Exploring Force & Potential Energy of a Donut-Shaped Planet

In summary: The potential energy of the central element of mass within the object would be...again, arbitrary. You could choose any point within the object as the reference point for potential energy.
  • #1
JLT
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TL;DR Summary
What happens when r is zero?
So, let's say you have a donut - shaped planet, so a second object can move right on top of the center of mass of the first object. Does force go to infinity? How about potential energy?

Or, just take one object, divide it into elements, what happens to the central element of mass within the object?
 
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  • #2
JLT said:
Summary: What happens when r is zero?

So, let's say you have a donut - shaped planet, so a second object can move right on top of the center of mass of the first object. Does force go to infinity?
The potential energy formula: ##PE=-\frac{k}{r}## holds for a point mass or (using Newton's spherical shell theorem) for the exterior of a spherically symmetric mass distribution. It does not hold for the interior of an arbitrary object. It does not even hold for the exterior of an arbitrary object.

It does hold approximately when one is far away from a small object. However, that does not help when trying to ask about a point inside the object.

If you want to know the potential at the center of an arbitrary mass distribution, you can add up (i.e. integrate) the potentials from all the component pieces of that mass distribution. If you do that for a doughnut-shaped distribution, the total will be finite.

Simpler than the case of a toroid is the case of a sphere of uniform density. In the region outside the sphere, the force formula for a unit mass takes the form ##F=\frac{GM}{r^2}##. Integrating yields the potential formula: ##F=-\frac{GM}{r}## [with the zero point at infinity]

In the interior, the relevant mass is the portion inside the radius r. The mass of the outside portion is irrelevant (by Newton's shell theorem). Let R be the object's radius. Then ##F=\frac{GMr^3}{R^3r^2}=\frac{GMr}{R^2}##. Integrating this yields ##\frac{GMr^2}{2R^2}## [with the zero point at the center].

The two formulas will not match up at the object's surface. To get them to match, we could add ##\frac{GM}{R}## to the exterior formula and subtract ##\frac{GM}{2}## from the interior formula so that both would be zero at the surface. Once the formulas are fitted to a single reference level, the delta PE between the center point and infinity can be extracted. It is clearly finite.

[I am not 100% confident in the formulas above]
 
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Well, let's see. The equation of motion for Newton's gravitational potential, ##\phi## reads
$$\Delta \phi=4 \pi G \rho.$$
Now consider a homogeneous sphere, where
$$\rho(r)=\begin{cases} M/V &\text{for} r<R \\
0 & \text{for} r \geq R, \end{cases}$$
where ##V=4 \pi R^3/3## is the volume of the sphere.

For symmetry reasons we can assume that also ##\phi=\phi(r)##. Then the equation simplifies to
$$\frac{1}{r} [r \phi(r)]''=4 \pi G M/V \quad \text{for} \quad r<R$$
This gives
$$\phi(r)=\frac{2 \pi G M r^2}{3V}+C_1+\frac{C_2}{r}.$$
Since there's no singularity at ##r=0## we must have ##C_2=0##.

For ##r>R## we have
$$\frac{1}{2} [r \phi(r)]''=0 \; \Rightarrow \; \phi(r)=C_1'+\frac{C_2'}{r}.$$
Since we want ##\phi(r) \rightarrow 0## for ##r \rightarrow \infty## we have ##C_1'=0##.

To get ##C_2'## we integrate the equation of motion over a sphere ##K## of radius ##a>R##. Then we find, using Gauss's Integral Theorem
$$4 \pi G M=\int_{\partial K} \mathrm{d}^2 \vec{f} \cdot \vec{\nabla} \phi.$$
Since
$$\vec{\nabla} \phi=-\frac{C_2'}{r^3} \vec{x},$$
the surface integral gets ##-4 \pi C_2'## and thus ##C_2'=-G M##.

Finally the potential must be continuous at ##r=R## leading to ##C_1=-3GM/(2R)## and thus
$$\phi(r)=\begin{cases} \frac{G M}{2 R^3} (r^2-3R^2) &\text{for} \quad r<R, \\
-\frac{GM}{r} & \text{for} \quad r \geq R.\end{cases}$$
 
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  • #4
JLT said:
Summary: What happens when r is zero?

So, let's say you have a donut - shaped planet, so a second object can move right on top of the center of mass of the first object. Does force go to infinity? How about potential energy?

Or, just take one object, divide it into elements, what happens to the central element of mass within the object?
You can easily show, invoking a symmetry argument, that the force at the center of a donut shaped planet will be zero, while the potential energy will be ##\int\frac{Gdm}{r}=\frac{G}{r}\int dm=G\frac{M}{r}## that is the sum of potentials of all the mass elements dm which are at constant radius r from the center.

Actually the above hold for a mass ring of uniform mass density but you get the idea and you can see that the force and the potential energy will be finite at the center of a donut shaped planet or any object whose center of mass lies outside of the mass volume.
 
  • #5
JLT said:
What happens when r is zero?

The expression ##F=G\frac{Mm}{r^2}## is not valid when ##r=0##. In other words, you can't use it to find a value for ##F## when ##r=0##.

That expression for ##F## makes use of the particle model. Such a particle has mass but no size. When we use that model the size of the object has to be small compared to ##r##, otherwise it won't produce valid results. Of course, if you make ##r## small enough you no longer satisfy the particle model approximation.

So, let's say you have a donut - shaped planet, so a second object can move right on top of the center of mass of the first object.
Do you mean that the object is located at the center of mass of the donut? The force is zero and the potential energy is arbitrary, but the using the usual convention of taking the potential energy to be zero at infinity, the potential energy at the center of the donut would not be zero, it would equal the work done to bring the particle from infinity to the donut center. You can model the donut as a collection of differential elements, and treat each differential element as a point particle.

Or, just take one object, divide it into elements, what happens to the central element of mass within the object?

The force on a differential element located at the center of mass would be zero.
 
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  • #6
JLT said:
Summary: What happens when r is zero?

So, let's say you have a donut - shaped planet, so a second object can move right on top of the center of mass of the first object. Does force go to infinity? How about potential energy?

Or, just take one object, divide it into elements, what happens to the central element of mass within the object?
https://www.mathpages.com/home/kmath402/kmath402.htm
 
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  • #7
Interesting, thank you for the replies!
 
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Related to Exploring Force & Potential Energy of a Donut-Shaped Planet

1. What is force and potential energy?

Force is a physical quantity that describes the interaction between objects and can cause a change in motion. Potential energy is the energy that an object possesses due to its position or configuration.

2. How does the shape of a planet affect its force and potential energy?

The shape of a planet can affect its force and potential energy as it determines the distribution of mass and the distance between objects. In a donut-shaped planet, the force of gravity and potential energy would be different from a spherical planet due to the varying distances between objects.

3. What factors contribute to the force and potential energy of a donut-shaped planet?

The force and potential energy of a donut-shaped planet are influenced by its mass, the distance between objects, and the shape of the planet. Additionally, the rotation and gravitational pull of the planet can also play a role in determining its force and potential energy.

4. How does the force and potential energy of a donut-shaped planet affect objects on its surface?

Objects on the surface of a donut-shaped planet would experience a different force of gravity and potential energy compared to objects on a spherical planet. This could impact the motion and stability of objects on the surface.

5. Are there any real-life examples of donut-shaped planets and their force and potential energy?

While there are no known donut-shaped planets in our solar system, there are exoplanets (planets outside of our solar system) that have been discovered with unusual shapes, such as the cigar-shaped planet WASP-12b. The force and potential energy of these planets would be affected by their unique shapes and could provide valuable insights into the dynamics of different planetary systems.

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