Why do you integrate to find the gravitational field?

AI Thread Summary
To find the gravitational force between a sphere and a rod, integration is necessary due to the varying distances of mass elements along the rod from the sphere's center. Each mass piece (dm) experiences a different gravitational force based on its location, necessitating the summation of these forces through integration. The sphere can be treated as a point mass because, for external points, the gravitational field behaves as if all mass is concentrated at the center, a property stemming from its symmetry. This principle applies even if the sphere's density varies with radius, as long as the test points are outside the sphere. Understanding these concepts clarifies why integration is essential for calculating gravitational forces in non-uniform fields.
Vitani11
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Homework Statement


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.

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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.
 
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Vitani11 said:
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?
Yes.
Vitani11 said:
why can you treat the sphere as a point charge
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.
 
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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.
 
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TJGilb said:
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.
I was, thank you.
 
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