<p>A planet following this law would either:</p>
<ol>
<li><p>Follow a circular orbit.  In this case the value of R would be constant and so the magnitude of the central force would also be constant, only the direction would change.  Any central force permits circular orbits, and circular orbits have eccentricity zero.  This kind of orbit has a zero probably of occurring if the initial velocity and separation are chosen randomly</p>
</li>
<li><p>Not have a closed orbit. Here is a picture of a bit of such an orbit:</p>
</li>
</ol>
<p><a href="https://i.sstatic.net/oYJa1.png" rel="noreferrer"><img src="https://i.sstatic.net/oYJa1.png" alt="enter image description here" /></a>
(generated by code adapted by <a href="https://towardsdatascience.com/use-python-to-create-two-body-orbits-a68aed78099c" rel="noreferrer">Zack Fizell</a></p>
<p>As such an orbit isn't an ellipse, the eccentricity is not defined. Such an orbit has probably 1 of occurring if the initial conditions are random.</p>
<p>Your question cannot be answered.  The eccentricity or more generally the shape of an orbit is not completely determined by the central force. Even with Newtonian gravity, you can't determine the eccentricity only from the law of gravitation.</p>

CC BY-SA 4.0
2023-04-01T22:07:54.093Z
Note, an orbit doesn’t need to be an ellipse for eccentricity to be defined, this is standard practice when discussing orbits in general relativity.
TimRias
CC BY-SA 4.0
2023-04-02T04:38:06.327Z
**-1** "Not have a closed orbit" is an **unsupported assertion** and would be okay for Quora but not Stack Exchange. The -5/2 power law central force problem [does have an exact solution](https://en.wikipedia.org/wiki/Exact_solutions_of_classical_central-force_problems). This suggests that whether it is closed or not is going to be worked out and easy to find a supporting link. **Just saying stuff without supporting it** is not how Stack Exchange works. It is however exactly how ChatGPT works, and we've gone to great lengths to differentiate Stack Exchange answers from chatterbot answers.
uhoh
CC BY-SA 4.0
2023-04-02T05:59:05.827Z
@TimRias do you mean likke https://arxiv.org/pdf/2206.13532.pdf "The concept of eccentricity is uniquely defined in Newtonian gravity. An extension to General Relativity is not
strictly uniquely or even well defined, "
James K
CC BY-SA 4.0
2023-04-02T23:05:17.727Z
I was more thinking of the limit of a test particle orbiting a black hole, where there is a canonical widely excepted definition of eccentricity. Full blown GR is more tricky, but even then you can find a well defined coordinate invariant definition of eccentricity:  https://arxiv.org/abs/2302.11257
TimRias
CC BY-SA 4.0
2023-04-01T10:27:15.683Z
2023-04-02T05:51:47.527Z
2023-04-02T05:51:47.527Z
James K
<p>How can I calculate the eccentricity of the orbit of a planet (with mass equal to that of the Earth) around a star (with mass equal to that of the Sun) assuming that the force of gravity is given by the formula:
<span class="math-container">$$
 F= -\frac{GMm}{R^{5/2}} \hat{u}_r 
$$</span></p>
<p>Actually it would be enough for me to demonstrate that the eccentricity is between 0 and 1 to say that the planet follows a closed orbit.</p>

CC BY-SA 4.0
2023-03-30T13:35:29.333Z
You can't calculate eccentricity from just a law of gravitation, an orbital trajectory depends on velocity. An earth-sized rock could be put into a circular orbit around the sun, or an eccentric orbit, or it could just fall straight into the sun, depending only on its initial velocity and regardless of what power (>1) you put into the gravity formula. Are you asking if orbits would still be stable/closed with different powers in the denominator?
Nuclear Hoagie
CC BY-SA 4.0
2023-03-30T13:41:14.227Z
Only potentials with $\phi \propto 1/r$ or $\phi \propto r^2$ allow for closed trajectories. If you can define the eccentricity of a non-closed trajectory, then maybe you'd want to ask that first.
AtmosphericPrisonEscape
CC BY-SA 4.0
2023-03-30T14:12:35.550Z
yes, the question is whether, despite the different force of attraction, the planet could have a circular orbit
LOGIC
CC BY-SA 4.0
2023-03-30T15:55:55.663Z
A circular orbit is always _mathematically_ possible on any center-seeking attractive force whose strength is based on distance from the central body,  because the circular orbit is a balanced-on-knife-point special case that doesn't actually occur in practice.
notovny
CC BY-SA 4.0
2023-03-30T21:20:40.007Z
What are $R,r$ and $u$ ?  You need to define these for us rather than hoping we guess right.  Different people and sources use different meanings for symbols.
StephenG - Help Ukraine
CC BY-SA 4.0
2023-03-31T05:18:46.320Z
R is the distance between the center of the star and the planet, ur is the unit vector indicating that the force is directed towards the center
LOGIC
CC BY-SA 4.0
2023-03-31T07:30:10.267Z
Given the form of the central potential and an initial state (position and velocity) for the orbiting body, you can straightforwardly evaluate proxies for the orbital eccentricity, like the pericenter-to-apocenter ratio. The eccentricity itself is not meaningful, because it's a property of an ellipse, and the orbit is in general not an ellipse.
Sten
CC BY-SA 4.0
2023-03-31T09:07:19.217Z
I’m voting to close this question because the question doesn't make sense. As @AtmosphericPrisonEscape noted, the only potential functions that yield elliptical orbits are inversely proportional to distance (e.g., gravity, static electricity) and proportional to the square of distance (e.g., springs).
David Hammen
CC BY-SA 4.0
2023-04-02T06:45:50.793Z
I'm voting to **leave the question open** because I don't make up custom close reasons all the time and none of the actual Stack Exchange reasons apply. The effect of closing is *answer prevention* and I think the best way for the OP to understand the flawed premise is for good answers to be posted. JamesK's answer is a great start, I'll try to add something soon as well.
uhoh
CC BY-SA 4.0
2023-04-02T06:47:38.737Z
@DavidHammen faulty premises are not grounds for closing questions and preventing answers that could explain the problem with the premise. The question certainly makes sense, it's just that the OP needs more information on what happens in non-inverse square central force fields. Blocking answers is the worst of all possible solutions here.
uhoh
<gravity><orbital-mechanics>
Eccentricity with a law of gravity different from the classical one
LOGIC
# Eccentricity with a law of gravity different from the classical one

(Asked by: username_0 on 2023-03-30)

How can I calculate the eccentricity of the orbit of a planet (with mass equal to that of the Earth) around a star (with mass equal to that of the Sun) assuming that the force of gravity is given by the formula:
$$
F= -\frac{GMm}{R^{5/2}} \hat{u}\_r
$$

Actually it would be enough for me to demonstrate that the eccentricity is between 0 and 1 to say that the planet follows a closed orbit.

**Tags:** gravity, orbital-mechanics

> **Comments:**
> **username_1**: You can't calculate eccentricity from just a law of gravitation, an orbital trajectory depends on velocity. An earth-sized rock could be put into a circular orbit around the sun, or an eccentric orbit, or it could just fall straight into the sun, depending only on its initial velocity and regardless of what power (>1) you put into the gravity formula. Are you asking if orbits would still be stable/closed with different powers in the denominator?  
> **username_2**: Only potentials with $\phi \propto 1/r$ or $\phi \propto r^2$ allow for closed trajectories. If you can define the eccentricity of a non-closed trajectory, then maybe you'd want to ask that first.  
> **username_0**: yes, the question is whether, despite the different force of attraction, the planet could have a circular orbit  
> **username_3**: A circular orbit is always _mathematically_ possible on any center-seeking attractive force whose strength is based on distance from the central body,  because the circular orbit is a balanced-on-knife-point special case that doesn't actually occur in practice.  
> **username_4**: What are $R,r$ and $u$ ?  You need to define these for us rather than hoping we guess right.  Different people and sources use different meanings for symbols.  
> **username_0**: R is the distance between the center of the star and the planet, ur is the unit vector indicating that the force is directed towards the center  
> **username_5**: Given the form of the central potential and an initial state (position and velocity) for the orbiting body, you can straightforwardly evaluate proxies for the orbital eccentricity, like the pericenter-to-apocenter ratio. The eccentricity itself is not meaningful, because it's a property of an ellipse, and the orbit is in general not an ellipse.  
> **username_6**: I’m voting to close this question because the question doesn't make sense. As @username_2 noted, the only potential functions that yield elliptical orbits are inversely proportional to distance (e.g., gravity, static electricity) and proportional to the square of distance (e.g., springs).  
> **username_7**: I'm voting to **leave the question open** because I don't make up custom close reasons all the time and none of the actual Stack Exchange reasons apply. The effect of closing is *answer prevention* and I think the best way for the OP to understand the flawed premise is for good answers to be posted. JamesK's answer is a great start, I'll try to add something soon as well.  
> **username_7**: @DavidHammen faulty premises are not grounds for closing questions and preventing answers that could explain the problem with the premise. The question certainly makes sense, it's just that the OP needs more information on what happens in non-inverse square central force fields. Blocking answers is the worst of all possible solutions here.  

---

## 1 answers

### Answer by username_8 on 2023-04-01. Score: 5

A planet following this law would either:

1. Follow a circular orbit. In this case the value of R would be constant and so the magnitude of the central force would also be constant, only the direction would change. Any central force permits circular orbits, and circular orbits have eccentricity zero. This kind of orbit has a zero probably of occurring if the initial velocity and separation are chosen randomly
2. Not have a closed orbit. Here is a picture of a bit of such an orbit:

[![enter image description here](https://i.sstatic.net/oYJa1.png)](https://i.sstatic.net/oYJa1.png)
(generated by code adapted by [Zack Fizell](https://towardsdatascience.com/use-python-to-create-two-body-orbits-a68aed78099c)

As such an orbit isn't an ellipse, the eccentricity is not defined. Such an orbit has probably 1 of occurring if the initial conditions are random.

Your question cannot be answered. The eccentricity or more generally the shape of an orbit is not completely determined by the central force. Even with Newtonian gravity, you can't determine the eccentricity only from the law of gravitation.

> **Comments:**
> **username_9**: Note, an orbit doesn’t need to be an ellipse for eccentricity to be defined, this is standard practice when discussing orbits in general relativity.  
> **username_7**: **-1** "Not have a closed orbit" is an **unsupported assertion** and would be okay for Quora but not Stack Exchange. The -5/2 power law central force problem [does have an exact solution](https://en.wikipedia.org/wiki/Exact_solutions_of_classical_central-force_problems). This suggests that whether it is closed or not is going to be worked out and easy to find a supporting link. **Just saying stuff without supporting it** is not how Stack Exchange works. It is however exactly how ChatGPT works, and we've gone to great lengths to differentiate Stack Exchange answers from chatterbot answers.  
> **username_8**: @username_9 do you mean likke https://arxiv.org/pdf/2206.13532.pdf "The concept of eccentricity is uniquely defined in Newtonian gravity. An extension to General Relativity is not strictly uniquely or even well defined, "  
> **username_9**: I was more thinking of the limit of a test particle orbiting a black hole, where there is a canonical widely excepted definition of eccentricity. Full blown GR is more tricky, but even then you can find a well defined coordinate invariant definition of eccentricity:  https://arxiv.org/abs/2302.11257  

---

