights Author
Gold Member
12,377
5,157
You need to think more carefully. Remember that force is a vector.
 
  • #11
Vir
18
0
Yea, ok. I was thinking since each point on the circle has a point on the opposite side where the perpendicular component of the forces would cancel out the net force would be parallell to the axis through the center of the circle and P. Not so?
 
  • #12
PeroK
Science Advisor
Homework Helper
Insights Author
Gold Member
12,377
5,157
Yes, the perpendicular forces cancel. But what about the magnitude of the remaining force?
 
  • #13
Vir
18
0
That would depend on the mass at P right? In any case it's
\begin{equation}
F = -\frac{GMm}{(r-R Cos \theta)}
\end{equation}
 
  • #14
PeroK
Science Advisor
Homework Helper
Insights Author
Gold Member
12,377
5,157
Take a look again at your diagram and make s a vector. That's the direction of the force. So, you have a "vertical" component and a "horizontal" component. The magnitude of the horizontal component is not the magnitude of the full force.
 
  • #15
Vir
18
0
Then I don't know. You asked about the magnitude of the remaining force, isn't that the horizontal component of the total force?
 
  • #16
PeroK
Science Advisor
Homework Helper
Insights Author
Gold Member
12,377
5,157
Then I don't know. You asked about the magnitude of the remaining force, isn't that the horizontal component of the total force?
Yes, but the vertical force is ##Fsin(φ)## and the horizotal force is ##Fcos(φ)##
 
  • #17
Vir
18
0
Right. So what do I do with this?
 
  • #18
PeroK
Science Advisor
Homework Helper
Insights Author
Gold Member
12,377
5,157
Right. So what do I do with this?
You need to factor ##cos(φ)## into your equation. That's all your missing, I think.
 
  • #19
Vir
18
0
I don't know. My integral gives me the right answer for R>r:
\begin{equation}
V(r>R) = \frac{-GM}{2rR} \int_{r-R}^{r+R} ds
\end{equation}

Simplifies to:

\begin{equation}
V(r>R) = \frac{-GM}{r}
\end{equation}

I think the point is that when r<R then the rings both to the left and right of P have a positive distance to P, since the potential is supposed to be negative. I guess one has to think more in terms of polar coordinates? Thus, when r<R you have to integrate from R-r(closest side of sphere) to R+r(far side). This gives

\begin{equation}
V(r<R) = \frac{-GM}{2rR} \int_{R-r}^{R+r} ds
\end{equation}

Simplifies:

\begin{equation}
V(r<R) = \frac{-GM}{R}
\end{equation}
 
  • #20
PeroK
Science Advisor
Homework Helper
Insights Author
Gold Member
12,377
5,157
I think you've got two mistakes cancelling out for the problem outside the sphere. But, they won't cancel inside the sphere.

You've got ##dV = -GdM/S## but that assumes that all the mass in the ring is concentrated at one point a distance S. You need to take cos(φ) into account. The potential must be:

##dV = (-GdM/S)cos(φ)##

If you imagine being very close to the ring, then there would be almost no potential. In fact, when φ = π/2, the potential will be zero.
 
  • #21
Vir
18
0
So I checked my book:
image082.gif

The potential dV(x) at P from dM, that is now a small part of the ring, at distance S is given by:

\begin{equation}
dV(x) = \frac{-GdM}{S}
\end{equation}

\begin{equation}
dM = \frac{M}{2 \pi R}Rd \theta
\end{equation}

So

\begin{equation}
V(x) = \frac{-GM}{2 \pi S}\int_0^{2 \pi}d \theta = \frac{-GM}{S}
\end{equation}

So when S = R the potential is not zero, it's just the lowest possible value for any point on the axis through the ring's center. However the gravitational field will be zero, since g(0) = -∇V(0) = 0.

I think you are right about the cancelling though. Gravitational potential doesn't cancel out. The potential is exactly the same for every dM on the ring so the potential from the ring is just the sum of potential from all the dMs. So for the hollow sphere the gravitational potential is just the sum of the potential from all the rings in the sphere. So since the inside of a sphere is a equipotential region(source: wikipedia) V = constant for any point inside the sphere. I believe this is correct.

Thanks for the help!
 
Last edited:

Related Threads on Integration constants, gravitational potential of sphere

Replies
3
Views
594
Replies
11
Views
8K
Replies
1
Views
6K
  • Last Post
Replies
10
Views
3K
  • Last Post
Replies
14
Views
18K
  • Last Post
Replies
1
Views
3K
Replies
23
Views
2K
  • Last Post
Replies
20
Views
1K
  • Last Post
Replies
2
Views
3K
Replies
1
Views
5K
Top