[SR] - Test Particle inside the Sun's Gravitational Field - Part 3

[Athenian]

Homework Statement

[Question Context: Consider the motion of a test particle of (constant) mass $m$ inside the gravitational field produced by the sun in the context of special relativity.

Consider the equations of motion for the test particle, which can be written as $$\frac{d(m\gamma c)}{dt} = \frac{\vec{v}}{c} \cdot \vec{F},$$

OR

$$\frac{d(m\gamma \vec{v})}{dt} = \vec{F},$$

where $\vec{v}$ is the speed of the test particle, $c$ is the (constant) speed of light, and by definition, $$\gamma \equiv \frac{1}{\sqrt{1- \frac{\vec{v}^2}{c^2}}} .$$

In addition, the gravitational force is given by $$\vec{F} \equiv -\frac{GMm}{r^2} \hat{e}_r$$

where $\hat{e}_r$ is the unit vector in the direction between the Sun (of mass M) and the test particle (of mass $m$).]
------------------------------------

Three-Part Question:
1. On studying the equations of motion - $\frac{d(m\gamma \vec{v})}{dt} = \vec{F}$ - in particular their component in the $\hat{e}_\theta$ direction (i.e. the direction perpendicular, at any time, to $\hat{e}_r$ and to the axis perpendicular to the plane of the orbit), find the special-relativistic analog of Kepler’s second law – the law of equal areas. What is the physical conserved quantity implied by this law?

2. On using the previous relativistic analog of Kepler’s second law, and assuming $\dot{\theta} \neq 0$ along the motion of the particle, write down the differential equation for the trajectory, that is instead of solving for $r=r(t)$, change the dependent variable as $r \equiv \frac{1}{u}$, and also change the independent variable from $t$ to $\theta$. In other words, find a differential equation for the variable $u$ as a function of $\theta$, i.e. $\frac{1}{r}=u=u(t(\theta))=u(\theta)$.

3. Solve the previously found differential equation for the trajectory, i.e. find the solution for $u(\theta)$ (for all $\theta$). What kind of trajectories do you find?

Relevant Equations

Refer to Attempted Solution Below $\longrightarrow$

Below, I have already solved - I assume - correctly for question 1. Question 2, I am nearing to what I believe is the solution. Question 3, I simply have no idea where I should begin considering that it is interconnected with question 2.

With that said, below is the lengthy and somewhat tedious mathematical calculation for the solution(s) to the problems.

Starting with Question 1:
Using the equation of motion as the problem instructed me to do so ...

$$\frac{d}{dt} (m \gamma \vec{v}) = -\frac{GMm}{r^2} \hat{e}_r$$

Canceling $m$ on both sides, I get:
$$\frac{d}{dt} (\gamma \vec{v}) = -\frac{GM}{r^2} \hat{e}_r$$

Continuing on:
$$\frac{d}{dt} (\gamma \vec{v}) = \frac{d \gamma}{dt} \vec{v} + \gamma \frac{d \vec{v}}{dt}$$
$$\implies \frac{d \gamma}{dt} (\dot{r} \hat{e}_r + r \dot{\theta} \hat{e}_\theta) + \gamma [(\ddot{r} - r \dot{\theta}^2) \hat{e}_r + (2 \dot{r} \dot{\theta} + r \ddot{\theta}) \hat{e}_\theta]$$
$$\implies \bigg( \frac{d \gamma}{dt} \dot{r} + \gamma(\ddot{r} - r \dot{\theta}^2) \bigg) \hat{e}_r + \bigg( \frac{d \gamma}{dt} r \dot{\theta} + \gamma (2 \dot{r} \dot{\theta} + r \ddot{\theta}) \bigg) \hat{e}_\theta$$

Thus, I can say:
$$-\frac{GM}{r^2} \hat{e}_r = \bigg( \frac{d \gamma}{dt} \dot{r} + \gamma(\ddot{r} - r \dot{\theta}^2) \bigg) \hat{e}_r + \bigg( \frac{d \gamma}{dt} r \dot{\theta} + \gamma (2 \dot{r} \dot{\theta} + r \ddot{\theta}) \bigg) \hat{e}_\theta$$

Multiplying $\hat{e}_\theta$ on both sides of the equation and utilizing the property $\hat{e}_r \cdot \hat{e}_r = 1$ and $\hat{e}_\theta \cdot \hat{e}_\theta = 0$ to my advantage, I can get:

$$0 = 0 +\bigg( \frac{d \gamma}{dt} r \dot{\theta} + \gamma (2 \dot{r} \dot{\theta} + r \ddot{\theta}) \bigg)$$
$$\implies \bigg( \frac{d \gamma}{dt} r \dot{\theta} + \gamma (2 \dot{r} \dot{\theta} + r \ddot{\theta}) \bigg) = 0$$

Next, multiply $r$ on both sides:
$$\implies \bigg( \frac{d \gamma}{dt} r^2 \dot{\theta} + \gamma (2r \dot{r} \dot{\theta} + r^2 \ddot{\theta}) \bigg) = 0$$

Since $2r \dot{r} \dot{\theta} + r^2 \ddot{\theta} = \frac{d}{dt} (r^2 \dot{\theta})$, I can get the following:
$$\implies \frac{d\gamma}{dt} r^2 \dot{\theta} + \gamma \frac{d}{dt} (r^2 \dot{\theta}) = 0$$

Integrating the equation, I could get:
$$\frac{d}{dt} (\gamma r^2 \dot{\theta}) = 0$$

Or simply,
$$\gamma r^2 \dot{\theta} = constant$$

Thus, through the above solution, I was able to find for the special-relativistic analog of Kepler’s second law – the law of equal areas - as well as show what is the physical conserved quantity implied by this law.

In this case, I would make that $constant = \ell$ to solve for future questions.
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Now, onto Question 2 (i.e. where my confusion begins to settle in):
To solve for question 2, I begin by setting up a few equations.

$$- \frac{GM}{r^2} = \frac{d \gamma}{dt} \dot{r} + \gamma (\ddot{r} - r\dot{\theta}^2)$$
$$\implies \dot{\gamma} \dot{r} + \gamma(\ddot{r} - r \dot{\theta}^2) \longrightarrow [1]$$

Note: Bracket numbers (e.g. [1]) stand for the equation number (e.g. equation 1).

$$\dot{r} = \frac{dr}{dt}$$
$$\implies \frac{d\theta}{dt} \frac{dr}{d\theta}$$
$$\implies - \dot{r} \dot{\theta} \frac{d}{d\theta} \bigg( \frac{1}{r} \bigg)$$

Using the constant $\ell$, I got in the previous question:
$$- \frac{\ell}{\gamma} \frac{d}{d\theta} \bigg( \frac{1}{r} \bigg)$$
$$ \therefore \gamma \dot{r} = -\ell \frac{d}{d\theta} \bigg( \frac{1}{r} \bigg) \longrightarrow [2]$$

To get the third equation, I will differentiate the above equation with respect to $t$ (i.e. $\frac{d}{dt}$):
$$\dot{\gamma}\dot{r} + \gamma \ddot{r} = -\ell \frac{d}{dt} \bigg( \frac{d}{d\theta} \bigg( \frac{1}{r} \bigg) \bigg)$$

Now, continuing on where I left off:
$$\implies -\ell \frac{d\theta}{dt} \frac{d}{d\theta} \bigg( \frac{d}{d\theta} \bigg( \frac{1}{r} \bigg) \bigg)$$
$$\implies -\ell \dot{\theta} \frac{d^2}{d\theta^2} \bigg( \frac{1}{r} \bigg)$$
$$\implies - \frac{\ell^2}{\gamma r^2} \frac{d^2}{d\theta^2} \bigg( \frac{1}{r} \bigg) \longrightarrow [3]$$

And, finally, for the last set of equations:

$$\frac{\ell^2}{\gamma r^3} = \gamma r \dot{\theta}^2 \longrightarrow [4]$$

Now, using $[3]$ and $[4]$, $[1]$ becomes the following:
$$\frac{GM}{r^2} = \frac{\ell^2}{\gamma r^2} \frac{d^2}{d\theta^2} \bigg( \frac{1}{r} \bigg) + \frac{\ell^2}{\gamma r^3}$$

Since, according to the question, $u \equiv \frac{1}{r}$, I could get the following:

$$GMu^2 = \frac{\ell^2}{\gamma}u^2 \frac{d^2}{d\theta^2}u + \frac{\ell^2}{\gamma}u^3$$
$$\implies GM = \frac{\ell^2}{\gamma} \frac{d^2}{d\theta^2}u + \frac{\ell^2}{\gamma}u$$
$$\implies GM\gamma = \ell^2 \bigg( \frac{d^2}{d\theta^2} u +u \bigg)$$
$$GM \frac{\ell}{r^2 \dot{\theta}} = \ell^2 \bigg( \frac{d^2}{d\theta^2} u +u \bigg) \longrightarrow (\because \gamma r^2 \dot{\theta} = constant = \ell)$$
$$\therefore GMu^2 = \ell \dot{\theta} \bigg( \frac{d^2}{d\theta^2} u +u \bigg)$$
$$u = \frac{GM\gamma}{\ell^2} (1+ e \: cos\theta)$$

However, Question 2 states that I need to "find a differential equation for the variable $u$ as a function of $\theta$, i.e. $\frac{1}{r}=u=u(t(\theta))=u(\theta)$". But, despite all that mathematical calculation, I am stuck with $GMu^2 = \ell \dot{\theta} \bigg( \frac{d^2}{d\theta^2} u +u \bigg)$. Or, to be more specific, I am stuck with $\dot{\theta}$ rather than the desired $\theta$. I tried to solve the issue with the below equation - that is $u = \frac{GM\gamma}{\ell^2} (1+ e \: cos\theta)$. However, I'm not sure how well that works. Any suggestions toward resolving this issue would be very much appreciated. Thank you!

Beyond that, I have no idea how to proceed with question 3. Perhaps I might get an idea once I solve question 2. But, I find that unlikely unless I have some mild assistance.

Once again, thank you for reading through this wall of text - or, more accurately, equations - as well as the offered kind assistance!

 

[Orodruin]

As with your previous question, it is impossible to make sense of this question because gravity cannot be incorporated like this into SR. The only way of actually making sense out of it is replacing gravity by electromagnetism. It is therefore also impossible to say that your work is the ”correct kind of wrong”.

 

[PeroK]

$$GM\gamma = \ell^2 \bigg( \frac{d^2}{d\theta^2} u +u \bigg)$$

This looks okay until here. Now, you have $\gamma$ as a function of $u = 1/r$. If you substitute that you get an equation in terms of $u$ and $\theta$.

 

[Athenian]

This looks okay until here. Now, you have $\gamma$ as a function of $u = 1/r$. If you substitute that you get an equation in terms of $u$ and $\theta$.

Here's a quote a classmate offered that I thought would help with the solution process:
"Kepler's 2nd law states that the radius vector of a planet $(r\vec{e}_r)$ sweeps equal area in unit time. Thus, one could write $\frac{d}{dt} (r^2 \dot{\theta}) = 0$, meaning that
$$A=r^2 \dot{\theta}$$
"

With the quote, I thought that since the $\gamma$ in my calculation process is equivalent to $\frac{1}{A}$, then I can conclude that $\gamma$ is a constant as well here.

With that in mind, continuing where @PeroK told me where the identified problem is, I took
$$GM\gamma = \ell^2 \bigg( \frac{d^2}{d\theta^2}u + u \bigg)$$

$$\implies \frac{GM\gamma}{\ell^2} = \bigg( \frac{d^2}{d\theta^2}u + u \bigg)$$

Here, I purposely brought all the constant to the left side.
And, somehow, I should be able to derive the below solution:
$$u=A cos(\theta - \theta_0) - \frac{GM\gamma}{\ell^2}$$

So, what does everybody think about this? Is this even correct? Any help will be much appreciated. At this point, I feel quite lost ...

Of course, if I have interpreted PeroK's statement incorrectly, please correct what I have done wrong.

Thank you very much!

 

[PeroK]

With the quote, I thought that since the $\gamma$ in my calculation process is equivalent to $\frac{1}{A}$, then I can conclude that $\gamma$ is a constant as well here.

If $\gamma$ is constant, then we have exactly the Newtonian equations with $\gamma = 1$.

You already have $\gamma$ as a function of $1/r$. That was the first thing you calculated.

Everything you have so far is practically a carbon copy of the Newtonian case, with a single $\gamma$ factor hanging on in there. That said, if you are working on this variation, then you can't suddenly decide that $\gamma$ is a constant.

 

[PeroK]

Here's a quote a classmate offered that I thought would help with the solution process:
"Kepler's 2nd law states that the radius vector of a planet (r→er)(re→r) sweeps equal area in unit time. Thus, one could write ddt(r2˙θ)=0ddt(r2θ˙)=0, meaning that

A=r2˙θA=r2θ˙​

"

Kepler's second law is technically the conservation of angular momentum. Kepler didn't know about that so "equal area in equal time" is what he called it. You have derived the equivalent conservation law in this case:

$\frac{d}{dt}(\gamma r^2 \dot \theta) = 0$

To put this problem in some sort of context. You could assume a simplistic modification of Newtonian gravity and see what comes out. In this case, everything seems to follow Newtonian lines. The end result appears to be that you get bound orbits that vary slightly from an ellipse - the factor is very small for planetary orbits round the Sun.

Anyway ... you appear to have done all the hard work only to get stuck at a relatively easy step! Hmm?

 

[Athenian]

If $\gamma$ is constant, then we have exactly the Newtonian equations with $\gamma = 1$.

You already have $\gamma$ as a function of $1/r$. That was the first thing you calculated.

Everything you have so far is practically a carbon copy of the Newtonian case, with a single $\gamma$ factor hanging on in there. That said, if you are working on this variation, then you can't suddenly decide that $\gamma$ is a constant.

I see. In that case, if I can't just decide that $\gamma$ is a constant even with the "proof", should I continue by doing the following solution?

Continuing where I left off:
$$GM\gamma = \ell^2 \bigg( \frac{d^2}{d\theta^2}u + u \bigg)$$
$$\implies \frac{GM\gamma}{\ell^2} = \bigg( \frac{d^2}{d\theta^2}u + u \bigg)$$

Noting that the polar form of $\gamma$ is $\frac{GM}{rc^2} + d$, where $d$ is a constant (I'll write how I got this polar form in an old thread I wrote), I could get the following:

$$\implies \frac{GM}{\ell^2} \bigg(\frac{GM}{rc^2} + d \bigg) = \ddot{u} + u$$
$$\implies \frac{GM}{\ell^2} \bigg(\frac{GM}{rc^2} + d \bigg) = \ddot{u} + u$$

where, according to the question, $u \equiv \frac{1}{r}$,
$$\implies \frac{GM^2}{\ell^2 c^2} u + \frac{GM}{\ell^2}d = \ddot{u} + u$$
$$\implies \ddot{u} + u \bigg( 1 - \frac{G^2 M^2}{\ell^2 c^2} \bigg) - \frac{GMd}{\ell^2} = 0$$

However, with this solution, I am unable to solve for the above differential as Question 3 requested. Beyond that, I am not sure whether $\bigg( 1 - \frac{G^2 M^2}{\ell^2 c^2} \bigg)$ is positive or negative.

Unless there is a clever substitution or method to solve for the above differential, I have no idea how to proceed with question 3 ...

Do you perhaps have any ideas on how to proceed? Of course, if I managed to get my calculation wrong again, please let me know. Thank you very much for your patience and assistance!

 

[Athenian]

Thank you for your explanation!

Anyway ... you appear to have done all the hard work only to get stuck at a relatively easy step! Hmm?

Haha. Perhaps so. However, this last part has been driving I and the study group (i.e. my classmates) crazy for a few hours. This is especially the case when we thought we found the answer only to get stumped on question 3.

If possible, do you mind giving us a "starting" equation to work with following $GM\gamma = \ell^2 \bigg( \frac{d^2}{d\theta^2} u +u \bigg)$ provided that my above solution did not work.

At this point, I am honestly having little to no idea on how to proceed, even with the below hint:

This looks okay until here. Now, you have $\gamma$ as a function of $u=\frac{1}{r}$. If you substitute that you get an equation in terms of $u$ and $\theta$.

Any extra "push" to move us along the right direction would be greatly appreciated. Even if we are stuck on a relatively "easy" part of the question, I think we're missing something in "plain sight" and thinking too complicatedly. Of course, this is assuming that my above solution does not work.

Thank you!

 

[Athenian]

That looks like the right equation. It's of the form:

¨u+u(1−α)−β=0u¨+u(1−α)−β=0​

It's not that much harder to solve than the Newtonian equivalent.

First, you could put some numbers in for the Earth's orbit round the Sun. You'll see that, as expected, αα is very small and the other factor d≈1d≈1. I.e. it's very close to the Newtonian case.

To get something different you would need rr to be very small. Much less than the Sun's radius. Eventually, if you make rr small enough the sign flips and you get an exponential inward spiral.

Okay, I posted my last message a tad bit too soon. Let me try solving for question 3 then and I'll follow you with my solution afterward. Thank you!

 

[PeroK]

⟹¨u+u(1−G2M2ℓ2c2)−GMdℓ2=0⟹u¨+u(1−G2M2ℓ2c2)−GMdℓ2=0​

That looks like the right equation. It's of the form:

¨u+u(1−α)−GMdℓ2=0u¨+u(1−α)−GMdℓ2=0​

It's not that much harder to solve than the Newtonian equivalent.

First, you could put some numbers in for the Earth's orbit round the Sun. You'll see that, as expected, αα is very small and the other factor d≈1d≈1. I.e. it's very close to the Newtonian case.

PS It seems to me that there is an orbital precession there. That's interesting!

To get something different you would need rr to be very small. Much less than the Sun's radius. Eventually, if you make rr small enough the sign flips and you get an exponential inward spiral.

 

[Athenian]

While I understand that I have managed to successfully find the equational form of $\ddot{u} + u(1-\alpha) - \frac{GMd}{\ell^2}$, isn't the $\ddot{u}$ here still a function of $t$ rather than $\theta$? Since that's the case, wouldn't I need to find what $\ddot{u}$ is in terms of $\theta$ first before beginning to solve for question 3 as question 2 asks the student to find a differential equation for the variable $u$ as a function of $\theta$?

Once I have managed to make the equation $u$ a function of $\theta$ completely, would that make the solution process for question 3 easier?
What are your thoughts regarding this?
Thank you!

 

[PeroK]

While I understand that I have managed to successfully find the equational form of $\ddot{u} + u(1-\alpha) - \frac{GMd}{\ell^2}$, isn't the $\ddot{u}$ here still a function of $t$ rather than $\theta$? Since that's the case, wouldn't I need to find what $\ddot{u}$ is in terms of $\theta$ first before beginning to solve for question 3 as question 2 asks the student to find a differential equation for the variable $u$ as a function of $\theta$?

Once I have managed to make the equation $u$ a function of $\theta$ completely, would that make the solution process for question 3 easier?
What are your thoughts regarding this?
Thank you!

Where you have $\ddot u$ that should be $\frac{d^2u}{d\theta^2}$.

 

[Athenian]

Where you have $\ddot u$ that should be $\frac{d^2u}{d\theta^2}$.

You're right. My brain got short-circuited there. Regardless, thanks for the confirmation!

To get something different you would need $r$ to be very small. Much less than the Sun's radius. Eventually, if you make $r$ small enough the sign flips and you get an exponential inward spiral.

As for this, let me give this a bit more thought. I'll see how I do after I get some rest. Thank you!

 

[Athenian]

@PeroK , after much back and forth discussion with my team, this is what we came up with:

Method 1:
$$\frac{d^2u}{d\theta^2} + u \bigg( 1- \frac{G^2 M^2}{\ell^2 c^2} \bigg) - \frac{GMd}{\ell^2} = 0$$

With the above equation in mind, I tried to solve this differential equation by simplifying first to the below equation:
$$\ddot{u} (\theta) + \alpha u(\theta) - \beta = 0$$
Continuing, I made the following substitution to help make the solution process easier.
$$y=\alpha u(\theta) - \beta$$

With the above solution in mind, I could take a step further and try to find $\ddot{u}(\theta)$ in terms of $y(\theta)$:
Thus,
$$\dot{y} = \alpha \dot{u}$$
$$\implies \ddot{y} = \alpha \ddot{u}$$
$$\implies \ddot{u} = \frac{\ddot{y}}{\alpha}$$

Bringing everything back together:
$$\ddot{u} (\theta) + \alpha u(\theta) - \beta = \frac{\ddot{y}}{\alpha} +y$$

Which also means that:
$$\frac{\ddot{y}}{\alpha} +y \Longrightarrow \ddot{y}(\theta) + \alpha y(\theta) = 0$$

Using the ever-handy tool called Wolfram Alpha, I got the below solution for the differential equation:
$$y(\theta) = c_2 sin(\sqrt{\alpha}\theta) + c_1 cos(\sqrt{\alpha}\theta)$$

Noting that $y(\theta) = \alpha u(\theta) - \beta$
$$\implies \alpha u(\theta) - \beta = c_2 sin(\sqrt{\alpha}\theta) + c_1 cos(\sqrt{\alpha}\theta)$$
$$\implies \alpha u(\theta) = c_2 sin(\sqrt{\alpha}\theta) + c_1 cos(\sqrt{\alpha}\theta) + \beta$$
$$\implies u(\theta) = \frac{c_2 sin(\sqrt{\alpha}\theta) + c_1 cos(\sqrt{\alpha}\theta) + \beta}{\alpha}$$

Substituting (or reverting) back the original values of $\alpha$ and $\beta$, I get:
$$\implies u(\theta) = \frac{c_2 sin(\sqrt{1- \frac{G^2 M^2}{\ell^2 c^2}}\theta) + c_1 cos(\sqrt{1- \frac{G^2 M^2}{\ell^2 c^2}}\theta) + \frac{GMd}{\ell^2}}{1- \frac{G^2 M^2}{\ell^2 c^2}}$$

And, there's the answer. With $c_1$ and $c_2$ being two more constants to add to the list of constants I already have.

Method 2:
Not much different from method 1, but with the exception that I substituted the original equation a bit differently:
$$\ddot{u} (\theta) + (1-\alpha) u(\theta) - \beta = 0$$

But, with this small change, $y$ then gets substituted as $(1-\alpha)u(\theta) - \beta$.

And, similarly, $\ddot{u}$ then now becomes $\frac{\ddot{y}}{1-\alpha}$.
Ultimately, giving me the following differential equation to solve:
$$\ddot{y}(\theta) + (1-\alpha)y(\theta) =0$$

Solving the above differential equation on WoframAlpha gives:
$$y(\theta) = c_1 \: e^{\sqrt{\alpha -1} \theta} + c_2 \: e^{-\sqrt{\alpha -1} \theta}$$
$$\implies (1-\alpha)u(\theta) - \beta = c_1 \: e^{\sqrt{\alpha -1} \theta} + c_2 \: e^{-\sqrt{\alpha -1} \theta}$$
$$\implies (1-\alpha)u(\theta)= c_1 \: e^{\sqrt{\alpha -1} \theta} + c_2 \: e^{-\sqrt{\alpha -1} \theta} + \beta$$
$$\implies u(\theta)= \frac{c_1 \: e^{\sqrt{\alpha -1} \theta} + c_2 \: e^{-\sqrt{\alpha -1} \theta} + \beta}{ (1-\alpha)}$$

Putting the original values of $\alpha$ and $\beta$ back:
$$\implies u(\theta)= \frac{c_1 \: e^{\sqrt{\frac{G^2 M^2}{\ell^2 c^2}-1} \theta} + c_2 \: e^{-\sqrt{\frac{G^2 M^2}{\ell^2 c^2}-1} \theta} + \frac{GMd}{\ell^2}}{ (1-\alpha)}$$

Likewise, $c_1$ and $c_2$ are constants here.

Beyond that, the answers in method 1 and method 2, while different, should technically be the same.

Method 3:
This is a solution proposed by a friend. However, I have a hard time making any sense out of it. Perhaps, what impresses me the most about it though is the lack of additional constants.

Using an ansatz, namely $u = a(1+ e \: cos(bt))$, for some $a$, $b$ and substitute it back in the differential equation, I could get:
$$u = \frac{GMd}{\ell^2} \bigg(1- \frac{G^2 M^2}{\ell^2 c^2} \bigg)^{-1} \bigg( 1 + e \: cos \bigg( \big( 1- \frac{G^2 M^2}{\ell^2 c^2} \big) \theta \bigg) \bigg)$$
(Note: $e$ is the eccentricity of the orbit, which is determined by the intial condition)

after solving for the original differential equation, i.e. $u(\theta)$ (for all $\theta$).

Note that, once again, this is the original equation I am trying to solve:
$$\frac{d^2u}{d\theta^2} + u \bigg( 1- \frac{G^2 M^2}{\ell^2 c^2} \bigg) - \frac{GMd}{\ell^2} = 0$$

---------------------------------------

Finally, with that said, are any of these methods correct? Personally, I think either the first and second methods are correct. However, the third one - given by a physics senior - intrigues me by the equation's lack of constants.

However, with that said though, I keep coming back to your comment here:

To get something different you would need $r$ to be very small. Much less than the Sun's radius. Eventually, if you make $r$ small enough the sign flips and you get an exponential inward spiral.

Even after discussing it with the team, we just can't figure out and apply what you mean to our differential equation and try to ultimately solve it. Provided that my solutions are incorrect, do you mind providing us more insight into the solution process?

Once again, thank you for helping!

 

[PeroK]

Your methods 1) and 2) are not technically wrong, but they don't make some convenient simplifications. And you haven't considered that the solution depends on the sign of $\alpha$.

I must admit, I just wrote down the solution with any calculation. First I took $\alpha$ to be positive, and used $\alpha^2$, so we have: $$\frac{d^2u}{d\theta^2} = -\alpha^2 u + \beta$$
We have an obvious solution to this: $$u = \frac{\beta}{\alpha^2}$$
For the homogeneous equation we have the solution:$$u = B\cos(\alpha \theta+ \phi)$$
And, we can take $\phi = 0$. (Physically this means taking $\theta = 0$ at the perihelion.) This gives us the general solution:
$$u(\theta) = B\cos(\alpha \theta) + \frac{\beta}{\alpha^2}$$
The interesting factor, regarding precession, is $\alpha$.

You could also solve the second case:$$\frac{d^2u}{d\theta^2} = \alpha^2 u + \beta$$ which will yield an exponential solution.

 

[Athenian]

Thank you very much for showing how to go about solving the differential equation with one of the cases.

Here's my team's solution for the whole question:
$u'' = \alpha^2 u + \beta \Longrightarrow u'' - \alpha^2 u = \beta$

[1] Homogenous: $u'' - \alpha^2 u =0$

Note that:
\begin{cases} u(\theta) = Ae^{-\beta \theta} \\ u'(\theta) = -A \beta e^{-\beta \theta} \\ u''(\theta) = A \beta^2 e^{-\beta \theta} \end{cases}

$$\implies A \beta^2 e^{-\beta \theta} - \alpha^2 A e^{-\beta \theta} = 0$$
$$\implies Ae^{-\beta \theta} (\beta^2 - \alpha ^2) =0$$
With the above equation in mind, I can thus deduce that $\beta = \alpha$ or $\beta = -\alpha$
$$\implies u(\theta) = Ae^{-\alpha \theta} + Ae^{\alpha \theta}$$

[2] Non-homogenous: $u'' - \alpha^2 u = \beta$

Noting that $u(\theta) = A_1 (\theta) \cdot e^{-\alpha \theta} + A_2 (\theta) e^{\alpha \theta}$,

I can continue by making the following "cases".

$$\begin{cases} A_1' (\theta) e^{-\alpha \theta} + A_2 ' (\theta) e^{\alpha \theta} =0 \;\;\;\;\;\; (1) \\ A_1 ' (\theta) \cdot (-\alpha) e^{-\alpha \theta} + A_2 ' (\theta) \cdot (\alpha) e^{\alpha \theta} = \beta \;\;\;\;\;\; (2) \end{cases}$$

$$(1): e^{\alpha \theta} (A_1' (\theta) e^{-1} + A_2 ' (\theta)) =0$$
Noting that the above equation applies as $e^{\alpha \theta} \neq 0$
$$\implies A_2 ' (\theta) = \frac{-A_1 ' (\theta)}{e} \;\;\;\;\;\; (3)$$

Placing $(3)$ into $(2)$, I can get:
$$A_1' (\theta) \cdot (-\alpha) e^{-1} \cdot e^{\alpha \theta} + (-\alpha) A_1 ' (\theta) e^{-1} e^{\alpha \theta} =\beta$$
$$\implies 2 \cdot A_1 ' (\theta) (-\alpha) e^{-\alpha \theta} = \beta$$
$$\implies A_1 ' (\theta) = \frac{-\beta}{2\alpha } e^{\alpha \theta} \;\;\;\;\;\; (4)$$

Integrating, I get:
$$A_1 (\theta) = -\frac{\beta}{2\alpha^2} e^{\alpha \theta} \;\;\;\;\;\; (5)$$

Placing equation $(4)$ into equation $(1)$, I get:
$$-\frac{\beta}{2\alpha^2} e^{\alpha \theta} \cdot e^{-\alpha \theta} + A_2 ' (\theta) e^{2\theta} =0$$
$$\implies A_2' (\theta) = \frac{\beta}{2\alpha} e^{-\alpha \theta} \;\;\;\;\;\; (6)$$

Integrating, I get:
$$\implies A_2 (\theta) = -\frac{\beta}{2\alpha^2} e^{-\alpha \theta} \;\;\;\;\;\; (7)$$

Adding equation $(6)$ and $(7)$ into $u(\theta) = A_1 (\theta) \cdot e^{-\alpha \theta} + A_2 (\theta) e^{\alpha \theta}$, I get:
$$u(\theta) = -\frac{\beta}{2\alpha^2} e^{\alpha \theta} \cdot e^{-\alpha \theta} -\frac{\beta}{2\alpha^2} e^{-\alpha \theta} \cdot e^{\alpha \theta}$$
$$\implies -\frac{\beta}{2\alpha^2} -\frac{\beta}{2\alpha^2} $$
$$\implies u(\theta) = -\frac{\beta}{\alpha^2}$$
$$\implies u(\theta) = Ae^{-\alpha \theta} + Ae^{\alpha \theta} - \frac{\beta}{\alpha^2}$$

And, there it is, the solution!
Thank you very much for your help. Beyond that, just to make sure, is the above solution correct?
Once again, thank you for your assistance.

 

[PeroK]

That is the long way round! Note that it should be:$$u(\theta) = Ae^{\alpha \theta} + Be^{-\alpha \theta} - \frac{\beta}{\alpha^2}$$
Physically, one of these exponentials represents an outward spiral ($r$ increasing) and one represents an inward spiral ($r$ decreasing).