# Derivation of Gravitational Potential

1. Feb 18, 2013

### AbsoluteZer0

Hi,

The derivation of the Gravitational Potential formula, as I understand, is:

$W = Fd$ (1)

$W = G \frac{M_1m_2}{r^2}d$ (2) Substituting the Gravitational Force formula

$W = - \int_R^∞G \frac{M_1m_2}{r^2} \, dr$ (3) Integrating within the boundaries of the initial distance (R) and Infinity

Which allows us to arrive at:

$E_p = - \frac{GM_2m_1}{R}$ (4)

However, what I don't understand is how we are able to proceed from step 3 to step 4.
What method must be used in order to proceed as such?

My proficiency with Calculus is still in the works.

Thanks,

2. Feb 18, 2013

### Staff: Mentor

Do you know how to do integrals like this one?

$$\int_a^b {x^n dx}$$

If so, here's a hint: $\frac{1}{r^2} = r^{-2}$.

If no, then you'd best develop your calculus up to that point.

3. Feb 18, 2013

### iRaid

$$\lim_{t \to \infty} \int_{R}^{t} \frac{Gm_{1}m{2}}{r^{2}} dr$$

Do you know how to solve that?

4. Feb 18, 2013

### AbsoluteZer0

I can use integrals like these, to an extent.

Unfortunately not.

5. Feb 18, 2013

### iRaid

$$\lim_{t \to \infty} -\int_{R}^{t} \frac{Gm_{1}m{2}}{r^{2}} dr$$
If you're taking an integral with respect to r and G, m1, and m2 are all constants, then what happens to the integral:
$$\lim_{t \to \infty} -Gm_{1}m_{2}\int_{R}^{t} \frac{1}{r^{2}}dr$$
Now take the fact that 1/r^2 = r^-2
So the integral is then solvable:
$$\lim_{t \to \infty} (-Gm_{1}m_{2} \frac{-1}{r})|_{t}^{R}$$
So then this becomes:
$$\lim_{t \to \infty} (\frac{-Gm_{1}m_{2}}{R}-\frac{-Gm_{1}m_{2}}{t})$$
Finally take the limit, anything over infinity tends to 0.
So you end up with:
$$\frac{-Gm_{1}m_{2}}{R}$$

Pretty sure the math is correct, someone might be able to fix any physics errors I have.

Last edited: Feb 18, 2013
6. Feb 18, 2013

### Bipolarity

Also, note that potential at infinity is conventionally defined to be 0. The convention, being what it is, isn't derivable mathematically, so you need to use it as a given in solving the problem when evaluating potential at infinity in your integral.

BiP

7. Feb 18, 2013

### iRaid

Quick question (for my knowledge), why are the limits of intergration from R to infinity?

8. Feb 18, 2013

### Bipolarity

The gravitationalpotential is defined as the work done by gravity to bring an object from infinity to a distance R from an object along a straight line, hence the limits of integration. A theorem from vector calculus shows that it does not matter what path the object travels, so the definition can be adjusted to an "arbitrary path from infinity to a distance R" from an object. But that is an offshoot of vector calculus.

BiP

9. Feb 21, 2013

### AbsoluteZer0

I think I figured it out, please correct me if I'm wrong.

$W = Fd$

$W = G\frac{M_1m_2}{r^2}d$

$W =- \int_R^∞ G\frac{M_1m_2}{r^2}\,dr$

$W = - G M_1m_2 \int_R^∞ r^{-2}\,dr$ (Initially I was uncertain about pulling $GM_1m_2$ out)

$W = - G M_1m_2 [\frac{1}{r}]^R_∞$

Am I right in assuming that this is the reason why $-G\frac{M_1m_2}{∞}$ produces zero?

$W = [-G \frac{M_1m_2}{r} - -G \frac{-GM_1m_2}{∞}]$

$E_p = -G\frac{M_1m_2}{r}$