# Correct statement about gravitational force, field and potential

• songoku
In summary: I note that the problem mentions "four identical objects" not "four point masses". If we assume them to be spheres of finite radius, the potential at the four centers will be finite and can be easily calculated.You are correct, the potential at the center is a minimum of the potential function ##U(z)## along the z-axis.
songoku
Homework Statement
4 identical objects are situated around a circle positioned at 12 o'clock, 3 o'clock, 6 o'clock and 9 o'clock. Which of the following statements is true?
a. The net gravitational potential at center of circle is minimum
b. The net force acting on any of the bodies is the same
c. The net gravitational field strength is non - zero at center of circle
d. The net acceleration on any body is directed towards center of circle
Relevant Equations
F = G m1 m2 / r^2

g = GM/r^2

gravitational potential = -GM/r
I think choice B is correct because when I draw the free body diagram of each object, there are three forces acting on each of them and the resultant force is towards the center.

Choice C is wrong because the net field at center is zero.

I think choice D is also correct because if the resultant force on each object is directed towards the center, the net acceleration is also towards the center.

For option A, there are four values of gravitational potential at the center (one from each object) and they are all the same but I don't know whether it is minimum or not.

My doubt:
1. I don't think there are more than one correct answer so the possibility is choice B and D are wrong, leaving choice A as the answer but I don't know why B and D are wrong (if they really are wrong)

2. How to know whether the minimum gravitational potential is at center or not?

Thanks

Here is a hint for determining which of (b) and (d) is wrong. "Force is a vector".

As for (a), can you find a point in space where the potential is less than at the center? If so, then it is not a minimum at the center.

songoku
kuruman said:
Here is a hint for determining which of (b) and (d) is wrong. "Force is a vector".
I get it. Choice B is wrong.

As for (a), can you find a point in space where the potential is less than at the center? If so, then it is not a minimum at the center.
I still don't know. If I take the point to be at one of the object, let say at 12 o'clock, can I say that the gravitational potential at that point due to the object at 12 o'clock is zero?

songoku said:
If I take the point to be at one of the object, let say at 12 o'clock, can I say that the gravitational potential at that point due to the object at 12 o'clock is zero?
What does the formula for gravitational potential say?

songoku
jbriggs444 said:
What does the formula for gravitational potential say?
If I put r = 0, gravitational potential = - infinity? What does this mean? The gravitational potential at center of an object due to that object is really small?

songoku said:
If I put r = 0, gravitational potential = - infinity? What does this mean? The gravitational potential at center of an object due to that object is really small?
The gravitational potential near a point mass gets very negative. The closer you get, the more negative it gets. The gravitational potential exactly at the position of a point mass is undefined.

The waters get deep here. See this Insights article. https://www.physicsforums.com/insights/struggles-continuum-part-1/

songoku
jbriggs444 said:
The gravitational potential near a point mass gets very negative. The closer you get, the more negative it gets. The gravitational potential exactly at the position of a point mass is undefined.
Ok so it means that the potential at center is not minimum.

Thank you very much for the help kuruman and jbriggs444

jbriggs444
songoku said:
Ok so it means that the potential at center is not minimum.
Strictly speaking, it just means that the potential at the center is not a global minimum.

One would likely need to use some calculus to demonstrate that it is not a local minimum either. For instance, find a trajectory starting at the center along which gravitational potential is strictly monotone decreasing.

jbriggs444 said:
Strictly speaking, it just means that the potential at the center is not a global minimum.

One would likely need to use some calculus to demonstrate that it is not a local minimum either. For instance, find a trajectory starting at the center along which gravitational potential is strictly monotone decreasing.
I think that it might be a funny kind (there is an element of tongue-in-cheek here) of global minimum. If the face of the clock is in the xy-plane with principal axes going through the masses, then the potential at the center is a minimum of the potential function ##U(z)## along the z-axis. I hasten to add that one can certainly find a point off the z-axis where the potential ##U(x,y,z)## is less than at the center. However, symmetry requires that there more than one points with the same potential. Therefore off-axis points cannot qualify as global minima because they are not unique. This leaves us with the choice of either accepting the center as a funny kind of global minimum because of its uniqueness or asserting that there can be no global minimum because there are off-axis points where the potential is lower than at the center.

I note that the problem mentions "four identical objects" not "four point masses". If we assume them to be spheres of finite radius, the potential at the four centers will be finite and can be easily calculated.

kuruman said:
I think that it might be a funny kind (there is an element of tongue-in-cheek here) of global minimum. If the face of the clock is in the xy-plane with principal axes going through the masses, then the potential at the center is a minimum of the potential function ##U(z)## along the z-axis. I hasten to add that one can certainly find a point off the z-axis where the potential ##U(x,y,z)## is less than at the center. However, symmetry requires that there more than one points with the same potential. Therefore off-axis points cannot qualify as global minima because they are not unique. This leaves us with the choice of either accepting the center as a funny kind of global minimum because of its uniqueness or asserting that there can be no global minimum because there are off-axis points where the potential is lower than at the center.

I note that the problem mentions "four identical objects" not "four point masses". If we assume them to be spheres of finite radius, the potential at the four centers will be finite and can be easily calculated.
I think (without carefully calculating) that the potential at the middle is in a kind of saddle point. It is a minimum among points "up" or "down" along the z axis but a maximum among points anywhere nearby in the "east"/"west"/"north"/"south" plane.

A crude argument for that proposition is to pick a direction on the plane. If one moves away from the center in that direction, the potential from two of the four objects increases and from the other two of the four objects decreases. But with a potential function that goes as ##\frac{1}{x}##, the decrease is greater than the increase.

Oh, point well taken about the potential within the four objects.

Last edited:
For whatever it's worth, here is a plot of the potential. The masses are at the corners of a square of side 2 units. The potential is displayed at z = 0.25 units and plotted from -4 to +4 units in each direction. At constant z there is no saddle point because of the fourfold symmetry.

However there is a typical saddle point if the potential is similarly calculated in a plane parallel to ##xz##.

kuruman said:
For whatever it's worth, here is a plot of the potential. The masses are at the corners of a square of side 2 units. The potential is displayed at z = 0.25 units and plotted from -4 to +4 units in each direction. At constant z there is no saddle point because of the fourfold symmetry.
I am considering a "saddle" in three dimensions rather than in a constant z slice with only two.

jbriggs444 said:
I am considering a "saddle" in three dimensions rather than in a constant z slice with only two.
OK, this is the last one. It is the best I can do with 3D equipotentials and concentrates near the origin. Note that the side of the plot box is 1/20th of the side of the square. The smallest value (blue) equipotential shows the saddle point shaped like the neck of a carafe.

## 1. What is gravitational force?

Gravitational force is a type of force that exists between any two objects with mass. It is an attractive force, meaning that it pulls objects towards each other.

## 2. What is a gravitational field?

A gravitational field is a region of space around an object where the gravitational force is present. The strength of the field depends on the mass of the object and decreases with distance from the object.

## 3. How does distance affect gravitational force?

The force of gravity decreases as the distance between two objects increases. This relationship is described by the inverse-square law, which states that the force is inversely proportional to the square of the distance between the objects.

## 4. What is gravitational potential?

Gravitational potential is a measure of the potential energy that an object has in a gravitational field. It is directly related to the distance from the object and decreases as the distance increases.

## 5. How is gravitational potential different from gravitational force?

Gravitational potential is a scalar quantity, meaning it only has magnitude, while gravitational force is a vector quantity, meaning it has both magnitude and direction. Additionally, while force describes the interaction between two objects, potential describes the energy associated with that interaction.

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