Register to reply 
Derivation of Gravitational Potential 
Share this thread: 
#1
Feb1813, 11:08 AM

P: 126

Hi,
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. Thanks, 


#2
Feb1813, 11:15 AM

Mentor
P: 11,864

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
Feb1813, 11:45 AM

P: 551

[tex]\lim_{t \to \infty} \int_{R}^{t} \frac{Gm_{1}m{2}}{r^{2}} dr[/tex]
Do you know how to solve that? 


#4
Feb1813, 12:23 PM

P: 126

Derivation of Gravitational Potential



#5
Feb1813, 01:49 PM

P: 551

[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: [tex]\frac{Gm_{1}m_{2}}{R}[/tex] Pretty sure the math is correct, someone might be able to fix any physics errors I have. 


#6
Feb1813, 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.
BiP 


#7
Feb1813, 06:04 PM

P: 551

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



#8
Feb1813, 10:00 PM

P: 783

BiP 


#9
Feb2113, 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] 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 


Register to reply 
Related Discussions  
Gravitational lensing derivation using equivalence principle  Special & General Relativity  33  
Linearised Gravitational Waves Derivation  Special & General Relativity  3  
Derivation of gravitational potential energy (Feynman Lectures on Physics Vol 1: 43)  General Physics  8  
Gravitational Potential Energy Derivation question?  Introductory Physics Homework  7  
Gravitational Force, the derivative of Gravitational Potential Energy?  Classical Physics  3 