- #1
jaydnul
- 558
- 15
When I was in introductory physics I remember being told that the net force (gravitational for example) inside a shell is always 0. I always felt that, intuitively, this would only be true at the center of the ring. Not sure what made me think of it today, by I decided to sit down and do the calculation.
The vector formulation for a force of gravity
F=\frac{Gm_1m_2}{|r|^2}\frac{r}{|r|}
m1 will be the mass of the particle inside the shell, m2 will be the mass of the shell. Inside a shell this becomes
F=\frac{Gm_1m_2}{|r|^2}<cos(θ),sin(θ)>
dF=\frac{Gm_1dm_2}{|r|^2}<cos(θ),sin(θ)>
where
dm_2=\frac{m_2}{2π}dθ
If r is changing as a function of θ, we can write it in this form
dF=\frac{Gm_1m_2}{2π}<\frac{cos(θ)}{r(θ)^2},\frac{sin(θ)}{r(θ)^2}>dθ
For the r(θ), I made up a simple function that satisfied my question. r(θ)=2+sin(θ-π/2) will put my particle somewhere in the right section (as long as it's not centered) of my oval shaped ring. (the biggest radius from my particle would be 3, the smallest would be 1). Finally we get
F=\frac{Gm_1m_2}{2π}∫<\frac{cos(θ)}{(2+sin(θ-π/2))^2},\frac{sin(θ)}{(2+sin(θ-π/2))^2}>dθ
The integral is from 0 to 2π (not sure how to do a definite). After doing the integrals in my calculator I get roughly
F=\frac{Gm_1m_2}{2π}<1.21,0>
There would be a pull to the right. So unless the particle was exactly centered, it would never just float, there would always be a force on it. Anyways, just thought I'd share. Why are we told to assume the net force at every point inside the shell is 0?
The vector formulation for a force of gravity
F=\frac{Gm_1m_2}{|r|^2}\frac{r}{|r|}
m1 will be the mass of the particle inside the shell, m2 will be the mass of the shell. Inside a shell this becomes
F=\frac{Gm_1m_2}{|r|^2}<cos(θ),sin(θ)>
dF=\frac{Gm_1dm_2}{|r|^2}<cos(θ),sin(θ)>
where
dm_2=\frac{m_2}{2π}dθ
If r is changing as a function of θ, we can write it in this form
dF=\frac{Gm_1m_2}{2π}<\frac{cos(θ)}{r(θ)^2},\frac{sin(θ)}{r(θ)^2}>dθ
For the r(θ), I made up a simple function that satisfied my question. r(θ)=2+sin(θ-π/2) will put my particle somewhere in the right section (as long as it's not centered) of my oval shaped ring. (the biggest radius from my particle would be 3, the smallest would be 1). Finally we get
F=\frac{Gm_1m_2}{2π}∫<\frac{cos(θ)}{(2+sin(θ-π/2))^2},\frac{sin(θ)}{(2+sin(θ-π/2))^2}>dθ
The integral is from 0 to 2π (not sure how to do a definite). After doing the integrals in my calculator I get roughly
F=\frac{Gm_1m_2}{2π}<1.21,0>
There would be a pull to the right. So unless the particle was exactly centered, it would never just float, there would always be a force on it. Anyways, just thought I'd share. Why are we told to assume the net force at every point inside the shell is 0?