Mistake in proof of gravitational potential equation?

Click For Summary

Homework Help Overview

The discussion revolves around deriving the equation for gravitational potential, specifically addressing the confusion regarding the distances involved in the derivation. The original poster attempts to understand why the distance from infinity to a point mass is set as 'r' in the textbook, while also grappling with potential typos in the text that complicate the understanding.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants explore the derivation of gravitational potential, questioning the use of distance in the equations. Some suggest using integration to clarify the relationship between force and distance, while others discuss the implications of moving from infinity to a final position and how that affects the variables in the equations.

Discussion Status

Participants are actively engaging with the original poster's confusion, providing insights into the mathematical principles involved, particularly the need for integration when dealing with variable forces. There is a recognition of the complexities introduced by the textbook's wording and potential errors, but no consensus has been reached on the exact nature of the misunderstanding.

Contextual Notes

There are mentions of potential typos in the textbook that may hinder understanding, as well as the original poster's uncertainty about their calculus knowledge, which could affect their grasp of the concepts being discussed.

dotcom
Messages
17
Reaction score
0
mistake in proof of gravitational potential equation?

Hi, I have a problem in deriving an equation for the gravitational potential.

Well, I think that the graviational potential is the work done by external forces in moving a unit mass Mp from infinity to that point through distance d, and so I converted the equation in this way:

when M is the mass of a planet creating the gravitational field and r is the distance between M and Mp,

V= (work done) / (mass)
=F.d / Mp
=(G*M*Mp/r^2)*d/ Mp

... and I got nowhere.

However, my text states that the distance from the infinity to the point mass Mp should be r as well. So the text says that

V= (work done) / (mass)
=F.r / Mp
=(G*M*Mp/r^2)*r/ Mp

and in this way the correct equation, V=-GM/r (negative sign is added afterwards) shows up.

But I don't understand why the text can set both the distance from the infinity to the point mass Mp and the distance between M and Mp as r.

To make the problem more complicated, my teacher has said that there might be a typo or two in the text, which makes the understanfing very difficult. So it might be included in this context, I don't know...

If anyone could explain me this, please help me by doing so...
I would really really appreciate your advise!
 
Physics news on Phys.org
You're on the right track and I think I see area of confusion,

it is much easier to understand using integration where force is expressed as function of distance and integrated from lower limit of infinity to upper limit of R.
But since at infinity Force is zero, this simplifies to

0-GMm/r^2*r see here if this is illegible:

http://hyperphysics.phy-astr.gsu.edu/hbase/gpot.html#ui
 
I was responding to this earlier and the electricty went down on me!

dotcom said:
Hi, I have a problem in deriving an equation for the gravitational potential.

Well, I think that the graviational potential is the work done by external forces in moving a unit mass Mp from infinity to that point through distance d, and so I converted the equation in this way:

when M is the mass of a planet creating the gravitational field and r is the distance between M and Mp,

V= (work done) / (mass)
=F.d / Mp
=(G*M*Mp/r^2)*d/ Mp
... and I got nowhere.
Work = F*d only for constant force. For a variable force, as here you need an integral. (If your textbook talks about the gravitational potential energy, surely it mentions integration? You don't say whether you know anything about Calculus.) Technically, you should integrate along the path the object actually takes. Fortunately, since gravitational force is "conservative", the work done moving from one point to another along ANY path depends only on the endpoints not on the particular path. By convention, gravitational potential energy is taken to be 0 at infinity so we can take the integral from infinity to R where R is the FINAL distance from M to Mp.

However, my text states that the distance from the infinity to the point mass Mp should be r as well. So the text says that

V= (work done) / (mass)
=F.r / Mp
=(G*M*Mp/r^2)*r/ Mp

and in this way the correct equation, V=-GM/r (negative sign is added afterwards) shows up.
Since the mass is MOVING, r, the distance from Mp to M changes. Use a different value, say R as I said abovee, to represent the distance between M and Mp in its FINAL position, the point at which you are finding the potential. In moving from "infinity" to its final distance from M, r changes from infinity to R.

But I don't understand why the text can set both the distance from the infinity to the point mass Mp and the distance between M and Mp as r.
It's not. r is always the distance between M and Mp but as the mass moves that distance changes from "infinity" to R.

To make the problem more complicated, my teacher has said that there might be a typo or two in the text, which makes the understanfing very difficult. So it might be included in this context, I don't know...

If anyone could explain me this, please help me by doing so...
I would really really appreciate your advise!
The force at any distance, r, from M is -GMpM/r2. If it moves a SHORT distance [itex]\delta r[/itex], so the force doesn't change much, the work done there is about [itex](-GMpM/r^2)\delta r[/itex]. To approximate the work over the entire path, add for each little step along the path (that's what's known as a Riemann sum). Taking the limit as you make each step smaller and smaller (and have more and more steps) that becomes the integral that gives the exact value:
[tex]-GM \int_{-\infty}^R \frac{dr}{r^2}[/tex]
 
thanks Hall for adding rigor, and doing a darn good job of explaining integration in a couple of paragraphs. Thats the problem of doing math w/o calculus, its like trying to build a house without blueprints.
 

Similar threads

Replies
8
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 5 ·
Replies
5
Views
1K
  • · Replies 9 ·
Replies
9
Views
2K
Replies
8
Views
2K
Replies
14
Views
5K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
15
Views
2K
  • · Replies 7 ·
Replies
7
Views
3K