Kepler's 3rd law: ratios don't fit data












3












$begingroup$


I have been looking at satellite orbits around the earth, or any object around any planet in fact, and am trying to find the orbital radius, or semi major length of a given satellite.



Kepler's third law gives the equation $P^2 = a^3$ where $P$ is the period of orbit and $a$ the distance.



I have a table of satellites currently orbiting the earth, as well as their altitude in the sky on their geosynchronous trajectory. One in particular is 99.9 and has an altitude of 705.



By solving the equation for $a$, I get $a = (P^2)^{1/3}$.



When I plug in the numbers, they don't correspond.



So my questions are:




  1. Are there unit standards I need for both $P$ and $a$? Currently $P$ is in minutes, $a$ in kilometres.

  2. Am I missing something, like Newton's universal gravitational constant? I get a page deriving Kepler's third law using this constant.






newtonian-mechanics newtonian-gravity orbital-motion celestial-mechanics satellites







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triple7 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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  • $begingroup$
    hyperphysics.phy-astr.gsu.edu/hbase/kepler.html#c6
    $endgroup$
    – Kyle Kanos
    10 hours ago










  • $begingroup$
    The equality only holds in certain units since its dimensionally inhomogeneous. In particular, if you use Earth years and the Earth-Sun distance (i.e. 1a.u.) then it's true, so it must be true in those specific units.
    $endgroup$
    – jacob1729
    7 hours ago






  • 1




    $begingroup$
    BTW, do you understand that $a$ is not altitude but rather distance from the center of the Earth?
    $endgroup$
    – G. Smith
    4 hours ago


















3












$begingroup$


I have been looking at satellite orbits around the earth, or any object around any planet in fact, and am trying to find the orbital radius, or semi major length of a given satellite.



Kepler's third law gives the equation $P^2 = a^3$ where $P$ is the period of orbit and $a$ the distance.



I have a table of satellites currently orbiting the earth, as well as their altitude in the sky on their geosynchronous trajectory. One in particular is 99.9 and has an altitude of 705.



By solving the equation for $a$, I get $a = (P^2)^{1/3}$.



When I plug in the numbers, they don't correspond.



So my questions are:




  1. Are there unit standards I need for both $P$ and $a$? Currently $P$ is in minutes, $a$ in kilometres.

  2. Am I missing something, like Newton's universal gravitational constant? I get a page deriving Kepler's third law using this constant.






newtonian-mechanics newtonian-gravity orbital-motion celestial-mechanics satellites







share|cite|improve this question









New contributor




triple7 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$












  • $begingroup$
    hyperphysics.phy-astr.gsu.edu/hbase/kepler.html#c6
    $endgroup$
    – Kyle Kanos
    10 hours ago










  • $begingroup$
    The equality only holds in certain units since its dimensionally inhomogeneous. In particular, if you use Earth years and the Earth-Sun distance (i.e. 1a.u.) then it's true, so it must be true in those specific units.
    $endgroup$
    – jacob1729
    7 hours ago






  • 1




    $begingroup$
    BTW, do you understand that $a$ is not altitude but rather distance from the center of the Earth?
    $endgroup$
    – G. Smith
    4 hours ago
















3












3








3





$begingroup$


I have been looking at satellite orbits around the earth, or any object around any planet in fact, and am trying to find the orbital radius, or semi major length of a given satellite.



Kepler's third law gives the equation $P^2 = a^3$ where $P$ is the period of orbit and $a$ the distance.



I have a table of satellites currently orbiting the earth, as well as their altitude in the sky on their geosynchronous trajectory. One in particular is 99.9 and has an altitude of 705.



By solving the equation for $a$, I get $a = (P^2)^{1/3}$.



When I plug in the numbers, they don't correspond.



So my questions are:




  1. Are there unit standards I need for both $P$ and $a$? Currently $P$ is in minutes, $a$ in kilometres.

  2. Am I missing something, like Newton's universal gravitational constant? I get a page deriving Kepler's third law using this constant.






newtonian-mechanics newtonian-gravity orbital-motion celestial-mechanics satellites







share|cite|improve this question









New contributor




triple7 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$




I have been looking at satellite orbits around the earth, or any object around any planet in fact, and am trying to find the orbital radius, or semi major length of a given satellite.



Kepler's third law gives the equation $P^2 = a^3$ where $P$ is the period of orbit and $a$ the distance.



I have a table of satellites currently orbiting the earth, as well as their altitude in the sky on their geosynchronous trajectory. One in particular is 99.9 and has an altitude of 705.



By solving the equation for $a$, I get $a = (P^2)^{1/3}$.



When I plug in the numbers, they don't correspond.



So my questions are:




  1. Are there unit standards I need for both $P$ and $a$? Currently $P$ is in minutes, $a$ in kilometres.

  2. Am I missing something, like Newton's universal gravitational constant? I get a page deriving Kepler's third law using this constant.






newtonian-mechanics newtonian-gravity orbital-motion celestial-mechanics satellites




newtonian-mechanics newtonian-gravity orbital-motion celestial-mechanics satellites






share|cite|improve this question









New contributor




triple7 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question









New contributor




triple7 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this question




share|cite|improve this question








edited 9 hours ago









Qmechanic♦

108k122001253




108k122001253






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Check out our Code of Conduct.









asked 10 hours ago









triple7triple7

183




183




New contributor




triple7 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





New contributor





triple7 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






triple7 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.












  • $begingroup$
    hyperphysics.phy-astr.gsu.edu/hbase/kepler.html#c6
    $endgroup$
    – Kyle Kanos
    10 hours ago










  • $begingroup$
    The equality only holds in certain units since its dimensionally inhomogeneous. In particular, if you use Earth years and the Earth-Sun distance (i.e. 1a.u.) then it's true, so it must be true in those specific units.
    $endgroup$
    – jacob1729
    7 hours ago






  • 1




    $begingroup$
    BTW, do you understand that $a$ is not altitude but rather distance from the center of the Earth?
    $endgroup$
    – G. Smith
    4 hours ago




















  • $begingroup$
    hyperphysics.phy-astr.gsu.edu/hbase/kepler.html#c6
    $endgroup$
    – Kyle Kanos
    10 hours ago










  • $begingroup$
    The equality only holds in certain units since its dimensionally inhomogeneous. In particular, if you use Earth years and the Earth-Sun distance (i.e. 1a.u.) then it's true, so it must be true in those specific units.
    $endgroup$
    – jacob1729
    7 hours ago






  • 1




    $begingroup$
    BTW, do you understand that $a$ is not altitude but rather distance from the center of the Earth?
    $endgroup$
    – G. Smith
    4 hours ago


















$begingroup$
hyperphysics.phy-astr.gsu.edu/hbase/kepler.html#c6
$endgroup$
– Kyle Kanos
10 hours ago




$begingroup$
hyperphysics.phy-astr.gsu.edu/hbase/kepler.html#c6
$endgroup$
– Kyle Kanos
10 hours ago












$begingroup$
The equality only holds in certain units since its dimensionally inhomogeneous. In particular, if you use Earth years and the Earth-Sun distance (i.e. 1a.u.) then it's true, so it must be true in those specific units.
$endgroup$
– jacob1729
7 hours ago




$begingroup$
The equality only holds in certain units since its dimensionally inhomogeneous. In particular, if you use Earth years and the Earth-Sun distance (i.e. 1a.u.) then it's true, so it must be true in those specific units.
$endgroup$
– jacob1729
7 hours ago




1




1




$begingroup$
BTW, do you understand that $a$ is not altitude but rather distance from the center of the Earth?
$endgroup$
– G. Smith
4 hours ago






$begingroup$
BTW, do you understand that $a$ is not altitude but rather distance from the center of the Earth?
$endgroup$
– G. Smith
4 hours ago












3 Answers
3






active

oldest

votes


















7












$begingroup$

that equality should be a proportional to sign. In particular, in SI, the squared period has units of seconds squared, and the semi-major radius of of the orbit cubed is in meters cubed, so they can't be equal.



Instead, I'd be checking whether $T^{2}/a^{3}$ is constant for different satellites orbiting the same object (Like the ISS and the moon, for example)






share|cite|improve this answer











$endgroup$









  • 6




    $begingroup$
    Ok, as I'm blind and use a screenreader, I didn't realise it was a proportion sign. And most sites only show images for formulas, which are inaccessible too. Also, I am assuming you use latec or math jacks for the symbols here, which also make the screen reader hang In a cycle. Could you give me a simple ASCII form of the distance given a period?
    $endgroup$
    – triple7
    10 hours ago






  • 3




    $begingroup$
    @triple7: Keplers law says that the square of the period divided by the cube of the distance is equal to a constant for every central body. so, t squared divided by a cubed should be the same for the ISS and for the moon. After Kepler, Newton was able to come up with a theoretical formula to predict what this constant should be, which is 4 * pi squared / (G * M), where G is Newton's constant, and M is the mass of the central body.
    $endgroup$
    – Jerry Schirmer
    9 hours ago






  • 1




    $begingroup$
    Yep, finally found it. Thanks
    $endgroup$
    – triple7
    9 hours ago



















2












$begingroup$

The general form of Kepler's period law is $T^2 = frac{4pi^2}{G(M+m)}a^3$. Often, we make the simplifying assumption that $M>>m$, so that $M+m approx M$.



Kepler's period law only takes the form $T^2 = a^3$ (forgetting about the units) when you use certain quantities- in this case, $M$ being solar mass, $T$ being an Earth year, and $a$ being an astronomical unit.



Try plugging into the equation for the mass of earth (and don't bother with the satellite mass) and use units of meters and seconds. See if you get the right result!






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    Yes, I used seconds and meters. R would be planet radius and satellite altitude, and the numbers seem to correspond. I needed this parameter to calculate the distance between two points by lat long alt, so i’m subtracting earth’s radius to the result of the kepler equation. I just wish math reading blind wasn’t so convoluted :/
    $endgroup$
    – triple7
    23 mins ago



















0












$begingroup$

Kepler's third law claims that $p^2 propto a^3$. The equality sign you use is incorrect.






share|cite|improve this answer









$endgroup$














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    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    7












    $begingroup$

    that equality should be a proportional to sign. In particular, in SI, the squared period has units of seconds squared, and the semi-major radius of of the orbit cubed is in meters cubed, so they can't be equal.



    Instead, I'd be checking whether $T^{2}/a^{3}$ is constant for different satellites orbiting the same object (Like the ISS and the moon, for example)






    share|cite|improve this answer











    $endgroup$









    • 6




      $begingroup$
      Ok, as I'm blind and use a screenreader, I didn't realise it was a proportion sign. And most sites only show images for formulas, which are inaccessible too. Also, I am assuming you use latec or math jacks for the symbols here, which also make the screen reader hang In a cycle. Could you give me a simple ASCII form of the distance given a period?
      $endgroup$
      – triple7
      10 hours ago






    • 3




      $begingroup$
      @triple7: Keplers law says that the square of the period divided by the cube of the distance is equal to a constant for every central body. so, t squared divided by a cubed should be the same for the ISS and for the moon. After Kepler, Newton was able to come up with a theoretical formula to predict what this constant should be, which is 4 * pi squared / (G * M), where G is Newton's constant, and M is the mass of the central body.
      $endgroup$
      – Jerry Schirmer
      9 hours ago






    • 1




      $begingroup$
      Yep, finally found it. Thanks
      $endgroup$
      – triple7
      9 hours ago
















    7












    $begingroup$

    that equality should be a proportional to sign. In particular, in SI, the squared period has units of seconds squared, and the semi-major radius of of the orbit cubed is in meters cubed, so they can't be equal.



    Instead, I'd be checking whether $T^{2}/a^{3}$ is constant for different satellites orbiting the same object (Like the ISS and the moon, for example)






    share|cite|improve this answer











    $endgroup$









    • 6




      $begingroup$
      Ok, as I'm blind and use a screenreader, I didn't realise it was a proportion sign. And most sites only show images for formulas, which are inaccessible too. Also, I am assuming you use latec or math jacks for the symbols here, which also make the screen reader hang In a cycle. Could you give me a simple ASCII form of the distance given a period?
      $endgroup$
      – triple7
      10 hours ago






    • 3




      $begingroup$
      @triple7: Keplers law says that the square of the period divided by the cube of the distance is equal to a constant for every central body. so, t squared divided by a cubed should be the same for the ISS and for the moon. After Kepler, Newton was able to come up with a theoretical formula to predict what this constant should be, which is 4 * pi squared / (G * M), where G is Newton's constant, and M is the mass of the central body.
      $endgroup$
      – Jerry Schirmer
      9 hours ago






    • 1




      $begingroup$
      Yep, finally found it. Thanks
      $endgroup$
      – triple7
      9 hours ago














    7












    7








    7





    $begingroup$

    that equality should be a proportional to sign. In particular, in SI, the squared period has units of seconds squared, and the semi-major radius of of the orbit cubed is in meters cubed, so they can't be equal.



    Instead, I'd be checking whether $T^{2}/a^{3}$ is constant for different satellites orbiting the same object (Like the ISS and the moon, for example)






    share|cite|improve this answer











    $endgroup$



    that equality should be a proportional to sign. In particular, in SI, the squared period has units of seconds squared, and the semi-major radius of of the orbit cubed is in meters cubed, so they can't be equal.



    Instead, I'd be checking whether $T^{2}/a^{3}$ is constant for different satellites orbiting the same object (Like the ISS and the moon, for example)







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited 9 hours ago

























    answered 10 hours ago









    Jerry SchirmerJerry Schirmer

    31.7k257107




    31.7k257107








    • 6




      $begingroup$
      Ok, as I'm blind and use a screenreader, I didn't realise it was a proportion sign. And most sites only show images for formulas, which are inaccessible too. Also, I am assuming you use latec or math jacks for the symbols here, which also make the screen reader hang In a cycle. Could you give me a simple ASCII form of the distance given a period?
      $endgroup$
      – triple7
      10 hours ago






    • 3




      $begingroup$
      @triple7: Keplers law says that the square of the period divided by the cube of the distance is equal to a constant for every central body. so, t squared divided by a cubed should be the same for the ISS and for the moon. After Kepler, Newton was able to come up with a theoretical formula to predict what this constant should be, which is 4 * pi squared / (G * M), where G is Newton's constant, and M is the mass of the central body.
      $endgroup$
      – Jerry Schirmer
      9 hours ago






    • 1




      $begingroup$
      Yep, finally found it. Thanks
      $endgroup$
      – triple7
      9 hours ago














    • 6




      $begingroup$
      Ok, as I'm blind and use a screenreader, I didn't realise it was a proportion sign. And most sites only show images for formulas, which are inaccessible too. Also, I am assuming you use latec or math jacks for the symbols here, which also make the screen reader hang In a cycle. Could you give me a simple ASCII form of the distance given a period?
      $endgroup$
      – triple7
      10 hours ago






    • 3




      $begingroup$
      @triple7: Keplers law says that the square of the period divided by the cube of the distance is equal to a constant for every central body. so, t squared divided by a cubed should be the same for the ISS and for the moon. After Kepler, Newton was able to come up with a theoretical formula to predict what this constant should be, which is 4 * pi squared / (G * M), where G is Newton's constant, and M is the mass of the central body.
      $endgroup$
      – Jerry Schirmer
      9 hours ago






    • 1




      $begingroup$
      Yep, finally found it. Thanks
      $endgroup$
      – triple7
      9 hours ago








    6




    6




    $begingroup$
    Ok, as I'm blind and use a screenreader, I didn't realise it was a proportion sign. And most sites only show images for formulas, which are inaccessible too. Also, I am assuming you use latec or math jacks for the symbols here, which also make the screen reader hang In a cycle. Could you give me a simple ASCII form of the distance given a period?
    $endgroup$
    – triple7
    10 hours ago




    $begingroup$
    Ok, as I'm blind and use a screenreader, I didn't realise it was a proportion sign. And most sites only show images for formulas, which are inaccessible too. Also, I am assuming you use latec or math jacks for the symbols here, which also make the screen reader hang In a cycle. Could you give me a simple ASCII form of the distance given a period?
    $endgroup$
    – triple7
    10 hours ago




    3




    3




    $begingroup$
    @triple7: Keplers law says that the square of the period divided by the cube of the distance is equal to a constant for every central body. so, t squared divided by a cubed should be the same for the ISS and for the moon. After Kepler, Newton was able to come up with a theoretical formula to predict what this constant should be, which is 4 * pi squared / (G * M), where G is Newton's constant, and M is the mass of the central body.
    $endgroup$
    – Jerry Schirmer
    9 hours ago




    $begingroup$
    @triple7: Keplers law says that the square of the period divided by the cube of the distance is equal to a constant for every central body. so, t squared divided by a cubed should be the same for the ISS and for the moon. After Kepler, Newton was able to come up with a theoretical formula to predict what this constant should be, which is 4 * pi squared / (G * M), where G is Newton's constant, and M is the mass of the central body.
    $endgroup$
    – Jerry Schirmer
    9 hours ago




    1




    1




    $begingroup$
    Yep, finally found it. Thanks
    $endgroup$
    – triple7
    9 hours ago




    $begingroup$
    Yep, finally found it. Thanks
    $endgroup$
    – triple7
    9 hours ago











    2












    $begingroup$

    The general form of Kepler's period law is $T^2 = frac{4pi^2}{G(M+m)}a^3$. Often, we make the simplifying assumption that $M>>m$, so that $M+m approx M$.



    Kepler's period law only takes the form $T^2 = a^3$ (forgetting about the units) when you use certain quantities- in this case, $M$ being solar mass, $T$ being an Earth year, and $a$ being an astronomical unit.



    Try plugging into the equation for the mass of earth (and don't bother with the satellite mass) and use units of meters and seconds. See if you get the right result!






    share|cite|improve this answer











    $endgroup$









    • 1




      $begingroup$
      Yes, I used seconds and meters. R would be planet radius and satellite altitude, and the numbers seem to correspond. I needed this parameter to calculate the distance between two points by lat long alt, so i’m subtracting earth’s radius to the result of the kepler equation. I just wish math reading blind wasn’t so convoluted :/
      $endgroup$
      – triple7
      23 mins ago
















    2












    $begingroup$

    The general form of Kepler's period law is $T^2 = frac{4pi^2}{G(M+m)}a^3$. Often, we make the simplifying assumption that $M>>m$, so that $M+m approx M$.



    Kepler's period law only takes the form $T^2 = a^3$ (forgetting about the units) when you use certain quantities- in this case, $M$ being solar mass, $T$ being an Earth year, and $a$ being an astronomical unit.



    Try plugging into the equation for the mass of earth (and don't bother with the satellite mass) and use units of meters and seconds. See if you get the right result!






    share|cite|improve this answer











    $endgroup$









    • 1




      $begingroup$
      Yes, I used seconds and meters. R would be planet radius and satellite altitude, and the numbers seem to correspond. I needed this parameter to calculate the distance between two points by lat long alt, so i’m subtracting earth’s radius to the result of the kepler equation. I just wish math reading blind wasn’t so convoluted :/
      $endgroup$
      – triple7
      23 mins ago














    2












    2








    2





    $begingroup$

    The general form of Kepler's period law is $T^2 = frac{4pi^2}{G(M+m)}a^3$. Often, we make the simplifying assumption that $M>>m$, so that $M+m approx M$.



    Kepler's period law only takes the form $T^2 = a^3$ (forgetting about the units) when you use certain quantities- in this case, $M$ being solar mass, $T$ being an Earth year, and $a$ being an astronomical unit.



    Try plugging into the equation for the mass of earth (and don't bother with the satellite mass) and use units of meters and seconds. See if you get the right result!






    share|cite|improve this answer











    $endgroup$



    The general form of Kepler's period law is $T^2 = frac{4pi^2}{G(M+m)}a^3$. Often, we make the simplifying assumption that $M>>m$, so that $M+m approx M$.



    Kepler's period law only takes the form $T^2 = a^3$ (forgetting about the units) when you use certain quantities- in this case, $M$ being solar mass, $T$ being an Earth year, and $a$ being an astronomical unit.



    Try plugging into the equation for the mass of earth (and don't bother with the satellite mass) and use units of meters and seconds. See if you get the right result!







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited 9 hours ago

























    answered 10 hours ago









    swickrotationswickrotation

    715




    715








    • 1




      $begingroup$
      Yes, I used seconds and meters. R would be planet radius and satellite altitude, and the numbers seem to correspond. I needed this parameter to calculate the distance between two points by lat long alt, so i’m subtracting earth’s radius to the result of the kepler equation. I just wish math reading blind wasn’t so convoluted :/
      $endgroup$
      – triple7
      23 mins ago














    • 1




      $begingroup$
      Yes, I used seconds and meters. R would be planet radius and satellite altitude, and the numbers seem to correspond. I needed this parameter to calculate the distance between two points by lat long alt, so i’m subtracting earth’s radius to the result of the kepler equation. I just wish math reading blind wasn’t so convoluted :/
      $endgroup$
      – triple7
      23 mins ago








    1




    1




    $begingroup$
    Yes, I used seconds and meters. R would be planet radius and satellite altitude, and the numbers seem to correspond. I needed this parameter to calculate the distance between two points by lat long alt, so i’m subtracting earth’s radius to the result of the kepler equation. I just wish math reading blind wasn’t so convoluted :/
    $endgroup$
    – triple7
    23 mins ago




    $begingroup$
    Yes, I used seconds and meters. R would be planet radius and satellite altitude, and the numbers seem to correspond. I needed this parameter to calculate the distance between two points by lat long alt, so i’m subtracting earth’s radius to the result of the kepler equation. I just wish math reading blind wasn’t so convoluted :/
    $endgroup$
    – triple7
    23 mins ago











    0












    $begingroup$

    Kepler's third law claims that $p^2 propto a^3$. The equality sign you use is incorrect.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      Kepler's third law claims that $p^2 propto a^3$. The equality sign you use is incorrect.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Kepler's third law claims that $p^2 propto a^3$. The equality sign you use is incorrect.






        share|cite|improve this answer









        $endgroup$



        Kepler's third law claims that $p^2 propto a^3$. The equality sign you use is incorrect.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 10 hours ago









        my2ctsmy2cts

        5,9642719




        5,9642719






















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