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

2

$begingroup$

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

edited 4 hours ago

Qmechanic♦

108k122001253

asked 5 hours ago

triple7triple7

133

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
  4 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
  2 hours ago
  

  
  
  
  

add a comment | 

2

$begingroup$

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

edited 4 hours ago

Qmechanic♦

108k122001253

asked 5 hours ago

triple7triple7

133

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
  4 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
  2 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

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

edited 4 hours ago

Qmechanic♦

108k122001253

asked 5 hours ago

triple7triple7

133

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.

$endgroup$

share|cite|improve this question

edited 4 hours ago

Qmechanic♦

108k122001253

asked 5 hours ago

triple7triple7

133

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

edited 4 hours ago

Qmechanic♦

108k122001253

asked 5 hours ago

triple7triple7

133

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.

edited 4 hours ago

Qmechanic♦

108k122001253

edited 4 hours ago

Qmechanic♦

108k122001253

asked 5 hours ago

triple7triple7

133

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.

asked 5 hours ago

triple7triple7

133

3 Answers
3

active

oldest

votes

3

$begingroup$

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

edited 4 hours ago

answered 4 hours ago

Jerry SchirmerJerry Schirmer

31.7k257107

$endgroup$

• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  $begingroup$
  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
  4 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
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

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

  
  
  
  

add a comment | 

1

$begingroup$

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

edited 4 hours ago

answered 4 hours ago

swickrotationswickrotation

615

$endgroup$

add a comment | 

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

answered 4 hours ago

my2ctsmy2cts

5,9342719

$endgroup$

add a comment | 

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3

$begingroup$

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

edited 4 hours ago

answered 4 hours ago

Jerry SchirmerJerry Schirmer

31.7k257107

$endgroup$

• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  $begingroup$
  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
  4 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
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

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

  
  
  
  

add a comment | 

3

$begingroup$

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

edited 4 hours ago

answered 4 hours ago

Jerry SchirmerJerry Schirmer

31.7k257107

$endgroup$

• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  $begingroup$
  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
  4 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
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

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

  
  
  
  

add a comment | 

$begingroup$

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

edited 4 hours ago

answered 4 hours ago

Jerry SchirmerJerry Schirmer

31.7k257107

$endgroup$

share|cite|improve this answer

edited 4 hours ago

answered 4 hours ago

Jerry SchirmerJerry Schirmer

31.7k257107

answered 4 hours ago

Jerry SchirmerJerry Schirmer

31.7k257107

answered 4 hours ago

Jerry SchirmerJerry Schirmer

31.7k257107

1

$begingroup$

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

edited 4 hours ago

answered 4 hours ago

swickrotationswickrotation

615

$endgroup$

add a comment | 

1

$begingroup$

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

edited 4 hours ago

answered 4 hours ago

swickrotationswickrotation

615

$endgroup$

add a comment | 

$begingroup$

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

edited 4 hours ago

answered 4 hours ago

swickrotationswickrotation

615

$endgroup$

share|cite|improve this answer

edited 4 hours ago

answered 4 hours ago

swickrotationswickrotation

615

answered 4 hours ago

swickrotationswickrotation

615

answered 4 hours ago

swickrotationswickrotation

615

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

answered 4 hours ago

my2ctsmy2cts

5,9342719

$endgroup$

add a comment | 

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

answered 4 hours ago

my2ctsmy2cts

5,9342719

$endgroup$

add a comment | 

$begingroup$

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

share|cite|improve this answer

answered 4 hours ago

my2ctsmy2cts

5,9342719

$endgroup$

share|cite|improve this answer

answered 4 hours ago

my2ctsmy2cts

5,9342719

answered 4 hours ago

my2ctsmy2cts

5,9342719

answered 4 hours ago

my2ctsmy2cts

5,9342719