Using Kepler's 3rd law for satellites

[aphh]

<p>Kepler says that P^2 / a^3 = 1, where P is the orbital period of the orbiter and a is the semi-major axis. </p><p>Using this ratio is easy for planets, if using Earth as a model, which has the semi-major axis of 1AU and orbital period of 1 Year. But how do you use the ratio for satellites, both natural and man-made?</p><p>Naturally a man-made satellite orbits the center of earth, which means the semi-major axis would be 6370 * 10^3m + satellite's distance from the ground in metres.</p><p>The orbital period would be one orbit in seconds or minutes.</p><p>So if I knew the other, how do I translate this into a equation, where Kepler's 3rd holds true and I will get the desired result? </p>

 

[UFmbutler]

<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>Kepler says that P^2 / a^3 = 1, where P is the orbital period of the orbiter and a is the semi-major axis. Using this ratio is easy for planets, if using Earth as a model, which has the semi-major axis of 1AU and orbital period of 1 Year. But how do you use the ratio for satellites, both natural and man-made?Naturally a man-made satellite orbits the center of earth, which means the semi-major axis would be 6370 * 10^3m + satellite's distance from the ground in metres.The orbital period would be one orbit in seconds or minutes.So if I knew the other, how do I translate this into a equation, where Kepler's 3rd holds true and I will get the desired result? <br /> Posted by aphh</DIV></p><p>The equation in that form isn't very useful for a problem like this.&nbsp; Try the full version: &nbsp;</p><p>(m1+m2)*P^2=4pi^2*G*a^3 </p> <div class="Discussion_UserSignature"> </div>

 

[aphh]

<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>The equation in that form isn't very useful for a problem like this.&nbsp; Try the full version: &nbsp;(m1+m2)*P^2=4pi^2*G*a^3 <br /> Posted by UFmbutler</DIV></p><p>Thanks, I will try that next.</p><p>However, I found calculators, that seem to use the short form for earth circling satellites (mass of the satellite is insignificant compared to earth's mass, same as earth's mass is insignificant compared to Sun's mass).</p><p>Conversion to SI units seems to be the root of the problem.&nbsp;</p>

 

[derekmcd]

<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>The equation in that form isn't very useful for a problem like this.&nbsp; Try the full version: &nbsp;(m1+m2)*P^2=4pi^2*G*a^3 <br /> Posted by UFmbutler</DIV></p><p>Of course, if you are figuring for a man-made satellite, you can just approximate m2 (satellite) to 1.&nbsp; Depending on how accurate your answer needs to be, you can also approximate G to equal 1.&nbsp;</p> <div class="Discussion_UserSignature"> <div> </div><br /><div><span style="color:#0000ff" class="Apple-style-span">"If something's hard to do, then it's not worth doing." - Homer Simpson</span></div> </div>

 

[aphh]

<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>Of course, if you are figuring for a man-made satellite, you can just approximate m2 (satellite) to 1.&nbsp; Depending on how accurate your answer needs to be, you can also approximate G to equal 1.&nbsp; <br /> Posted by derekmcd</DIV></p><p>Could you provide an example using SI units (m/s)? Let's say the orbiter orbits the earth at 91.4 minutes at a mean distance of 340 km?</p><p>If I just plug these into the equation at m/s, I surely get a ratio, but it still needs conversion? Or shall I use this ratio to calculate either mean distance or orbital period to some other satellite?</p><p>I think the ratio that we get this way is constant?&nbsp;</p>

 

[derekmcd]

<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>Could you provide an example using SI units (m/s)? Let's say the orbiter orbits the earth at 91.4 minutes at a mean distance of 340 km?If I just plug these into the equation at m/s, I surely get a ratio, but it still needs conversion? Or shall I use this ratio to calculate either mean distance or orbital period to some other satellite?I think the ratio that we get this way is constant?&nbsp; <br /> Posted by aphh</DIV></p><p>I would use meter and seconds... 3.4 * 10^5 meters and 5.5 * 10^3 seconds. </p> <div class="Discussion_UserSignature"> <div> </div><br /><div><span style="color:#0000ff" class="Apple-style-span">"If something's hard to do, then it's not worth doing." - Homer Simpson</span></div> </div>

 

[aphh]

<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>I would use meter and seconds... 3.4 * 10^5 meters and 5.5 * 10^3 seconds. <br /> Posted by derekmcd</DIV></p><p>Thanks, I also browsed some literature and it seems that indeed constant G is what results from plugging these numbers into the equation. <br /><br />Now knowing the constant, it is possible to calculate any orbital period or mean distance for any satellite, if one of the factors is known.</p><p>I also saw it is possible to define the orbit very accurately using vector math (vector projection and vector sum), but I'm not there yet.<br />Orbital mech seems quite addictive. <img src="http://sitelife.space.com/ver1.0/content/scripts/tinymce/plugins/emotions/images/smiley-smile.gif" border="0" alt="Smile" title="Smile" /> </p>

 

[DrRocket]

<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>Thanks, I also browsed some literature and it seems that indeed constant G is what results from plugging these numbers into the equation. Now knowing the constant, it is possible to calculate any orbital period or mean distance for any satellite, if one of the factors is known.I also saw it is possible to define the orbit very accurately using vector math (vector projection and vector sum), but I'm not there yet.Orbital mech seems quite addictive. <br />Posted by aphh</DIV></p><p>I think you will find that most treatments use vector analysis and that it makes the task easier in the long run.&nbsp; There are some good books on the subject.</p><p><em>Orbital Mechanics</em> by John Prussing and Bruce Conway is a nice basic engineering text.</p><p><em>An Introduction to the Mathematics and Methods of Astrodynamics</em> by Richard Battin is more complete and more complicated.</p><p>But a budding astrophysicist ought to at least take a look at <em>The Sheer Joy of Celestial Mechanics</em> by Nathaniel Grossman.</p> <div class="Discussion_UserSignature"> </div>

 

[aphh]

<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>I think you will find that most treatments use vector analysis and that it makes the task easier in the long run.&nbsp; There are some good books on the subject.Orbital Mechanics by John Prussing and Bruce Conway is a nice basic engineering text.An Introduction to the Mathematics and Methods of Astrodynamics by Richard Battin is more complete and more complicated.But a budding astrophysicist ought to at least take a look at The Sheer Joy of Celestial Mechanics by Nathaniel Grossman. <br /> Posted by DrRocket</DIV></p><p>Thanks for the suggestions.<br /><br />It appears, that using vector analysis one could accurately predict the position and velocity of the orbiter at any given moment plus the shape of the ellipse.</p><p>Circular orbit is the result of horizontal acceleration, whereas ellipse is the result of angle from the horizontal velocity during acceleration. The greater the angle, the more eccentric the orbit. Too much acceleration, and the orbiter will escape and be lost in space forever. <img src="http://sitelife.space.com/ver1.0/content/scripts/tinymce/plugins/emotions/images/smiley-smile.gif" border="0" alt="Smile" title="Smile" /></p><p>If the orbit is very eccentric, the orbiter spends most of the time far away from the primary, moving slowly, and then accelerating rapidly when swinging past the body that is being orbited.</p><p>The math looks complicated anyway. But Kepler didn't have integrals and vectors, so we're in much better position.&nbsp; &nbsp;</p>

 

[DrRocket]

<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>Thanks for the suggestions.It appears, that using vector analysis one could accurately predict the position and velocity of the orbiter at any given moment plus the shape of the ellipse.Circular orbit is the result of horizontal acceleration, whereas ellipse is the result of angle from the horizontal velocity during acceleration. The greater the angle, the more eccentric the orbit. Too much acceleration, and the orbiter will escape and be lost in space forever. If the orbit is very eccentric, the orbiter spends most of the time far away from the primary, moving slowly, and then accelerating rapidly when swinging past the body that is being orbited.The math looks complicated anyway. But Kepler didn't have integrals and vectors, so we're in much better position.&nbsp; &nbsp; <br />Posted by aphh</DIV></p><p>You don't actually get an expression that gives the position and velocity at a given point in time.&nbsp; In fact no such expression is known.&nbsp; The necessary integrals don't have closed-form expressions.&nbsp; There is a pretty good discussion in Prussing and Conway.</p><p>The mathematics is not particularly complicated.&nbsp; It is actually pretty straightforward and relatively simple.&nbsp; But you do need to understand vector analysis in 3 dimensions and be able to apply basic calculus.</p><p>No, Kepler did not have integrals.&nbsp; In fact, it was Newton's interest in understanding Kepler's laws from a more fundamental viewpoint that led him to invent calculus.</p><p>A slightly, but only slightly, simplified history runs something like this.&nbsp; Tycho Brahe made a lot of careful observations of planetary motion.&nbsp; His assistant, Johannes Kepler, studied that data and made some empirical curve fits to come up with 3 empirical laws of planetary motion that could be used to predict the position of the planets.&nbsp; Newton, in ordre to uderstand Kepler's laws invented the law of universal gravitation, and his 3 laws of motion.&nbsp;&nbsp;&nbsp;&nbsp;To use those laws he developed what would become the theory of differential equations and the calculus necessary to formulate and solve them. </p> <div class="Discussion_UserSignature"> </div>

 

[aphh]

<p>Here's how I verified Kepler's 3rd in short form for earth-orbiting satellites;</p><p>I checked orbital period and semi-major axis for ISS from wikipedia, 5441s and 6732 * 10^3m.</p><p>This gives orbital speed of 2pi * 6732 * 10^3m / 5441s = 7774m/s.</p><p>Using Kepler's 3rd in short form gives constant G for SI units = (5.441 * 10^3s)^2 / (6732 * 10^3m)^3 = 9.703 * 10^-14.</p><p>Now I can use the constant to calculate semi-major axis in SI units for a geosynchronous satellite with an orbital period of 23hrs 56min:</p><p>(86.164 * 10^3s)^2 / 9.703 * 10^-14 = (7.651 * 10^22m)^(1/3) = 42.453 * 10^3m.</p><p>If you subtract earth's mean radius 6375 * 10^3m the altitude of the geosynchronous satellite would be 36 078 kilometers, which is very close to being within 0.7% margin of error for Kepler's 3rd law (when mass2 is considered insignificant compared to mass1).</p><p>Wikipedia tells the altitude for a geosynchronous satellite to be 35 786km.</p><p>So you can use the short form to quickly get some orbital parameters of a satellite. I think. <img src="http://sitelife.space.com/ver1.0/content/scripts/tinymce/plugins/emotions/images/smiley-smile.gif" border="0" alt="Smile" title="Smile" />&nbsp; </p>