Calculating the Earth's Elliptical Orbit around the Sun

In summary: Then in summer, say Jun 21, the Earth is further from the sun tilted more towards the Northern Hemisphere. This is because the Earth's orbit is elliptical, not a perfect circle.
  • #1
Philosophaie
462
0
To help me you will need tho know Excel VBA very well.

Earth has an elliptical orbit around the sun. JPL's website has the data to calculate this orbit:
http://ssd.jpl.nasa.gov/txt/p_elem_t1.txt
In the spreadsheet the parameters are calculated from widely used equations and some of my making. The major discrepencies are in how to calculate the distance between the Earth and the Sun,r. By my calculations:
r<1AU and a minimum @True Anomaly=0
r>1AU and a maximum @True Anomaly=180
This is not the case when directly calculated from the table(In Green). My calculations of what they should be are in yellow.

Any suggestions would be deeply appreciated!
 

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  • Earth's Orbit.xls
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  • #2
Your equation for calculating the eccentric anomaly is incorrect:
Code:
  Do
     E1 = E0 - (E0 - e(k) * 180 / Pi * Math.Sin(E0) - MA(k)) / (1 - e(k) * Math.Cos(E0))
   Loop While E1 = E0
   EA(k) = E1

Your input values for the true anomaly (column M) are completely wrong. True anomaly is an increasing function of time, not decreasing, and it is not uniform in time.
 
  • #3
If I change the values of the True Anomaly to increase by starting with 210,240,270,300,330,0,30,60,90,120,150,180 that would be increasing and will still prove my point.
I got the Eccentric Anomaly from this website:
http://www.stjarnhimlen.se/comp/tutorial.html
If I may, does anyone have a better equation for Eccentric Anomally?
 
  • #4
Here is the corrected True Anomaly with increasing degrees in E and M. The True, Mean and Eccentric Anomaly increase each year by about 360 degrees making a wrap around function. I think that the True Anomaly v=0 should start @Jun 21 the Summer Solstice not at v=168.0600083 as calculated from the table. I think the calculations of M and E from the table (in Green) are wrong.
 

Attachments

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  • #5
Philosophaie said:
If I change the values of the True Anomaly to increase by starting with 210,240,270,300,330,0,30,60,90,120,150,180 that would be increasing
What makes you think the initial value is 210? (Hint: It isn't.) What makes you think the true anomaly increases by 30 degrees per month? (Hint: It doesn't.)

... and will still prove my point.
What is your point, exactly? That you posted this in the "Beyond the Standard Model" section makes me think your point might be some crackpot theory. I hope that that is not the case.

I got the Eccentric Anomaly from this website:
http://www.stjarnhimlen.se/comp/tutorial.html
If I may, does anyone have a better equation for Eccentric Anomally?
There are two things wrong with your Newton's method implementation. (1) You are treating the mean anomaly as if it were in radians. (2) Your loop condition is wrong. As written, the loop will only execute once.
 
  • #6
Philosophaie said:
I think that the True Anomaly v=0 should start @Jun 21 the Summer Solstice not at v=168.0600083 as calculated from the table. I think the calculations of M and E from the table (in Green) are wrong.
You're wrong. In 2008, for example, perihelion occurred on January 3. True anomaly does not increase by exactly 30 degrees per month, for several reasons.
  1. No month is 365.25/12 days long.
  2. You are confusing the tropical year (solstice to solstice) and the sidereal year (apsis to apsis). The sidereal year is 20 minutes and 24 seconds longer than the tropical year.
  3. True anomaly is not a uniform function of time. Think of it in terms of a comet's orbit. Near apohelion, the true anomaly changes very little from day to day, while near perihelion, the true anomaly changes very rapidly.
 
  • #7
My equations (in Yellow) are just my way of trying to understand, PLEASE DISREGARD. The main equations (in Green) are what need to be correct then comes understanding. Are these equations correct?
'Argument of the Perihelion
PA(k) = LP(k) - LAN(k)
'Mean Longitude of Earth at a particular time
ML(k) = .Cells(7, 5) + .Cells(8, 5) * JT(k)
ML(k) = ML(k) - 360 * Int(ML(k) / 360)
'Mean Anomaly of Earth at a particular time
MA(k) = ML(k) - LP(k)
MA(k) = MA(k) - 360 * Int(MA(k) / 360)
'Eccentric Anomaly of Earth at a particular time from above website
EA(k) = MA(k) + 180 / Pi * e(k) * Math.Sin(MA(k) * Pi / 180) * (1 + e(k) * Math.Cos(MA(k) * Pi / 180))
EA(k) = EA(k) - 360 * Int(EA(k) / 360)
 
Last edited:
  • #8
So in the winter, say Jan 3, the Earth is closer to the sun just tilted away from the Northern Hemisphere.
 

What is the Earth's elliptical orbit around the Sun?

The Earth's elliptical orbit around the Sun is the path that the Earth follows as it revolves around the Sun. It is not a perfect circle, but rather an elliptical shape with the Sun at one of its foci.

How is the Earth's orbit calculated?

The Earth's orbit is calculated using Kepler's laws of planetary motion. This involves measuring the distance between the Earth and the Sun, as well as the Earth's speed and direction of motion.

What is the shape of the Earth's orbit?

The shape of the Earth's orbit is an ellipse, which is a closed curve with two focal points. The Sun is located at one of these focal points, and the Earth moves along the ellipse, with the Sun being closer to one end of the ellipse than the other.

What causes the Earth's orbit to be elliptical?

The Earth's orbit is elliptical due to the gravitational pull of the Sun. This pull is not constant throughout the orbit, causing the Earth to speed up or slow down at different points, resulting in an elliptical shape rather than a perfect circle.

How long does it take for the Earth to complete one orbit around the Sun?

The Earth takes approximately 365.24 days to complete one orbit around the Sun. This is known as a sidereal year and is the basis for our calendar system. However, due to slight variations in the Earth's speed and orbit, a tropical year (based on the seasons) is approximately 20 minutes shorter than a sidereal year.

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