Gravity at an arbitrary location near a disc

[bumpkin]

Homework Statement

calculate the gravity acceleration at an arbitrary location due to a disc of thickness h, radius r and density p

Homework Equations

g=Gm/r^2

The Attempt at a Solution

define r in terms of the vector magnitude from the measurement point to some point on the disc, then hit it with a volume integral? Is there an easier way, say using symmetry?

 

[Hootenanny]

Homework Statement

calculate the gravity acceleration at an arbitrary location due to a disc of thickness h, radius r and density p

Homework Equations

g=Gm/r^2

The Attempt at a Solution

define r in terms of the vector magnitude from the measurement point to some point on the disc, then hit it with a volume integral? Is there an easier way, say using symmetry?

Welcome to Physics Forums.

I'm assuming that rho and h are constant.

It may well be easier to tackle this problem in two separate cases: (a) When the point of interest is outside the body; and (b) when the point of interest is inside the body. For the former case, the gravitational field of the disc is identical to that of a point source of equivalent mass, located at the centre of the disc.

 

[bumpkin]

Welcome to Physics Forums.

I'm assuming that rho and h are constant.

It may well be easier to tackle this problem in two separate cases: (a) When the point of interest is outside the body; and (b) when the point of interest is inside the body. For the former case, the gravitational field of the disc is identical to that of a point source of equivalent mass, located at the centre of the disc.

rho and h are constant. I hadn't even thought of b. But for a, I would assume the acceleration at the edge of the disc would be different to the gravity long the axis of the disc? Ie if the disc was in the xy plane, the gravity at (r,0,h) would be different to (0,0,sqrt(r^2+h^2))?

 

[Doc Al]

For the former case, the gravitational field of the disc is identical to that of a point source of equivalent mass, located at the centre of the disc.

That would be true for a uniform sphere, but not for a disk.

 

[Hootenanny]

rho and h are constant. I hadn't even thought of b. But for a, I would assume the acceleration at the edge of the disc would be different to the gravity long the axis of the disc? Ie if the disc was in the xy plane, the gravity at (r,0,h) would be different to (0,0,sqrt(r^2+h^2))?

Oh, sorry. I assumed that you were working in 2D, in the plane of the disc. My bad.

The best option then is to transform into cylindrical coordinates. Apologies for the confusion.