Gravitational force - point mass and circular platform

Click For Summary

Homework Help Overview

The discussion revolves around calculating the gravitational force between a point mass and a circular platform, with participants exploring the implications of integrating over an angle and the use of area density in the context of infinite geometries.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the role of cosine in vector components, question the definition of area density in relation to the mass of the disc, and explore the implications of an infinite radius on the gravitational force calculation.

Discussion Status

Some participants have provided clarifications regarding the use of cosine and area density, while others are investigating the relationship between the mass of the disc and the area density formula. There is an acknowledgment of differing interpretations of the problem, particularly concerning infinite geometries.

Contextual Notes

There is mention of a potential discrepancy in the source material regarding the formula for gravitational force and area density, with participants referencing different editions of the same text. The language of one participant is noted as Slovenian.

redivider
Messages
8
Reaction score
2

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.
 
Physics news on Phys.org
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?
 
  • Like
Likes   Reactions: Simon Bridge
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. language is Slovenian. Heres a quick snap of the newer: http://i.imgur.com/m0j9s9q.png ("neskončno velika plošča" literally means "infinitely big plate").
 
  • Like
Likes   Reactions: Simon Bridge

Similar threads

Replies
2
Views
2K
  • · Replies 28 ·
Replies
28
Views
2K
Replies
5
Views
2K
Replies
8
Views
2K
  • · Replies 1 ·
Replies
1
Views
5K
  • · Replies 12 ·
Replies
12
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
2
Views
6K
Replies
10
Views
4K
Replies
1
Views
2K