# Mistake in proof of gravitational potential equation?

1. Apr 11, 2007

### dotcom

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...

2. Apr 11, 2007

### denverdoc

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

3. Apr 11, 2007

### HallsofIvy

Staff Emeritus
I was responding to this earlier and the electricty went down on me!

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.

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.

It's not. r is always the distance between M and Mp but as the mass moves that distance changes from "infinity" to R.

The force at any distance, r, from M is -GMpM/r2. If it moves a SHORT distance $\delta r$, so the force doesn't change much, the work done there is about $(-GMpM/r^2)\delta r$. 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:
$$-GM \int_{-\infty}^R \frac{dr}{r^2}$$

4. Apr 11, 2007

### denverdoc

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.