# Why do you integrate to find the gravitational field?

1. Dec 5, 2016

### Vitani11

1. The problem statement, all variables and given/known data
I know that in general it is when the gravitational field is not constant, so that part is fine. To find the gravitational force between a sphere with mass M and a radius R and a thin rod of length L and mass m which has one end a distance x away from the center of the sphere, you need to integrate.

2. Relevant equations
Unnecessary

3. The attempt at a solution
My issue is not the integration, it's just the concept. Is it because each piece of mass (dm) at a length dl on the rod experiences a difference magnitude of gravitational force since each piece of mass is at different locations (and a certain distance away from the spheres center) along the rod? Also, why can you treat the sphere as a point charge in this case? I know it is because it is a sphere, but I don't understand how being a sphere allows it to be treated as a point charge.

2. Dec 5, 2016

### haruspex

Yes.
It's a bit more general, applying to a uniform spherical shell. So the sphere as a whole does not need to be uniform - you can have the density depending on radius.
if you assume validity of the lines of flux model for forces obeying an inverse square law, you can put a Gaussian surface around the sphere and argue i) that the total flux through the surface is proportional to the mass (or charge) enclosed, and ii) by symmetry, the force must be radial and spherically symmetric. It follows that it is independent of the radius of the sphere, so that can be reduced to a point.
Alternatively, you can just do the integration and discover the result.

Of course, this is only for test points outside the spherical shell. There is no field inside.

3. Dec 5, 2016

### TJGilb

From the way you worded that, it sounds like the rod is oriented so that its long axis is radially outwards from the spherical mass. So yes, your reasoning is correct. Break down the rod into pieces, and you'll see that each one experiences a different force due to the equation $\vec F = G \frac {m_1 m_2} {r^2}$. So to find the overall force, you need to break it down into infinite infinitesimal pieces and sum those forces.

4. Dec 5, 2016

### Vitani11

I was, thank you.