Gravitational field in a hollow sphere

AI Thread Summary
The gravitational field inside a uniform hollow sphere is zero due to the symmetrical distribution of mass. According to Gauss's law for gravity, any point inside the sphere experiences equal gravitational forces from opposing mass portions, which cancel each other out. This cancellation occurs because the gravitational force is inversely proportional to the square of the distance from the mass, leading to equal but opposite forces at any point inside the sphere. Visualizing this with cones extending from a point inside the sphere helps clarify that the forces from the surface areas do not result in a net gravitational pull. Therefore, the total gravitational force at any point within the hollow sphere is zero.
maddys_daddy
Messages
2
Reaction score
0
Can someone please give me a qualitative justification for the gravitational field inside a uniform hollow sphere being zero? I'm having a lot of trouble understanding this. Prof. said (in class) not to worry about the higher order polynomials involved, just be concerned with it "qualitatively" at this point. I'm lost.
Thanks,
m_d
 
Physics news on Phys.org
maddys_daddy said:
Can someone please give me a qualitative justification for the gravitational field inside a uniform hollow sphere being zero? I'm having a lot of trouble understanding this. Prof. said (in class) not to worry about the higher order polynomials involved, just be concerned with it "qualitatively" at this point. I'm lost.
Thanks,
m_d

You did not indicate what level of physics you already have. So, I have no idea if what I will say here makes any sense to you.

There is a Gauss's law equivalent for gravitational field. So if you have seen gauss's law applied to electrostatic, you should understand the identical principle applied to gravitational field. So apply that.

http://scienceworld.wolfram.com/physics/GausssLaw.html

Zz.
 
Qualitatively, think of it like this: At any given point inside the sphere, there is x amount of mass to the right in the shape of a bowl, and 1-x amount of mass to the left in the shape of a shperical shell missing a bowl shaped cap. These two complementary portions have centers of mass at let's say rx and r1-x. Then, qualitatively, you can imagine that rx < r1-x by just such an amount that:

x/rx2 = (1-x)/r1-x2.

It would probably help to draw a picture.

Qualitatively, you're supposed to realize that Fgrav is larger for larger amounts of mass and smaller at larger distances away from the mass. So, qualitatively, you can imagine that the effect of more mass (1-x) is canceled by the effect of further away (r1-x).
 

Attachments

  • qualitative_g_shell.gif
    qualitative_g_shell.gif
    1.5 KB · Views: 674
Last edited:
turin,
Thanks, I think I can understand that. However, when I click on the link to your .gif, I get a "you need to login to view this" message--well, I am! Oh, well, I'll take that up with the admins.
ZapperZ, FWIW, I'm in first year college physics for engineers (PHYS141 @ UoArizona). Next semester is my Electricity and Magnetism. So I have no idea what Gauss' law is, but I guess I'll find out.
Thanks for the help!
 
From any point inside the sphere, imagine a cone extending to a portion of the surface. Now extend that cone back to the opposite side of the sphere. If your point is not exactly at the center, your two cones will not intersect equal areas (and, so, not equal masses) because area is proportional to the square of the distance. HOWEVER, since gravitational force is inversely proportional to the square of the distance, the gravitational force from each of those portions of the surface will be the same. Since they are in opposite directions, the two equal but opposite forces cancel. The total gravitational force is 0.

(The same thing is true of magnetic force as well.)
 
I like HallsofIvy's explanation better than mine. It more clearly incorporates the inverse square dependence.
 
Thread ''splain this hydrostatic paradox in tiny words'
This is (ostensibly) not a trick shot or video*. The scale was balanced before any blue water was added. 550mL of blue water was added to the left side. only 60mL of water needed to be added to the right side to re-balance the scale. Apparently, the scale will balance when the height of the two columns is equal. The left side of the scale only feels the weight of the column above the lower "tail" of the funnel (i.e. 60mL). So where does the weight of the remaining (550-60=) 490mL go...
Consider an extremely long and perfectly calibrated scale. A car with a mass of 1000 kg is placed on it, and the scale registers this weight accurately. Now, suppose the car begins to move, reaching very high speeds. Neglecting air resistance and rolling friction, if the car attains, for example, a velocity of 500 km/h, will the scale still indicate a weight corresponding to 1000 kg, or will the measured value decrease as a result of the motion? In a second scenario, imagine a person with a...
Scalar and vector potentials in Coulomb gauge Assume Coulomb gauge so that $$\nabla \cdot \mathbf{A}=0.\tag{1}$$ The scalar potential ##\phi## is described by Poisson's equation $$\nabla^2 \phi = -\frac{\rho}{\varepsilon_0}\tag{2}$$ which has the instantaneous general solution given by $$\phi(\mathbf{r},t)=\frac{1}{4\pi\varepsilon_0}\int \frac{\rho(\mathbf{r}',t)}{|\mathbf{r}-\mathbf{r}'|}d^3r'.\tag{3}$$ In Coulomb gauge the vector potential ##\mathbf{A}## is given by...
Back
Top