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Kepler's first law states that planets revolve around the sun in an ellipse with the sun at one focus of this ellipse. (a special case would be a circular orbit with the sun at the center). The second law states that the areal velocity is a constant. Thus we can write ($dA = c dt$). If we integrate over one complete cycle we find that the area of the orbit, which is proportional to the square of the radius, is proportional to the time period. The third law states that the square of the time period is proportional to the cube of the semi-major axis of the elliptical orbit. My question is, should we swap out "semi-major axis" and replace it with "radius", or is there something missing? if we can, that leads to a contradiction with the second law. however, how can the result obtained from the second law be wrong for circular orbits? What am I missing?

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  • $\begingroup$ Constant means doesn’t change with time; your $c$ can be different for different radii. $\endgroup$
    – G. Smith
    Mar 5, 2021 at 22:59
  • $\begingroup$ it's the same constant for the single orbit we're concerned with in the problem. it actually has a particular value related to the angular momentum of the system and the reduced mass of the objects involved. $\endgroup$
    – user626542
    Mar 6, 2021 at 4:31
  • $\begingroup$ You’re missing the point and confusing yourself. When you say the square of the period is proportional to the cube of the radius, you are talking about different radii having different periods. $\endgroup$
    – G. Smith
    Mar 6, 2021 at 5:18
  • $\begingroup$ It is meaningless to say that (some power of) one period is proportional to (some power of) one radius. That’s like saying 5 is proportional to 3. $\endgroup$
    – G. Smith
    Mar 6, 2021 at 5:30

2 Answers 2

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For circular orbits, $$\frac{dA}{dt}=\frac{L}{2\mu}\Rightarrow \pi R^2=\frac{L}{2\mu}T$$ Further $$E=\frac{\mu C^2}{2L^2}\ \ \ \text{For circular orbits}$$ From now on I'm just going to keep track of proportionality. $$R^2\propto LT\propto\frac{1}{\sqrt{E}}T$$ But further we known that $R\propto 1/E$ $$R^{3/2}\propto T\Rightarrow \boxed{R^3\propto T^2}$$

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  • $\begingroup$ What is $C$? .. $\endgroup$
    – G. Smith
    Mar 5, 2021 at 22:56
  • $\begingroup$ It's a constant which is used for $Gm_1m_2$. $\endgroup$ Mar 5, 2021 at 22:59
  • $\begingroup$ Thanks. It’s not common in textbooks in my country. $\endgroup$
    – G. Smith
    Mar 5, 2021 at 23:00
  • $\begingroup$ But the angular momentum is conserved. It's a constant. so R squared is proportional to a constant times T which means it's proportional to T. Something is off about this. $\endgroup$
    – user626542
    Mar 6, 2021 at 4:33
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My question is, should we swap out "semi-major axis" and replace it with "radius"...

We can always do that. As you noted, circle is a special case of ellipse.

The second law states that the areal velocity is a constant. Thus we can write (dA=cdt).

Correct. Here c is areal speed.

If we integrate over one complete cycle we find that the area of the orbit, which is proportional to the square of the radius, is proportional to the time period.

Integrating we will get $\pi r^2 = cT$

From this we cannot conclude that $T$ is proportional to $r^2$. The reason is areal velocity $c$ is also dependent on $r$. I hope the source of confusion is clear.

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