Calculate Earth's distance travelled in a year











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This is probably more of a math question than astronomy, but I'd like to know the distance the Earth travels in one revolution around the sun (i.e. a year).



According to the data in Wikipedia, I think there may be several ways for me to calculate this but I'm not sure whether they are correct or how to do them:




  1. Calculate by velocity and time: the average velocity is given as $107200 mathrm{km/h}$ (a suspiciously round number; imprecise?) and one revolution takes $365.256363004$ days. That gives $text{distance} = 107,200 mathrm{km/h} times (365.256363004 times 24 mathrm h) = 939,731,570.736691 mathrm{km} approx 939.731 mathrm{Gm}$

  2. Calculate by semi-major axis and eccentricity: my math is too weak for that. I searched the web and didn't find a solution for that (calculating the circumference of an ellipse exactly seems to be a problem?)

  3. Any other way using the orbital data given in the Wikipedia article?


In essence, this can be made into a pretty generic question: How do I calculate the distance an object travels in one revolution using the usual orbital parameters?










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  • Are you interested in only the distance traveled wrt. the Sun, or are you interested in the total distance traveled wrt. some other reference (e.g. wrt. the center of the Milky Way; or wrt. some other extra-galactic reference (i.e. including the motion of the Milky Way))?
    – Makyen
    Nov 10 at 23:32






  • 1




    @Makyen: The distance travelled with respect to the Sun is what I'm interested in, yes. We can also ignore the barycenter motion.
    – DarkDust
    Nov 11 at 9:50

















up vote
4
down vote

favorite
1












This is probably more of a math question than astronomy, but I'd like to know the distance the Earth travels in one revolution around the sun (i.e. a year).



According to the data in Wikipedia, I think there may be several ways for me to calculate this but I'm not sure whether they are correct or how to do them:




  1. Calculate by velocity and time: the average velocity is given as $107200 mathrm{km/h}$ (a suspiciously round number; imprecise?) and one revolution takes $365.256363004$ days. That gives $text{distance} = 107,200 mathrm{km/h} times (365.256363004 times 24 mathrm h) = 939,731,570.736691 mathrm{km} approx 939.731 mathrm{Gm}$

  2. Calculate by semi-major axis and eccentricity: my math is too weak for that. I searched the web and didn't find a solution for that (calculating the circumference of an ellipse exactly seems to be a problem?)

  3. Any other way using the orbital data given in the Wikipedia article?


In essence, this can be made into a pretty generic question: How do I calculate the distance an object travels in one revolution using the usual orbital parameters?










share|improve this question
























  • Are you interested in only the distance traveled wrt. the Sun, or are you interested in the total distance traveled wrt. some other reference (e.g. wrt. the center of the Milky Way; or wrt. some other extra-galactic reference (i.e. including the motion of the Milky Way))?
    – Makyen
    Nov 10 at 23:32






  • 1




    @Makyen: The distance travelled with respect to the Sun is what I'm interested in, yes. We can also ignore the barycenter motion.
    – DarkDust
    Nov 11 at 9:50















up vote
4
down vote

favorite
1









up vote
4
down vote

favorite
1






1





This is probably more of a math question than astronomy, but I'd like to know the distance the Earth travels in one revolution around the sun (i.e. a year).



According to the data in Wikipedia, I think there may be several ways for me to calculate this but I'm not sure whether they are correct or how to do them:




  1. Calculate by velocity and time: the average velocity is given as $107200 mathrm{km/h}$ (a suspiciously round number; imprecise?) and one revolution takes $365.256363004$ days. That gives $text{distance} = 107,200 mathrm{km/h} times (365.256363004 times 24 mathrm h) = 939,731,570.736691 mathrm{km} approx 939.731 mathrm{Gm}$

  2. Calculate by semi-major axis and eccentricity: my math is too weak for that. I searched the web and didn't find a solution for that (calculating the circumference of an ellipse exactly seems to be a problem?)

  3. Any other way using the orbital data given in the Wikipedia article?


In essence, this can be made into a pretty generic question: How do I calculate the distance an object travels in one revolution using the usual orbital parameters?










share|improve this question















This is probably more of a math question than astronomy, but I'd like to know the distance the Earth travels in one revolution around the sun (i.e. a year).



According to the data in Wikipedia, I think there may be several ways for me to calculate this but I'm not sure whether they are correct or how to do them:




  1. Calculate by velocity and time: the average velocity is given as $107200 mathrm{km/h}$ (a suspiciously round number; imprecise?) and one revolution takes $365.256363004$ days. That gives $text{distance} = 107,200 mathrm{km/h} times (365.256363004 times 24 mathrm h) = 939,731,570.736691 mathrm{km} approx 939.731 mathrm{Gm}$

  2. Calculate by semi-major axis and eccentricity: my math is too weak for that. I searched the web and didn't find a solution for that (calculating the circumference of an ellipse exactly seems to be a problem?)

  3. Any other way using the orbital data given in the Wikipedia article?


In essence, this can be made into a pretty generic question: How do I calculate the distance an object travels in one revolution using the usual orbital parameters?







orbital-mechanics






share|improve this question















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edited Nov 11 at 11:14









Loong

1033




1033










asked Nov 10 at 17:22









DarkDust

1829




1829












  • Are you interested in only the distance traveled wrt. the Sun, or are you interested in the total distance traveled wrt. some other reference (e.g. wrt. the center of the Milky Way; or wrt. some other extra-galactic reference (i.e. including the motion of the Milky Way))?
    – Makyen
    Nov 10 at 23:32






  • 1




    @Makyen: The distance travelled with respect to the Sun is what I'm interested in, yes. We can also ignore the barycenter motion.
    – DarkDust
    Nov 11 at 9:50




















  • Are you interested in only the distance traveled wrt. the Sun, or are you interested in the total distance traveled wrt. some other reference (e.g. wrt. the center of the Milky Way; or wrt. some other extra-galactic reference (i.e. including the motion of the Milky Way))?
    – Makyen
    Nov 10 at 23:32






  • 1




    @Makyen: The distance travelled with respect to the Sun is what I'm interested in, yes. We can also ignore the barycenter motion.
    – DarkDust
    Nov 11 at 9:50


















Are you interested in only the distance traveled wrt. the Sun, or are you interested in the total distance traveled wrt. some other reference (e.g. wrt. the center of the Milky Way; or wrt. some other extra-galactic reference (i.e. including the motion of the Milky Way))?
– Makyen
Nov 10 at 23:32




Are you interested in only the distance traveled wrt. the Sun, or are you interested in the total distance traveled wrt. some other reference (e.g. wrt. the center of the Milky Way; or wrt. some other extra-galactic reference (i.e. including the motion of the Milky Way))?
– Makyen
Nov 10 at 23:32




1




1




@Makyen: The distance travelled with respect to the Sun is what I'm interested in, yes. We can also ignore the barycenter motion.
– DarkDust
Nov 11 at 9:50






@Makyen: The distance travelled with respect to the Sun is what I'm interested in, yes. We can also ignore the barycenter motion.
– DarkDust
Nov 11 at 9:50












3 Answers
3






active

oldest

votes

















up vote
7
down vote













You need to find the perimeter of an ellipse. This might be accurate enough for you (recalling that an ellipse is only an accurate description of an orbit for a two-body system.



An ellipse has no simple formula to find the perimeter. There appear to be several commonly used approximations based on infinite series (see for example "Maths is Fun").



e.g. Define
$$ h = frac{(a-b)^2}{(a+b)^2},$$
where $a$ and $b$ are the semi-major and semi-minor axes and $b = asqrt{1-e^2}$, where $e$ is the eccentricity.



Then the result you want is
$$ p = pi(a+b)left[1 + frac{h}{4} + frac{h^2}{64} + frac{h^3}{256} ...right]$$



All this is general, but might be overkill for your problem involving the Earth. If my calculator skills are working, then for the Earth's orbit around the Sun, $a = 1.00000011$ au, $e = 0.01671022$, $b = 0.99986048$ au and $h = 7times 10^{-5}$.



Using this and just the first 2 terms in the series, I get a perimeter of $1.999895pi$ au. So hardly a correction from $2pi a$.



However, if you were serious about accuracy then you would have to add to this about 12.5 orbits around the barycentre of the Earth-Moon system and a further correction for the motion of the Sun around the solar system barycentre.






share|improve this answer























  • (I guess you mean $b$ to be the semi-minor axis.) Deriving the semi-minor axis via the the semi-major axis and eccentricity seems to be the missing piece, thanks. Will try this later and post my results.
    – DarkDust
    Nov 10 at 18:31












  • Plugging in this gives about $939,902 Gm$, which matches the time-based solution at least in the integer part. You're right, this ignores the barycenter but is good enough for me since it seems the available base values are not 100% precise to begin with.
    – DarkDust
    Nov 11 at 8:58










  • Nice! Considering that a Kelplerian orbit isn't just any ellipse, but governed by some constraints of motion, it doesn't necessarily follow that "An ellipse has no simple formula to find the perimeter." means the integral of $ds/dt$ has no simple formula. Maybe it doesn't, but that would have to be demonstrated separately. Since $t(theta)$ can be addressed analytically, as can the relationship between $r$ and $theta$, is there really no exact analytical solution for the integral of $ds/dt$ over one orbit?
    – uhoh
    Nov 11 at 10:02












  • @uhoh The orbit is in fact any ellipse with the barycenter in one of its focal points and we still want to find its perimeter. There is no simple formula in terms of the axes etc. if you integrate at constant speed or at orbital mechanic speed. We are just lucky that the excentrity is very small and that the influx by other planets is very small and that the Sun essentially is the barycenter and that relativistic effects can be ignored. Then again, if either of these lucky coincidences were not true, we'd hardly be here to discuss this
    – Hagen von Eitzen
    Nov 11 at 14:12








  • 1




    @StephenG I've posted where I got it, but no original source is given mathsisfun.com/geometry/ellipse-perimeter.html
    – Rob Jeffries
    Nov 11 at 18:04


















up vote
3
down vote













I pondered this myself a couple of weeks ago. For Earth's orbital circumference Ramanujan's (q.v.) formula for the perimeter of an ellipse is good enough. (Actually, Jean Meeus says so in his book, but I can't find the reference. Sorry.) The Excel formula I used is:



=PI() * (F4 + F6) * (1 + 3 * ((F4 - F6)^2/(F4 + F6)^2)/(10 + (4 - 3 * ((F4 - F6)^2/(F4 + F6)^2))^1/2))


Where F4 and F6 are the semi-major and semi-minor axes, respectively.



Using this formula I get a circumference of 939.8855 X 106 km.






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    up vote
    3
    down vote



    accepted










    The distance travelled by Earth in one revolution around the Sun as calculated by the data from Wikipedia's entry about Earth is about $936.013 Gm$ or $939.886Gm$, depending on the approach used for calculation. Since my math-fu is not very strong please point out any mistakes I've made.



    Integral approach



    It seems the exact way to calculate the perimeter/circumference is via the formula $C = 4aE(e)$ where $E$ is the complete elliptic integral of the second kind.



    I've found a solution using Python. I tried using the numbers from Wikipedia here for consistency:



    import numpy as np
    from scipy.special import ellipe
    a = 149598023000 # semi-major in meter
    e = 0.0167086 # eccentricity
    pe = 4 * a * ellipe(e)
    print(pe) # 936013392131.0


    So about $936.013 Gm$. That's quite off from the $approx 940Gm$ given by the other answers. I'm wondering why. I know the other methods are approximations but the error seems rather large to me.



    Ramanujan's first approximation



    Ramanujan has given two approximations. The first is $C approx pi[3(a+b) - sqrt{10ab + 3(a^2+b^2)} ]$. In Python:



    import math
    a = 149598023000
    e = 0.0167086
    b = a * math.sqrt(1 - e**2) # derive semi-minor
    pe = math.pi * ( 3*(a+b) - math.sqrt( (3*a + b) * (a + 3*b) ) )
    print(pe) # 9.39886493337e+11


    Here we are at $939.886Gm$ which is closer to the solution from the other answers and the $velocity*time$ approach.



    Ramanujan's second approximation



    The formula used by @BillDOe is Ramanujan's second approximation: $C = pi(a+b)(1 + frac{3h}{10 + sqrt{4-3h}})$ with $h = frac{(a-b)^2}{(a+b)^2}$. In Python:



    a = 149598023000
    e = 0.0167086
    b = a * math.sqrt(1 - e**2) # derive semi-minor
    h = (a-b)**2 / (a+b)**2
    pe = math.pi * (a+b) * (1 + ((3*h) / (10 + math.sqrt(4 - 3*h))))
    print(pe) # 9.39886493337e+11


    That's the same result as the first approximation. The error for Ramanujan's approximations is stated as $h^3$ and $h^5$ respectively. That's about $1.157^{-25}$ and $2.747^{-42}$. I'm not a math geek, so these sound pretty low for me and I wonder why the solution is so different than the integral approach.



    @RobJeffries's approach



    I don't know how this is called. The formula is:



    $$ C = pi(a+b)left[1 + frac{h}{4} + frac{h^2}{64} + frac{h^3}{256} ...right]$$



    In Python:



    a = 149598023000
    e = 0.0167086
    b = a * math.sqrt(1 - e**2) # derive semi-minor
    h = (a-b)**2 / (a+b)**2
    pe = math.pi * (a+b) * (1 + (h/4) + (h**2/64) + (h**3/256))
    print(pe) # 9.39886493337e+11


    Again, $939.886Gm$






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      3 Answers
      3






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      3 Answers
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      up vote
      7
      down vote













      You need to find the perimeter of an ellipse. This might be accurate enough for you (recalling that an ellipse is only an accurate description of an orbit for a two-body system.



      An ellipse has no simple formula to find the perimeter. There appear to be several commonly used approximations based on infinite series (see for example "Maths is Fun").



      e.g. Define
      $$ h = frac{(a-b)^2}{(a+b)^2},$$
      where $a$ and $b$ are the semi-major and semi-minor axes and $b = asqrt{1-e^2}$, where $e$ is the eccentricity.



      Then the result you want is
      $$ p = pi(a+b)left[1 + frac{h}{4} + frac{h^2}{64} + frac{h^3}{256} ...right]$$



      All this is general, but might be overkill for your problem involving the Earth. If my calculator skills are working, then for the Earth's orbit around the Sun, $a = 1.00000011$ au, $e = 0.01671022$, $b = 0.99986048$ au and $h = 7times 10^{-5}$.



      Using this and just the first 2 terms in the series, I get a perimeter of $1.999895pi$ au. So hardly a correction from $2pi a$.



      However, if you were serious about accuracy then you would have to add to this about 12.5 orbits around the barycentre of the Earth-Moon system and a further correction for the motion of the Sun around the solar system barycentre.






      share|improve this answer























      • (I guess you mean $b$ to be the semi-minor axis.) Deriving the semi-minor axis via the the semi-major axis and eccentricity seems to be the missing piece, thanks. Will try this later and post my results.
        – DarkDust
        Nov 10 at 18:31












      • Plugging in this gives about $939,902 Gm$, which matches the time-based solution at least in the integer part. You're right, this ignores the barycenter but is good enough for me since it seems the available base values are not 100% precise to begin with.
        – DarkDust
        Nov 11 at 8:58










      • Nice! Considering that a Kelplerian orbit isn't just any ellipse, but governed by some constraints of motion, it doesn't necessarily follow that "An ellipse has no simple formula to find the perimeter." means the integral of $ds/dt$ has no simple formula. Maybe it doesn't, but that would have to be demonstrated separately. Since $t(theta)$ can be addressed analytically, as can the relationship between $r$ and $theta$, is there really no exact analytical solution for the integral of $ds/dt$ over one orbit?
        – uhoh
        Nov 11 at 10:02












      • @uhoh The orbit is in fact any ellipse with the barycenter in one of its focal points and we still want to find its perimeter. There is no simple formula in terms of the axes etc. if you integrate at constant speed or at orbital mechanic speed. We are just lucky that the excentrity is very small and that the influx by other planets is very small and that the Sun essentially is the barycenter and that relativistic effects can be ignored. Then again, if either of these lucky coincidences were not true, we'd hardly be here to discuss this
        – Hagen von Eitzen
        Nov 11 at 14:12








      • 1




        @StephenG I've posted where I got it, but no original source is given mathsisfun.com/geometry/ellipse-perimeter.html
        – Rob Jeffries
        Nov 11 at 18:04















      up vote
      7
      down vote













      You need to find the perimeter of an ellipse. This might be accurate enough for you (recalling that an ellipse is only an accurate description of an orbit for a two-body system.



      An ellipse has no simple formula to find the perimeter. There appear to be several commonly used approximations based on infinite series (see for example "Maths is Fun").



      e.g. Define
      $$ h = frac{(a-b)^2}{(a+b)^2},$$
      where $a$ and $b$ are the semi-major and semi-minor axes and $b = asqrt{1-e^2}$, where $e$ is the eccentricity.



      Then the result you want is
      $$ p = pi(a+b)left[1 + frac{h}{4} + frac{h^2}{64} + frac{h^3}{256} ...right]$$



      All this is general, but might be overkill for your problem involving the Earth. If my calculator skills are working, then for the Earth's orbit around the Sun, $a = 1.00000011$ au, $e = 0.01671022$, $b = 0.99986048$ au and $h = 7times 10^{-5}$.



      Using this and just the first 2 terms in the series, I get a perimeter of $1.999895pi$ au. So hardly a correction from $2pi a$.



      However, if you were serious about accuracy then you would have to add to this about 12.5 orbits around the barycentre of the Earth-Moon system and a further correction for the motion of the Sun around the solar system barycentre.






      share|improve this answer























      • (I guess you mean $b$ to be the semi-minor axis.) Deriving the semi-minor axis via the the semi-major axis and eccentricity seems to be the missing piece, thanks. Will try this later and post my results.
        – DarkDust
        Nov 10 at 18:31












      • Plugging in this gives about $939,902 Gm$, which matches the time-based solution at least in the integer part. You're right, this ignores the barycenter but is good enough for me since it seems the available base values are not 100% precise to begin with.
        – DarkDust
        Nov 11 at 8:58










      • Nice! Considering that a Kelplerian orbit isn't just any ellipse, but governed by some constraints of motion, it doesn't necessarily follow that "An ellipse has no simple formula to find the perimeter." means the integral of $ds/dt$ has no simple formula. Maybe it doesn't, but that would have to be demonstrated separately. Since $t(theta)$ can be addressed analytically, as can the relationship between $r$ and $theta$, is there really no exact analytical solution for the integral of $ds/dt$ over one orbit?
        – uhoh
        Nov 11 at 10:02












      • @uhoh The orbit is in fact any ellipse with the barycenter in one of its focal points and we still want to find its perimeter. There is no simple formula in terms of the axes etc. if you integrate at constant speed or at orbital mechanic speed. We are just lucky that the excentrity is very small and that the influx by other planets is very small and that the Sun essentially is the barycenter and that relativistic effects can be ignored. Then again, if either of these lucky coincidences were not true, we'd hardly be here to discuss this
        – Hagen von Eitzen
        Nov 11 at 14:12








      • 1




        @StephenG I've posted where I got it, but no original source is given mathsisfun.com/geometry/ellipse-perimeter.html
        – Rob Jeffries
        Nov 11 at 18:04













      up vote
      7
      down vote










      up vote
      7
      down vote









      You need to find the perimeter of an ellipse. This might be accurate enough for you (recalling that an ellipse is only an accurate description of an orbit for a two-body system.



      An ellipse has no simple formula to find the perimeter. There appear to be several commonly used approximations based on infinite series (see for example "Maths is Fun").



      e.g. Define
      $$ h = frac{(a-b)^2}{(a+b)^2},$$
      where $a$ and $b$ are the semi-major and semi-minor axes and $b = asqrt{1-e^2}$, where $e$ is the eccentricity.



      Then the result you want is
      $$ p = pi(a+b)left[1 + frac{h}{4} + frac{h^2}{64} + frac{h^3}{256} ...right]$$



      All this is general, but might be overkill for your problem involving the Earth. If my calculator skills are working, then for the Earth's orbit around the Sun, $a = 1.00000011$ au, $e = 0.01671022$, $b = 0.99986048$ au and $h = 7times 10^{-5}$.



      Using this and just the first 2 terms in the series, I get a perimeter of $1.999895pi$ au. So hardly a correction from $2pi a$.



      However, if you were serious about accuracy then you would have to add to this about 12.5 orbits around the barycentre of the Earth-Moon system and a further correction for the motion of the Sun around the solar system barycentre.






      share|improve this answer














      You need to find the perimeter of an ellipse. This might be accurate enough for you (recalling that an ellipse is only an accurate description of an orbit for a two-body system.



      An ellipse has no simple formula to find the perimeter. There appear to be several commonly used approximations based on infinite series (see for example "Maths is Fun").



      e.g. Define
      $$ h = frac{(a-b)^2}{(a+b)^2},$$
      where $a$ and $b$ are the semi-major and semi-minor axes and $b = asqrt{1-e^2}$, where $e$ is the eccentricity.



      Then the result you want is
      $$ p = pi(a+b)left[1 + frac{h}{4} + frac{h^2}{64} + frac{h^3}{256} ...right]$$



      All this is general, but might be overkill for your problem involving the Earth. If my calculator skills are working, then for the Earth's orbit around the Sun, $a = 1.00000011$ au, $e = 0.01671022$, $b = 0.99986048$ au and $h = 7times 10^{-5}$.



      Using this and just the first 2 terms in the series, I get a perimeter of $1.999895pi$ au. So hardly a correction from $2pi a$.



      However, if you were serious about accuracy then you would have to add to this about 12.5 orbits around the barycentre of the Earth-Moon system and a further correction for the motion of the Sun around the solar system barycentre.







      share|improve this answer














      share|improve this answer



      share|improve this answer








      edited Nov 11 at 18:03

























      answered Nov 10 at 18:23









      Rob Jeffries

      50.7k4101155




      50.7k4101155












      • (I guess you mean $b$ to be the semi-minor axis.) Deriving the semi-minor axis via the the semi-major axis and eccentricity seems to be the missing piece, thanks. Will try this later and post my results.
        – DarkDust
        Nov 10 at 18:31












      • Plugging in this gives about $939,902 Gm$, which matches the time-based solution at least in the integer part. You're right, this ignores the barycenter but is good enough for me since it seems the available base values are not 100% precise to begin with.
        – DarkDust
        Nov 11 at 8:58










      • Nice! Considering that a Kelplerian orbit isn't just any ellipse, but governed by some constraints of motion, it doesn't necessarily follow that "An ellipse has no simple formula to find the perimeter." means the integral of $ds/dt$ has no simple formula. Maybe it doesn't, but that would have to be demonstrated separately. Since $t(theta)$ can be addressed analytically, as can the relationship between $r$ and $theta$, is there really no exact analytical solution for the integral of $ds/dt$ over one orbit?
        – uhoh
        Nov 11 at 10:02












      • @uhoh The orbit is in fact any ellipse with the barycenter in one of its focal points and we still want to find its perimeter. There is no simple formula in terms of the axes etc. if you integrate at constant speed or at orbital mechanic speed. We are just lucky that the excentrity is very small and that the influx by other planets is very small and that the Sun essentially is the barycenter and that relativistic effects can be ignored. Then again, if either of these lucky coincidences were not true, we'd hardly be here to discuss this
        – Hagen von Eitzen
        Nov 11 at 14:12








      • 1




        @StephenG I've posted where I got it, but no original source is given mathsisfun.com/geometry/ellipse-perimeter.html
        – Rob Jeffries
        Nov 11 at 18:04


















      • (I guess you mean $b$ to be the semi-minor axis.) Deriving the semi-minor axis via the the semi-major axis and eccentricity seems to be the missing piece, thanks. Will try this later and post my results.
        – DarkDust
        Nov 10 at 18:31












      • Plugging in this gives about $939,902 Gm$, which matches the time-based solution at least in the integer part. You're right, this ignores the barycenter but is good enough for me since it seems the available base values are not 100% precise to begin with.
        – DarkDust
        Nov 11 at 8:58










      • Nice! Considering that a Kelplerian orbit isn't just any ellipse, but governed by some constraints of motion, it doesn't necessarily follow that "An ellipse has no simple formula to find the perimeter." means the integral of $ds/dt$ has no simple formula. Maybe it doesn't, but that would have to be demonstrated separately. Since $t(theta)$ can be addressed analytically, as can the relationship between $r$ and $theta$, is there really no exact analytical solution for the integral of $ds/dt$ over one orbit?
        – uhoh
        Nov 11 at 10:02












      • @uhoh The orbit is in fact any ellipse with the barycenter in one of its focal points and we still want to find its perimeter. There is no simple formula in terms of the axes etc. if you integrate at constant speed or at orbital mechanic speed. We are just lucky that the excentrity is very small and that the influx by other planets is very small and that the Sun essentially is the barycenter and that relativistic effects can be ignored. Then again, if either of these lucky coincidences were not true, we'd hardly be here to discuss this
        – Hagen von Eitzen
        Nov 11 at 14:12








      • 1




        @StephenG I've posted where I got it, but no original source is given mathsisfun.com/geometry/ellipse-perimeter.html
        – Rob Jeffries
        Nov 11 at 18:04
















      (I guess you mean $b$ to be the semi-minor axis.) Deriving the semi-minor axis via the the semi-major axis and eccentricity seems to be the missing piece, thanks. Will try this later and post my results.
      – DarkDust
      Nov 10 at 18:31






      (I guess you mean $b$ to be the semi-minor axis.) Deriving the semi-minor axis via the the semi-major axis and eccentricity seems to be the missing piece, thanks. Will try this later and post my results.
      – DarkDust
      Nov 10 at 18:31














      Plugging in this gives about $939,902 Gm$, which matches the time-based solution at least in the integer part. You're right, this ignores the barycenter but is good enough for me since it seems the available base values are not 100% precise to begin with.
      – DarkDust
      Nov 11 at 8:58




      Plugging in this gives about $939,902 Gm$, which matches the time-based solution at least in the integer part. You're right, this ignores the barycenter but is good enough for me since it seems the available base values are not 100% precise to begin with.
      – DarkDust
      Nov 11 at 8:58












      Nice! Considering that a Kelplerian orbit isn't just any ellipse, but governed by some constraints of motion, it doesn't necessarily follow that "An ellipse has no simple formula to find the perimeter." means the integral of $ds/dt$ has no simple formula. Maybe it doesn't, but that would have to be demonstrated separately. Since $t(theta)$ can be addressed analytically, as can the relationship between $r$ and $theta$, is there really no exact analytical solution for the integral of $ds/dt$ over one orbit?
      – uhoh
      Nov 11 at 10:02






      Nice! Considering that a Kelplerian orbit isn't just any ellipse, but governed by some constraints of motion, it doesn't necessarily follow that "An ellipse has no simple formula to find the perimeter." means the integral of $ds/dt$ has no simple formula. Maybe it doesn't, but that would have to be demonstrated separately. Since $t(theta)$ can be addressed analytically, as can the relationship between $r$ and $theta$, is there really no exact analytical solution for the integral of $ds/dt$ over one orbit?
      – uhoh
      Nov 11 at 10:02














      @uhoh The orbit is in fact any ellipse with the barycenter in one of its focal points and we still want to find its perimeter. There is no simple formula in terms of the axes etc. if you integrate at constant speed or at orbital mechanic speed. We are just lucky that the excentrity is very small and that the influx by other planets is very small and that the Sun essentially is the barycenter and that relativistic effects can be ignored. Then again, if either of these lucky coincidences were not true, we'd hardly be here to discuss this
      – Hagen von Eitzen
      Nov 11 at 14:12






      @uhoh The orbit is in fact any ellipse with the barycenter in one of its focal points and we still want to find its perimeter. There is no simple formula in terms of the axes etc. if you integrate at constant speed or at orbital mechanic speed. We are just lucky that the excentrity is very small and that the influx by other planets is very small and that the Sun essentially is the barycenter and that relativistic effects can be ignored. Then again, if either of these lucky coincidences were not true, we'd hardly be here to discuss this
      – Hagen von Eitzen
      Nov 11 at 14:12






      1




      1




      @StephenG I've posted where I got it, but no original source is given mathsisfun.com/geometry/ellipse-perimeter.html
      – Rob Jeffries
      Nov 11 at 18:04




      @StephenG I've posted where I got it, but no original source is given mathsisfun.com/geometry/ellipse-perimeter.html
      – Rob Jeffries
      Nov 11 at 18:04










      up vote
      3
      down vote













      I pondered this myself a couple of weeks ago. For Earth's orbital circumference Ramanujan's (q.v.) formula for the perimeter of an ellipse is good enough. (Actually, Jean Meeus says so in his book, but I can't find the reference. Sorry.) The Excel formula I used is:



      =PI() * (F4 + F6) * (1 + 3 * ((F4 - F6)^2/(F4 + F6)^2)/(10 + (4 - 3 * ((F4 - F6)^2/(F4 + F6)^2))^1/2))


      Where F4 and F6 are the semi-major and semi-minor axes, respectively.



      Using this formula I get a circumference of 939.8855 X 106 km.






      share|improve this answer



























        up vote
        3
        down vote













        I pondered this myself a couple of weeks ago. For Earth's orbital circumference Ramanujan's (q.v.) formula for the perimeter of an ellipse is good enough. (Actually, Jean Meeus says so in his book, but I can't find the reference. Sorry.) The Excel formula I used is:



        =PI() * (F4 + F6) * (1 + 3 * ((F4 - F6)^2/(F4 + F6)^2)/(10 + (4 - 3 * ((F4 - F6)^2/(F4 + F6)^2))^1/2))


        Where F4 and F6 are the semi-major and semi-minor axes, respectively.



        Using this formula I get a circumference of 939.8855 X 106 km.






        share|improve this answer

























          up vote
          3
          down vote










          up vote
          3
          down vote









          I pondered this myself a couple of weeks ago. For Earth's orbital circumference Ramanujan's (q.v.) formula for the perimeter of an ellipse is good enough. (Actually, Jean Meeus says so in his book, but I can't find the reference. Sorry.) The Excel formula I used is:



          =PI() * (F4 + F6) * (1 + 3 * ((F4 - F6)^2/(F4 + F6)^2)/(10 + (4 - 3 * ((F4 - F6)^2/(F4 + F6)^2))^1/2))


          Where F4 and F6 are the semi-major and semi-minor axes, respectively.



          Using this formula I get a circumference of 939.8855 X 106 km.






          share|improve this answer














          I pondered this myself a couple of weeks ago. For Earth's orbital circumference Ramanujan's (q.v.) formula for the perimeter of an ellipse is good enough. (Actually, Jean Meeus says so in his book, but I can't find the reference. Sorry.) The Excel formula I used is:



          =PI() * (F4 + F6) * (1 + 3 * ((F4 - F6)^2/(F4 + F6)^2)/(10 + (4 - 3 * ((F4 - F6)^2/(F4 + F6)^2))^1/2))


          Where F4 and F6 are the semi-major and semi-minor axes, respectively.



          Using this formula I get a circumference of 939.8855 X 106 km.







          share|improve this answer














          share|improve this answer



          share|improve this answer








          edited Nov 10 at 22:28

























          answered Nov 10 at 20:41









          BillDOe

          793410




          793410






















              up vote
              3
              down vote



              accepted










              The distance travelled by Earth in one revolution around the Sun as calculated by the data from Wikipedia's entry about Earth is about $936.013 Gm$ or $939.886Gm$, depending on the approach used for calculation. Since my math-fu is not very strong please point out any mistakes I've made.



              Integral approach



              It seems the exact way to calculate the perimeter/circumference is via the formula $C = 4aE(e)$ where $E$ is the complete elliptic integral of the second kind.



              I've found a solution using Python. I tried using the numbers from Wikipedia here for consistency:



              import numpy as np
              from scipy.special import ellipe
              a = 149598023000 # semi-major in meter
              e = 0.0167086 # eccentricity
              pe = 4 * a * ellipe(e)
              print(pe) # 936013392131.0


              So about $936.013 Gm$. That's quite off from the $approx 940Gm$ given by the other answers. I'm wondering why. I know the other methods are approximations but the error seems rather large to me.



              Ramanujan's first approximation



              Ramanujan has given two approximations. The first is $C approx pi[3(a+b) - sqrt{10ab + 3(a^2+b^2)} ]$. In Python:



              import math
              a = 149598023000
              e = 0.0167086
              b = a * math.sqrt(1 - e**2) # derive semi-minor
              pe = math.pi * ( 3*(a+b) - math.sqrt( (3*a + b) * (a + 3*b) ) )
              print(pe) # 9.39886493337e+11


              Here we are at $939.886Gm$ which is closer to the solution from the other answers and the $velocity*time$ approach.



              Ramanujan's second approximation



              The formula used by @BillDOe is Ramanujan's second approximation: $C = pi(a+b)(1 + frac{3h}{10 + sqrt{4-3h}})$ with $h = frac{(a-b)^2}{(a+b)^2}$. In Python:



              a = 149598023000
              e = 0.0167086
              b = a * math.sqrt(1 - e**2) # derive semi-minor
              h = (a-b)**2 / (a+b)**2
              pe = math.pi * (a+b) * (1 + ((3*h) / (10 + math.sqrt(4 - 3*h))))
              print(pe) # 9.39886493337e+11


              That's the same result as the first approximation. The error for Ramanujan's approximations is stated as $h^3$ and $h^5$ respectively. That's about $1.157^{-25}$ and $2.747^{-42}$. I'm not a math geek, so these sound pretty low for me and I wonder why the solution is so different than the integral approach.



              @RobJeffries's approach



              I don't know how this is called. The formula is:



              $$ C = pi(a+b)left[1 + frac{h}{4} + frac{h^2}{64} + frac{h^3}{256} ...right]$$



              In Python:



              a = 149598023000
              e = 0.0167086
              b = a * math.sqrt(1 - e**2) # derive semi-minor
              h = (a-b)**2 / (a+b)**2
              pe = math.pi * (a+b) * (1 + (h/4) + (h**2/64) + (h**3/256))
              print(pe) # 9.39886493337e+11


              Again, $939.886Gm$






              share|improve this answer



























                up vote
                3
                down vote



                accepted










                The distance travelled by Earth in one revolution around the Sun as calculated by the data from Wikipedia's entry about Earth is about $936.013 Gm$ or $939.886Gm$, depending on the approach used for calculation. Since my math-fu is not very strong please point out any mistakes I've made.



                Integral approach



                It seems the exact way to calculate the perimeter/circumference is via the formula $C = 4aE(e)$ where $E$ is the complete elliptic integral of the second kind.



                I've found a solution using Python. I tried using the numbers from Wikipedia here for consistency:



                import numpy as np
                from scipy.special import ellipe
                a = 149598023000 # semi-major in meter
                e = 0.0167086 # eccentricity
                pe = 4 * a * ellipe(e)
                print(pe) # 936013392131.0


                So about $936.013 Gm$. That's quite off from the $approx 940Gm$ given by the other answers. I'm wondering why. I know the other methods are approximations but the error seems rather large to me.



                Ramanujan's first approximation



                Ramanujan has given two approximations. The first is $C approx pi[3(a+b) - sqrt{10ab + 3(a^2+b^2)} ]$. In Python:



                import math
                a = 149598023000
                e = 0.0167086
                b = a * math.sqrt(1 - e**2) # derive semi-minor
                pe = math.pi * ( 3*(a+b) - math.sqrt( (3*a + b) * (a + 3*b) ) )
                print(pe) # 9.39886493337e+11


                Here we are at $939.886Gm$ which is closer to the solution from the other answers and the $velocity*time$ approach.



                Ramanujan's second approximation



                The formula used by @BillDOe is Ramanujan's second approximation: $C = pi(a+b)(1 + frac{3h}{10 + sqrt{4-3h}})$ with $h = frac{(a-b)^2}{(a+b)^2}$. In Python:



                a = 149598023000
                e = 0.0167086
                b = a * math.sqrt(1 - e**2) # derive semi-minor
                h = (a-b)**2 / (a+b)**2
                pe = math.pi * (a+b) * (1 + ((3*h) / (10 + math.sqrt(4 - 3*h))))
                print(pe) # 9.39886493337e+11


                That's the same result as the first approximation. The error for Ramanujan's approximations is stated as $h^3$ and $h^5$ respectively. That's about $1.157^{-25}$ and $2.747^{-42}$. I'm not a math geek, so these sound pretty low for me and I wonder why the solution is so different than the integral approach.



                @RobJeffries's approach



                I don't know how this is called. The formula is:



                $$ C = pi(a+b)left[1 + frac{h}{4} + frac{h^2}{64} + frac{h^3}{256} ...right]$$



                In Python:



                a = 149598023000
                e = 0.0167086
                b = a * math.sqrt(1 - e**2) # derive semi-minor
                h = (a-b)**2 / (a+b)**2
                pe = math.pi * (a+b) * (1 + (h/4) + (h**2/64) + (h**3/256))
                print(pe) # 9.39886493337e+11


                Again, $939.886Gm$






                share|improve this answer

























                  up vote
                  3
                  down vote



                  accepted







                  up vote
                  3
                  down vote



                  accepted






                  The distance travelled by Earth in one revolution around the Sun as calculated by the data from Wikipedia's entry about Earth is about $936.013 Gm$ or $939.886Gm$, depending on the approach used for calculation. Since my math-fu is not very strong please point out any mistakes I've made.



                  Integral approach



                  It seems the exact way to calculate the perimeter/circumference is via the formula $C = 4aE(e)$ where $E$ is the complete elliptic integral of the second kind.



                  I've found a solution using Python. I tried using the numbers from Wikipedia here for consistency:



                  import numpy as np
                  from scipy.special import ellipe
                  a = 149598023000 # semi-major in meter
                  e = 0.0167086 # eccentricity
                  pe = 4 * a * ellipe(e)
                  print(pe) # 936013392131.0


                  So about $936.013 Gm$. That's quite off from the $approx 940Gm$ given by the other answers. I'm wondering why. I know the other methods are approximations but the error seems rather large to me.



                  Ramanujan's first approximation



                  Ramanujan has given two approximations. The first is $C approx pi[3(a+b) - sqrt{10ab + 3(a^2+b^2)} ]$. In Python:



                  import math
                  a = 149598023000
                  e = 0.0167086
                  b = a * math.sqrt(1 - e**2) # derive semi-minor
                  pe = math.pi * ( 3*(a+b) - math.sqrt( (3*a + b) * (a + 3*b) ) )
                  print(pe) # 9.39886493337e+11


                  Here we are at $939.886Gm$ which is closer to the solution from the other answers and the $velocity*time$ approach.



                  Ramanujan's second approximation



                  The formula used by @BillDOe is Ramanujan's second approximation: $C = pi(a+b)(1 + frac{3h}{10 + sqrt{4-3h}})$ with $h = frac{(a-b)^2}{(a+b)^2}$. In Python:



                  a = 149598023000
                  e = 0.0167086
                  b = a * math.sqrt(1 - e**2) # derive semi-minor
                  h = (a-b)**2 / (a+b)**2
                  pe = math.pi * (a+b) * (1 + ((3*h) / (10 + math.sqrt(4 - 3*h))))
                  print(pe) # 9.39886493337e+11


                  That's the same result as the first approximation. The error for Ramanujan's approximations is stated as $h^3$ and $h^5$ respectively. That's about $1.157^{-25}$ and $2.747^{-42}$. I'm not a math geek, so these sound pretty low for me and I wonder why the solution is so different than the integral approach.



                  @RobJeffries's approach



                  I don't know how this is called. The formula is:



                  $$ C = pi(a+b)left[1 + frac{h}{4} + frac{h^2}{64} + frac{h^3}{256} ...right]$$



                  In Python:



                  a = 149598023000
                  e = 0.0167086
                  b = a * math.sqrt(1 - e**2) # derive semi-minor
                  h = (a-b)**2 / (a+b)**2
                  pe = math.pi * (a+b) * (1 + (h/4) + (h**2/64) + (h**3/256))
                  print(pe) # 9.39886493337e+11


                  Again, $939.886Gm$






                  share|improve this answer














                  The distance travelled by Earth in one revolution around the Sun as calculated by the data from Wikipedia's entry about Earth is about $936.013 Gm$ or $939.886Gm$, depending on the approach used for calculation. Since my math-fu is not very strong please point out any mistakes I've made.



                  Integral approach



                  It seems the exact way to calculate the perimeter/circumference is via the formula $C = 4aE(e)$ where $E$ is the complete elliptic integral of the second kind.



                  I've found a solution using Python. I tried using the numbers from Wikipedia here for consistency:



                  import numpy as np
                  from scipy.special import ellipe
                  a = 149598023000 # semi-major in meter
                  e = 0.0167086 # eccentricity
                  pe = 4 * a * ellipe(e)
                  print(pe) # 936013392131.0


                  So about $936.013 Gm$. That's quite off from the $approx 940Gm$ given by the other answers. I'm wondering why. I know the other methods are approximations but the error seems rather large to me.



                  Ramanujan's first approximation



                  Ramanujan has given two approximations. The first is $C approx pi[3(a+b) - sqrt{10ab + 3(a^2+b^2)} ]$. In Python:



                  import math
                  a = 149598023000
                  e = 0.0167086
                  b = a * math.sqrt(1 - e**2) # derive semi-minor
                  pe = math.pi * ( 3*(a+b) - math.sqrt( (3*a + b) * (a + 3*b) ) )
                  print(pe) # 9.39886493337e+11


                  Here we are at $939.886Gm$ which is closer to the solution from the other answers and the $velocity*time$ approach.



                  Ramanujan's second approximation



                  The formula used by @BillDOe is Ramanujan's second approximation: $C = pi(a+b)(1 + frac{3h}{10 + sqrt{4-3h}})$ with $h = frac{(a-b)^2}{(a+b)^2}$. In Python:



                  a = 149598023000
                  e = 0.0167086
                  b = a * math.sqrt(1 - e**2) # derive semi-minor
                  h = (a-b)**2 / (a+b)**2
                  pe = math.pi * (a+b) * (1 + ((3*h) / (10 + math.sqrt(4 - 3*h))))
                  print(pe) # 9.39886493337e+11


                  That's the same result as the first approximation. The error for Ramanujan's approximations is stated as $h^3$ and $h^5$ respectively. That's about $1.157^{-25}$ and $2.747^{-42}$. I'm not a math geek, so these sound pretty low for me and I wonder why the solution is so different than the integral approach.



                  @RobJeffries's approach



                  I don't know how this is called. The formula is:



                  $$ C = pi(a+b)left[1 + frac{h}{4} + frac{h^2}{64} + frac{h^3}{256} ...right]$$



                  In Python:



                  a = 149598023000
                  e = 0.0167086
                  b = a * math.sqrt(1 - e**2) # derive semi-minor
                  h = (a-b)**2 / (a+b)**2
                  pe = math.pi * (a+b) * (1 + (h/4) + (h**2/64) + (h**3/256))
                  print(pe) # 9.39886493337e+11


                  Again, $939.886Gm$







                  share|improve this answer














                  share|improve this answer



                  share|improve this answer








                  edited Nov 11 at 9:55









                  uhoh

                  4,88021456




                  4,88021456










                  answered Nov 11 at 9:48









                  DarkDust

                  1829




                  1829






























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