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Kepler's 3rd law relationship

  • Thread starter patcho
  • Start date
5
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Hi, just wondering if anyone could help me with this.

With a plot of log(T) against log(a) where 'T' is the period of satellite and 'a' the distance, I'm not sure what the y-intercept represents, something to do with the constant (4pi^2/GMm) I think, but not sure, any ideas?

Also, the y-intercept for me was -6.5, mean anything?

Thanks in advance

Nb, M>>m and M=ME :smile:
 

HallsofIvy

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Given a and T as average radius and orbit, Kepler's third law says that T2 is proportional to a3. That is, T2= Ka3.

If your graph of log(T) against log(a) is a straight line, then y= mx+ b or
log(T)= m log(a)+ b= log(am)+ b so that T= eb am.
The most important you should be able to derive from that is that your slope, m, should be, approximately, 3/2.

If you are really interested in what that "K" is, here is what I just did:

Since elliptic orbits are a nuisance, let's approximate by a circular orbit: imagine a satellite of mass m orbiting a body of mass M, in a circular orbit of radius R with angular velocity ω. Taking the center of the orbit as the origin of coordinate system, we can write the position vector as x= <R cos(ω t), R sin(ω t)>.
Differentiating gives the speed vector: v= x'= <-Rω sin(ω t), Rω cos(ωt)>.
Differentiating again gives the acceleration vector a= v'= <-Rω2 cos(ωt), -Rω2 sin(ω t) .
That is, the scalar value for acceleration is Rω2.
Since the force of gravity is [tex]\frac{-GmM}{R^2}[/tex] and F= ma
we must have [tex]a= R\omega^2= \frac{GM}{R^2}[/tex] or
[tex]\omega^2= \frac{GM}{R^3}[/tex].

Since sine and cosine both have period [tex]2\pi[/tex], if T is the period of the orbit, [tex]\omega T= 2\pi[/tex] so [tex]\omega = \frac{2\pi}{T}[/tex] and [tex]\omega^2= \frac{GM}{R^3}[/tex] becomes [tex]\frac{T^2}{4\pi^2}= \frac{R^3}{GM}[/tex]. That is the "K" in T2= Ka3 is [tex]\frac{4\pi^2}{GM}[/tex]. Since you graphed log(T) against log(a), your y-intercept is 1/2 the logarithm of that:
[tex]\frac{1}{2}log(\frac{4\pi^2}{GM})[/tex]!

(What I did here, really, was use Newton's gravity formula to derive Kepler's third law. Newton did it the other way around: used Kepler's third law to derive the formula for gravity. Of course, he had to develop calculus in order to be able to do that!)
 
Last edited by a moderator:
5
0
Thanks HallsofIvy! It makes complete sense now, your derivation of kepler 3 really helps. Yeh, I got the slope as 3/2 so I knew I was doing something right :smile:
 

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