How to calculate the gravitational field outside of a sphere

Homework Statement

A uniform solid sphere of mass M and radius R is fixed at a distance h above a thin infinite sheet of mass density σ. Obtain the magnitude of the force between the sphere and the sheet.

The Attempt at a Solution

I've found the gravitational field from the sheet which turned out to be a constant. Now I need the gravitational field of the sphere outside of its radius at a point h. Wouldn't that just be -GM/(R+h)2?

I've found the gravitational field from the sheet which turned out to be a constant. Now I need the gravitational field of the sphere outside of its radius at a point h.

Why do you need that? How would you calculate the gravitational force on a body?

• Vitani11
I don't know. I'm not good at physics at all - a lot of conceptual gaps here. Due to Newtons third law I can say that if I were to find the gravitational field due to the infinite sheet I can just multiply it by the mass of the sphere and then I have it, right? Here is the way I found the gravitational field of the infinite sheet:
∫g⋅dA = -4πGM
g∫dA = -4πGσA
g = -4πGσ

So I can then say the force between the two objects is

F = -4πGσM.

Thoughts?

Obviously these are vectors.

Close, but you need to make it clear what you're doing. For example, if you're trying to use Gauss's law for the entire plane, you haven't got a closed surface to integrate over. And the area and mass of the sheet are both infinite, so your first equation can't apply to the whole plane, and it's not clear exactly what you're trying to do when you cancel A from both sides--a procedure that is meaningless when the area is infinite.

The calculation can be done by treating the sheet as a set of elementary masses and adding up the resulting forces, but it needs some familiarity with multivariable integration.
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• Vitani11
Okay. So it would be better to not use Gauss law for this because this is not a closed surface? dF= -GMdm/(r2+h2). I changed dm to σ2πrdr so this became ∫dF = -GMσ2π∫r\(r2+h2)dr where the limits are from 0 to infinity. this makes sense to me, but when evaluating an integral a natural log comes up and its limit goes to infinity.

I changed dm to σ2πrdr so this became ∫dF = -GMσ2π∫r\(r2+h2)dr

You need to take account of the fact that the force from your ring of matter is acting at an angle. By symmetry there are no horizontal forces in the resultant; so what you want is the vertical component of the force you have now--you need another factor of
h/(r2 + h2)1/2
inside the integral.

If need be, you could probably convince yourself about this by taking the vertical component of the force from a single element of your ring and then integrating round the circle.

Incidentally, you've obviously done this without much multivariable integration. I was thinking of the limit of a square plane, rather than a circular one. But of course they both give the same answer in the limit.

• Vitani11
Solved. Thank you for the help - I'm aware that the horizontal components cancel I just was iffy on how to approach this problem because it seemed a lot more complicated than it is.

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