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Derivation of Gravitational Potential

by AbsoluteZer0
Tags: derivation, gravitational, potential
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Feb18-13, 11:08 AM
P: 126

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

[itex] W = Fd [/itex] (1)

[itex] W = G \frac{M_1m_2}{r^2}d [/itex] (2) Substituting the Gravitational Force formula

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

Which allows us to arrive at:

[itex] E_p = - \frac{GM_2m_1}{R}[/itex] (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.

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Feb18-13, 11:15 AM
jtbell's Avatar
P: 11,622
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.
Feb18-13, 11:45 AM
P: 550
[tex]\lim_{t \to \infty} \int_{R}^{t} \frac{Gm_{1}m{2}}{r^{2}} dr[/tex]

Do you know how to solve that?

Feb18-13, 12:23 PM
P: 126
Derivation of Gravitational Potential

Do you know how to do integrals like this one?
I can use integrals like these, to an extent.

Do you know how to solve that?
Unfortunately not.
Feb18-13, 01:49 PM
P: 550
[tex]\lim_{t \to \infty} -\int_{R}^{t} \frac{Gm_{1}m{2}}{r^{2}} dr[/tex]
If you're taking an integral with respect to r and G, m1, and m2 are all constants, then what happens to the integral:
[tex]\lim_{t \to \infty} -Gm_{1}m_{2}\int_{R}^{t} \frac{1}{r^{2}}dr[/tex]
Now take the fact that 1/r^2 = r^-2
So the integral is then solvable:
[tex]\lim_{t \to \infty} (-Gm_{1}m_{2} \frac{-1}{r})|_{t}^{R}[/tex]
So then this becomes:
[tex]\lim_{t \to \infty} (\frac{-Gm_{1}m_{2}}{R}-\frac{-Gm_{1}m_{2}}{t})[/tex]
Finally take the limit, anything over infinity tends to 0.
So you end up with:

Pretty sure the math is correct, someone might be able to fix any physics errors I have.
Feb18-13, 05:02 PM
P: 783
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.

Feb18-13, 06:04 PM
P: 550
Quick question (for my knowledge), why are the limits of intergration from R to infinity?
Feb18-13, 10:00 PM
P: 783
Quote Quote by iRaid View Post
Quick question (for my knowledge), why are the limits of intergration from R to infinity?
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.

Feb21-13, 10:47 AM
P: 126
I think I figured it out, please correct me if I'm wrong.

[itex]W = Fd[/itex]

[itex]W = G\frac{M_1m_2}{r^2}d[/itex]

[itex]W =- \int_R^∞ G\frac{M_1m_2}{r^2}\,dr[/itex]

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

[itex]W = - G M_1m_2 [\frac{1}{r}]^R_∞[/itex]

anything over infinity tends to 0.

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

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

Which leads to

[itex] E_p = -G\frac{M_1m_2}{r} [/itex]

Thanks for the help

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