Gravitational Potential: How to know where the max GP is located?

[tiffany chan]

By using the equation for the Gravitational Potential -GM/R. It is understand that the max Gravitation Potential would be at infinity point.
- G(100M)/22R-(-GM/22R) would be the maximum Gravitational Potential... I guess...

Then what would be the next step to find the distance?

 

[Orodruin]

By using the equation for the Gravitational Potential -GM/R. It is understand that the max Gravitation Potential would be at infinity point.

Yes. This is a quite clearly an error in the formulation of the problem. The intended problem is probably to find the point of maximal gravitational potential on the line between the centers of the moon and the planet.

- G(100M)/22R-(-GM/22R) would be the maximum Gravitational Potential... I guess...

Sorry, but it is very unclear what you mean by this.

 

[tiffany chan]

Yes. This is a quite clearly an error in the formulation of the problem. The intended problem is probably to find the point of maximal gravitational potential on the line between the centers of the moon and the planet.Sorry, but it is very unclear what you mean by this.

Thanks for your reply.
So do you mean -GM/R is not the right equation to be used to solve this question?

 

[jbriggs444]

Thanks for your reply.
So do you mean -GM/R is not the right equation to be used to solve this question?

-GM/R is a formula, not an equation. If G is Newton's universal gravitational constant, R is the distance of a test point from the center of the moon and M is the mass of the moon then it is a formula giving the moon's gravitational potential at that test point. So yes, it is a useful part of a correct approach to solve the problem.

You had proposed

- G(100M)/22R-(-GM/22R)

Like @Orodruin, I have no idea what this formula is supposed to denote. Taken at face value, it is the difference between the potential of a test point at the planet in the gravitational field of the moon and the potential of a test point at the moon in the gravitational field of the planet. Physically meaningless.

You should be trying to find a formula for the total gravitational potential at a given test point in the combined fields of planet plus moon. The position of the test point will be present in such a formula.

 

[tiffany chan]

1. Assume the maximum distance happened at a distance x from the centre of the planet.
2. Total Gravitational Potential would be: (-100GM/x) - (-GM/1-x) = (what should be the max GP equal to)??

 

[jbriggs444]

View attachment 240596

1. Assume the maximum distance gravitational potential happened at a distance x from the centre of the planet.
2. Total Gravitational Potential would be: (-100GM/x) - (-GM/1-x) = (what should be the max GP equal to)??

You write "1-x" to represent the distance between the moon and the test point. But the distance between moon and planet is not 1.

[Also, be aware of parentheses. by "-GM/1-x" you presumably mean $-\frac{GM}{1-x}$ but without proper parentheses, what you have written really means $-\frac{GM}{1}-x$]

 

[tiffany chan]

You write "1-x" to represent the distance between the moon and the test point. But the distance between moon and planet is not 1.

[Also, be aware of parentheses. by "-GM/1-x" you presumably mean $-\frac{GM}{1-x}$ but without proper parentheses, what you have written really means $-\frac{GM}{1}-x$]

I use x because i considered it as ratio.
So correct way would be
(-100GM/X) + (-GM/(22R-x)) = total GP?
But I’m confuse of what value to be put for the total GP in the equation

 

[jbriggs444]

Look at the question. You are not asked for total GP. You are asked for the position where total GP is maximized.

You are trying to find the place where a function is maximized of a function. If you have been exposed to calculus, there is a good way to use that. Otherwise, you could think about what has to happen to gravitational force at the point where potential is maximized.

 

[tiffany chan]

Look at the question. You are not asked for total GP. You are asked for the position where total GP is maximized.

You are trying to find the place where a function is maximized of a function. If you have been exposed to calculus, there is a good way to use that. Otherwise, you could think about what has to happen to gravitational force at the point where potential is maximized.

When potential is maximised, the gravitational force should be zero?

 

[jbriggs444]

When potential is maximised, the gravitational force should be zero?

Correct.

 

[tiffany chan]

Correct.

If the force = 0 when the GP is maximised. That means, gravitational field strength at that point = 0 as well.
So, the equation will be: 100GM/x^2 - GM/(22R-x)^2 = 0, then solve for x?

 

[haruspex]

If the force = 0 when the GP is maximised. That means, gravitational field strength at that point = 0 as well.
So, the equation will be: 100GM/x^2 - GM/(22R-x)^2 = 0, then solve for x?

Yes.

 

[tiffany chan]

Got it! Its 20R!
Thank you very much~

 

[epenguin]

If the force = 0 when the GP is maximised. That means, gravitational field strength at that point = 0 as well.
So, the equation will be: 100GM/x^2 - GM/(22R-x)^2 = 0, then solve for x?

Isn't this a long way around, negating the usefulness of the potential concept?

 

[Orodruin]

Isn't this a long way around, negating the usefulness of the potential concept?

Are you suggesting he finds the maximum by differentiating the potential and setting the derivative to zero? That is equivalent since the force is (minus) the derivative.