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Astrophysics: Calculating the circumference of an ellipse

  • Thread starter Andy1200
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  • #1
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Homework Statement:

I'm new to this website. Could someone explain how to solve this equation, its the formula for an ellipse circumference :

Homework Equations:

P = 2a(pi){1-(1/2)^2[(sqrta^2b^2)/a]^2- [(1*3)/(2*4)]^2{[(sqrta^2b^2)/a]/3}^4.....}
Substituting :
a = (9.15x10^7 mi)+(9.45x10^7mi) = 1.86x10^8 mi
b = ( a/2 ) = 9.3x10^7 mi

For this, I used six terms and got :

1.075x10^9 miles

Is my math wrong?














;
 

Answers and Replies

  • #2
mjc123
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There seem to be a few mistakes in your formula. If I've got it right, "sqrta^2b^2" should be sqrt(a^2 - b^2). (Maybe just a typo on your part.)
{[(sqrta^2b^2)/a]/3}^4 should be {[(sqrt(a^2-b^2))/a]^4}/3
And I don't know where you're getting your a and b from. In this formula, a is the semi-major axis and b the semi-minor axis. Not (as you are using?) the sum and average of these. That's why you're getting an answer about a factor of 2 high.
 
  • #3
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Thanks for the feedback. Yes, I'm still learning how to type formulas, but what you wrote is what I meant. I was using the formula to find the length of earth's orbit around the sun w/out using Google. My apologies on the a and b numbers. I've found it extremely difficult to find the semi-minor axis of the orbit, given that the farthest and closest distances make a 180-degree angle.
 
  • #4
mjc123
Science Advisor
Homework Helper
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That's because the sun is at a focus of the ellipse, not the centre. In this case, the semi-major axis is the average of the closest and farthest distances. You can't get the semi-minor axis from these (it is independently variable), but you should be able to find it easily, or the value of the eccentricity, from which you can work it out, on the internet. (Actually the eccentricity is all you need for the circumference equation, if you have the semi-major axis.)
 

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