Derivation of Gravitational Potential

by AbsoluteZer0
Tags: derivation, gravitational, potential
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 P: 126 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,
 Mentor P: 11,619 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.
 P: 550 $$\lim_{t \to \infty} \int_{R}^{t} \frac{Gm_{1}m{2}}{r^{2}} dr$$ Do you know how to solve that?
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.
 P: 550 $$\lim_{t \to \infty} -\int_{R}^{t} \frac{Gm_{1}m{2}}{r^{2}} dr$$ 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.
 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. BiP
 P: 550 Quick question (for my knowledge), why are the limits of intergration from R to infinity?
P: 783
 Quote by iRaid 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.

BiP
P: 126
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)

$W = - G M_1m_2 [\frac{1}{r}]^R_∞$

 anything over infinity tends to 0.

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

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

Which leads to

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

Thanks for the help

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