Derivation of Gravitational Potential

In summary, the conversation discusses the derivation of the Gravitational Potential formula using integrals and the concept of work done by gravity. The limits of integration are from infinity to a distance R, and the potential at infinity is defined to be zero. The conversation also clarifies the steps involved in the derivation and the reason behind certain mathematical operations. Overall, the conversation provides an explanation of the Gravitational Potential formula and its implications.
  • #1
AbsoluteZer0
125
1
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,
 
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  • #2
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
[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
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.
 
  • #5
[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.
 
Last edited:
  • #6
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
Quick question (for my knowledge), why are the limits of intergration from R to infinity?
 
  • #8
iRaid said:
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
 
  • #9
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
 
Last edited:

1. What is gravitational potential?

Gravitational potential is the measure of the potential energy that an object has due to its position in a gravitational field. It is a scalar quantity and is usually measured in joules per kilogram.

2. How is gravitational potential calculated?

The gravitational potential at a point is calculated by dividing the gravitational potential energy at that point by the mass of the object. The gravitational potential energy is calculated using the formula mgh, where m is the mass of the object, g is the acceleration due to gravity, and h is the height of the object above the reference point.

3. What is the relationship between gravitational potential and gravitational force?

Gravitational potential is directly related to gravitational force. The force of gravity between two objects is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. This means that as the gravitational potential increases, the gravitational force between the two objects also increases.

4. How does gravitational potential change with distance?

Gravitational potential decreases as the distance from the source of gravity increases. This is because the force of gravity weakens with distance, and therefore the potential energy of an object also decreases as it moves away from the source of gravity.

5. Why is gravitational potential important in understanding the behavior of celestial bodies?

Gravitational potential plays a crucial role in understanding the motion and interactions of celestial bodies. It helps us understand why objects orbit around each other, and it also plays a role in determining the shape and structure of galaxies. Additionally, studying the gravitational potential of different objects can provide insights into the mass and composition of those objects.

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