What does the y-intercept represent in Kepler's 3rd law relationship?

In summary, the conversation was about a graph of log(T) against log(a) where 'T' represents the period of a satellite and 'a' represents the distance. The y-intercept was discussed, with some confusion about its significance. The conversation then delved into the mathematical derivation of Kepler's third law, which states that T2 is proportional to a3. The y-intercept was determined to be 1/2 the logarithm of the constant 'K' in the equation T2= Ka3. The conversation ended with gratitude for the explanation and confirmation that the slope was correctly calculated.
  • #1
patcho
5
0
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:
 
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  • #2
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!)
 
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  • #3
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:
 

1. How is Kepler's 3rd law related to planetary motion?

Kepler's 3rd law states that the square of the orbital period of a planet is directly proportional to the cube of its semi-major axis. This means that the farther a planet is from the sun, the longer its orbital period will be.

2. What is the significance of Kepler's 3rd law?

Kepler's 3rd law provides a mathematical relationship between a planet's orbital period and its distance from the sun. This allows us to predict the orbital periods of planets and better understand the structure of our solar system.

3. How did Kepler's 3rd law contribute to the development of modern astronomy?

Kepler's 3rd law was a major breakthrough in understanding planetary motion and paved the way for Isaac Newton's law of universal gravitation. It also helped to disprove the widely accepted geocentric model of the universe and contributed to the development of the heliocentric model.

4. Can Kepler's 3rd law be applied to other celestial bodies besides planets?

Yes, Kepler's 3rd law can be applied to any two objects orbiting each other, as long as their masses are much smaller compared to the distance between them. This includes moons orbiting planets and even binary star systems.

5. How does Kepler's 3rd law relate to the concept of orbital stability?

Kepler's 3rd law is important in understanding the stability of an orbit. If a planet's orbital period is too short or too long compared to its distance from the sun, it may not be in a stable orbit and may eventually collide with the sun or be flung out of the solar system.

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