Kepler´s 2nd law -- Do any two planets sweep out equal area in equal time?

[LuisBabboni]

TL;DR

Any planet sweeps equal area in equal time?

I mean. For example, Earth in one month sweep the same area* than Jupiter in one month?

*The line joining the Earth with the Sun than the line joining Jupiter with the Sun.

I think yes, but is not what 2nd law says. I think in the fact that the aceleration just depends on the distance to the Sun.

Thanks!

 

[PeroK]

TL;DR Summary: Any planet sweeps equal area in equal time?

I mean. For example, Earth in one month seep the same area* than Jupiter in one month?

*The line joining the Earth with the Sun than the line joining Jupiter with the Sun.

I think yes, but is not what 2nd law says. I think in the fact that the aceleration just depends on the distance to the Sun.

Thanks!

Each planet sweeps out the same area every month. But not the same as every other planet!

 

[LuisBabboni]

OK. I understand now why.
At the same distance, the aceleration is the same, but not necesary the velocity, so the orbits are not the same. And being different the velocities, the sweeped areas are not the same!

Thanks!

 

[PeroK]

OK. I understand now why.
At the same distance, the aceleration is the same, but not necesary the velocity, so the orbits are not the same. And being different the velocities, the sweeped areas are not the same!

Thanks!

Kepler's second law is really conservation of angular momentum. The angular momentum of each planet is conserved (which means it is constant over time). But, each planet has a different angular momentum. And, geometrically, angular momentum is proportional to the rate at which the planet sweeps out the area of its circular or elliptical orbit. See, for example:

http://burro.case.edu/Academics/Astr221/Gravity/kep2rev.htm

 

[Orodruin]

But, each planet has a different angular momentum. And, geometrically, angular momentum is proportional to the rate at which the planet sweeps out the area of its circular or elliptical orbit.

It is worth it to point out that you could have different angular momentum and still sweep different areas per time - if the masses are different - so the important thing is not that the angular momentum differs. The important thing is that the angular momentum per mass differs.

 

[LuisBabboni]

I do not understand why the mass is important.
I think that if any planet, no matter its mass, is at any point at any velocity, the orbit is just defined. I´m wrong?

 

[LuisBabboni]

In what you linked, the result have L/m; L/m made m disapear. Im right?

 

[PeroK]

In what you linked, the result have L/m; L/m made m disapear. Im right?

Yes. The orbit and area are independent of the mass of the planet. So, angular momentum per unit mass ($L/m$) is the important quantity.

 

[Orodruin]

I do not understand why the mass is important.
I think that if any planet, no matter its mass, is at any point at any velocity, the orbit is just defined. I´m wrong?

It is not, that was my point. (As long as the primary has significantly larger mass)

 

[Trollfaz]

Kepler's second law is really conservation of angular momentum. The angular momentum of each planet is conserved (which means it is constant over time). But, each planet has a different angular momentum. And, geometrically, angular momentum is proportional to the rate at which the planet sweeps out the area of its circular or elliptical orbit. See, for example:

http://burro.case.edu/Academics/Astr221/Gravity/kep2rev.htm

Mathematically speaking, each orbit can be modelled in polar coordinates with sun in the center. So r, distance from the Sun is $f(\theta)$. So
$$\frac{da}{dt}=k$$
$$\frac{da}{dt}=\int^{\theta+\frac{d\theta}{dt}}_{\theta}\int^{r=f(\theta)}_{0} r dr d\theta =k$$

 

[Orodruin]

$$\frac{da}{dt}=\int^{\theta+\frac{d\theta}{dt}}_{\theta}\int^{r=f(\theta)}_{0} r dr d\theta =k$$

There is (at least) one dimensional inconsistency in this expression.

 

[Ibix]

Mathematically speaking, each orbit can be modelled in polar coordinates with sun in the center. So r, distance from the Sun is $f(\theta)$. So
$$\frac{da}{dt}=k$$
$$\frac{da}{dt}=\int^{\theta+\frac{d\theta}{dt}}_{\theta}\int^{r=f(\theta)}_{0} r dr d\theta =k$$

The area swept out by a line of length $r$ moving through a small angle $d\theta$ is $da=\frac 12 r^2d\theta$. Using the chain rule to write $d\theta=\frac{d\theta}{dt}dt$ we can write $da=\frac 12 r^2\frac{d\theta}{dt}dt$, and obviously the angle swept out in a finite time is the integral of that quantity with appropriate limits. If the area swept out is to be a constant but $r$ may vary as a function of time then we must require that $r^2\frac{d\theta}{dt}=\mathrm{const}$, which is just the law of conservation of angular momentum if the constant is $L/m$.