1

I think that I need to know how to determine the angle between two points on the Earth's elliptical orbit around the Sun in order to calculate the Sun's distance from Earth.

I found the main equations at this page This is a good reference but I am still lost because I don't see how to calculate $\theta$ there, nor $\psi$ (eccentric anomaly)

There is a another website at "How can I find the Distance to the Sun on any given day." That is my ultimate goal btw, to have a spreadsheet with an entire year in 4 hour increments including distance to the sun for each 4 hour period. For a novice person with no math background, this website explanation seems a little easier to understand for me. Alas, she also did not explain how to determine the radius angle between two points on the ellipse. (Unless I'm overlooking it - which could be the problem)

enter image description here

This is the formula I ultimately derived:

$a$ = semi-major axis is $149.6$ million km

$e$ = eccentrity of the ellipse = $0.017$

$\theta$ = Is supposed to be angle between two points on the ellipse But I used number of days since last perihelion This is where I get lost

$X$ = number of days since last perihelion

$$\text{distance} = \frac{a(1-e^2)}{1+e\cdot\cos\theta}$$

$$\text{distance} = \frac{149.6\cdot(1-0.017^2)}{1+0.017\cdot\cos\big(X\cdot\frac{365.25}{360}\big)}$$

Will someone please help me understand how to correct my formula?

John Muggins
  • 135
  • 1
    I edited your question to be more readable (and added $\LaTeX$). I normally wouldn't bother but it is obvious to me that you've put in the time to do your homework, so good luck! – Rushabh Mehta Sep 16 '19 at 15:44
  • 1
    You should correct your definition of $a$: it is not the distance from Sun to perihelion, which is $a(1-e)$ as you can see in the figure. Then, unfortunately, angle $\theta$ is NOT simply proportional to time. – Intelligenti pauca Sep 16 '19 at 16:33
  • Thank you Aretino. Actually I had assumed that the semi-major axis was the distance from the Sun to the point of perihelion. Most probably and as I said earlier I'm not familiar with the jargon. So what you are saying is that the semi-major axis is the point from the center of the ellipse to the point of perihelion - and not from the Sun's position? – John Muggins Sep 16 '19 at 19:49
  • @JohnMuggins Yes, it's the distance from the center of the ellipse to perihelion or aphelion. – Intelligenti pauca Sep 16 '19 at 20:28

1 Answers1

1

You are on the right track, however you cannot directly use the number of days since perhelion as θ, which requires an angle.

Technically it requires calculus to go from a time of orbit, to an angle/position of an elliptical orbit, but for barely eccentric orbits like earths, you can cheat by doing the following

Use the number of days divided by the period (365.25 days in a year) and multiply that by your full circle (2pi, if using radians, 360 if using degrees) like so.... $$ \theta = \frac DP * 2\pi $$ where $D=$ days since perhelion, $P=$ Period in days and $2\pi=$ circle constant (use 360 instead if doing calculations in degrees instead of radians)

For Earth this gives a good approximation that will be between $0 \pm \mathcal{E}$ radians from the actual value (less than 1 degree max)

That should make it more clear that the angle required is related to the percentage of the orbit completed since perhelion

Nox Ferocia
  • 36