Gravitational force - point mass and circular platform

AI Thread Summary
The discussion focuses on calculating the gravitational force between a point mass and a circular platform, highlighting confusion over the integration process and the use of cosine in the calculations. Participants clarify the area density formula, addressing the discrepancy of the factor of 4 in the expression σ = M/4πR^2, which is related to the context of the problem involving an infinite plane. The correct force formula for the interaction with an infinite plane is confirmed to be F = 2πGmσ, correcting earlier misunderstandings. The conversation also touches on the importance of understanding the geometric context when applying gravitational equations. Overall, the thread emphasizes the need for clarity in applying physical concepts to solve gravitational problems.
redivider
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Homework Statement


Calculate the gravitational force between a point mass and a circular platform.

Homework Equations


G=m1*m2/r^2

The Attempt at a Solution


http://i.imgur.com/dfZf9GK.jpg

The actual solution is different. They integrated by the angle between a/r (alpha) but i do not understand the cos(alpha) at the start, check this out:

http://i.imgur.com/nwxhCEb.jpg

Can anyone explain this please? Thanks.
 
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Force is a vector... the cosine is one of the components.
 
Yes, my mistake, thanks! What happens if the radius of the disc is infinite? The part I don't understand: σ(area density)=M/4πR^2, but isn't it just M/πR^2 (without "4" at the bottom)? I mean the whole mass of the disc would be M=πR^2σ, since πR^2 is the area of a cricle, with the 4 added its a sphere, but there is no sphere here just an infinite circle so where does the 4 come from?
 
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redivider said:
Yes, my mistake, thanks! What happens if the radius of the disc is infinite? The part I don't understand: σ(area density)=M/4πR^2, but isn't it just M/πR^2 (without "4" at the bottom)? I mean the whole mass of the disc would be M=πR^2σ, since πR^2 is the area of a cricle, with the 4 added its a sphere, but there is no sphere here just an infinite circle so where does the 4 come from?
I do not know the language (what is it, by the way?) but are you sure that the expression ##\sigma = M/4 \pi R^2 ## is referring to the same question? All I see used for that question is

$$ dM = 2 \pi x \rho dx h $$ which, if you integrate over x from 0 to R gives indeed ## M = \pi R^2 h \rho## as you expected.
 
It is not the same question but it is related to the first one (just with x going to infinity, not to R). Anyways, I have managed to use my google-fo and found a newer edition (1996); the one I have is 1988. This part is wrong. It states that F=8πGm (force between the particle and an infinite plane. The solution states specifically that you have to limit R towards infinity BUT with σ=M/4πR^2 being a constant), but the newer one is corrected to F=2πGmσ, which is what I got as well, so this is solved, thank you. Langauge is Slovenian. Heres a quick snap of the newer: http://i.imgur.com/m0j9s9q.png ("neskončno velika plošča" literally means "infinitely big plate").
 
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