Gravitational force - point mass and circular platform

In summary, the student attempted to solve the homework problem but mistakenly used an incorrect equation. The newer edition of the textbook has been corrected to use the correct equation.
  • #1
redivider
8
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
  • #2
Force is a vector... the cosine is one of the components.
 
  • #3
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 Simon Bridge
  • #4
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.
 
  • #5
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").
 
  • Like
Likes Simon Bridge

1. What is the definition of gravitational force?

The gravitational force is a natural phenomenon by which all objects with mass are attracted to one another. It is one of the four fundamental forces of nature and is responsible for the motion of planets, stars, and galaxies.

2. How is gravitational force calculated for a point mass?

The gravitational force between two point masses is calculated using the formula F = G * (m1 * m2) / r^2, where G is the gravitational constant, m1 and m2 are the masses of the two objects, and r is the distance between them.

3. What is a circular platform and how does it affect gravitational force?

A circular platform is a flat, circular surface on which an object can move in a circular path around a central point. In the context of gravitational force, a circular platform can affect the force acting on an object by altering its direction and magnitude.

4. How does the mass of a point mass affect the gravitational force?

The greater the mass of a point mass, the stronger its gravitational force will be. This is because the more massive an object is, the more it bends the fabric of space-time, causing other objects to be attracted to it with greater force.

5. Can gravitational force be cancelled out on a circular platform?

No, gravitational force cannot be cancelled out on a circular platform. However, it can be balanced by other forces, such as centripetal force, which keeps an object moving in a circular path without falling into the center of the platform.

Similar threads

  • Introductory Physics Homework Help
Replies
2
Views
859
  • Introductory Physics Homework Help
Replies
5
Views
810
  • Introductory Physics Homework Help
Replies
28
Views
1K
  • Introductory Physics Homework Help
Replies
8
Views
950
  • Introductory Physics Homework Help
Replies
1
Views
200
  • Introductory Physics Homework Help
Replies
12
Views
1K
  • Introductory Physics Homework Help
Replies
7
Views
101
  • Introductory Physics Homework Help
Replies
10
Views
681
  • Introductory Physics Homework Help
Replies
1
Views
836
  • Introductory Physics Homework Help
Replies
15
Views
361
Back
Top