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Which type of potentials lead to Kepler's second law "same area in same time"?

$$dA=\frac{1}{2} \vec{r} \times \vec{dr}.$$

$$\frac{dA}{dt}=c=\vec{r} \times \frac{\vec{dr}}{dt}=\vec{r} \times \dot{\vec{r}}. $$

Differentiating both sides we can get a statement on the force field

$\dot{\vec{r}} \times \ddot{\vec{r}}=0, m\dot{\vec{r}} \times m\ddot{\vec{r}}=0,m\ddot{\vec{r}} \times m\dddot{\vec{r}}=0$

I do not know how to proceed from here, however, I believe we should be able to say something on the potential field now since we have a statement on the force field. Perhaps something along the lines of

$\begin{bmatrix} F_x \\ F_y \\ F_z \end{bmatrix} \times \begin{bmatrix} \frac{d}{dt}F_x \\ \frac{d}{dt} F_y \\ \frac{d}{dt} F_z \end{bmatrix} = \begin{bmatrix} \frac{\partial \phi}{\partial x} \\ \frac{\partial \phi}{\partial y}\\ \frac{\partial \phi}{\partial z} \end{bmatrix} \times \begin{bmatrix} \frac{d}{dt}\frac{\partial \phi}{\partial x} \\ \frac{d}{dt} \frac{\partial \phi}{\partial y} \\ \frac{d}{dt} \frac{\partial \phi}{\partial z} \end{bmatrix} =0 $

Edit:

It was brought to my attention that the identity for dA should include a norm $dA=\frac{1}{2}||\vec{r} \times \vec{dr}||$

Applying the same derivative process

$\frac{d^2A}{dt^2}=\frac{1}{2}\frac{1}{||(\vec{r} \times \dot{\vec{r}}||} \frac{d}{dt}(\vec{r} \times \dot{\vec{r}})\cdot (\vec{r} \times \dot{\vec{r}})$

$\frac{d^2A}{dt^2}=\frac{1}{2||(\vec{r} \times \dot{\vec{r}}||} (\dot{\vec{r}} \times \dot{\vec{r}} + \vec{r} \times \ddot{\vec{r}}) \cdot (\vec{r} \times \dot{\vec{r}})=\frac{1}{2||(\vec{r} \times \dot{\vec{r}}||} (\vec{r} \times \ddot{\vec{r}}) \cdot (\vec{r} \times \dot{\vec{r}})=0$

$(\vec{r} \times \ddot{\vec{r}}) \perp (\vec{r} \times \dot{\vec{r}})$

I do not know how to proceed from here

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  • $\begingroup$ The potential is just the gravitational potential. This may be helpful: en.wikipedia.org/wiki/Areal_velocity $\endgroup$
    – PM 2Ring
    Commented Feb 20 at 18:49
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    $\begingroup$ Your mathematical manipulation is wrong. These days, it is usual to relate the areal velocity to angular momentum, because it is angular momentum conservation that gives you this behaviour. But even if you wanted to start with $\frac{\mathrm dA}{\mathrm dt}=\vec r\times\dot{\vec r}$, then $\frac{\mathrm d^2A}{\mathrm dt^2}=\vec r\times\ddot{\vec r}$ and not what you have. Then it is easy to see that any force law that is everywhere radially oriented would have worked. i.e. any force law that comes from a scalar potential solely dependent upon the radial coördinate will have this behaviour. $\endgroup$ Commented Feb 20 at 18:54
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    $\begingroup$ your first equation $dA=\frac12 \vec r\times d\vec r$ cannot be right as written as the RHS is a vector while the LHS is not. So you have at least a typo there. $\endgroup$ Commented Feb 22 at 15:57
  • $\begingroup$ @ZeroTheHero you are right, I should have put a norm on the RHS I am unsure as to how that would effect taking the derivative. Trransofrming the left side into a normal vector of the area dA would not work I believe. I will edit my post accordingly $\endgroup$
    – 16π Cent
    Commented Feb 23 at 11:26
  • $\begingroup$ Related: physics.stackexchange.com/a/517502/219131 $\endgroup$
    – xzd209
    Commented Mar 10 at 22:23

3 Answers 3

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Kepler's second law is just conservation of angular momentum. Angular momentum is conserved for any spherically symmetric potential.


In particular, as you note, the geometrical interpretation of the cross product implies that as the orbit crosses some differential $\mathrm{d}\vec r$, the differential area that it sweeps is $$\mathrm{d}\vec A=\frac{1}{2}\vec r\times\mathrm{d}\vec r$$ (where I let $\vec A$ be a vector area, but that's not important). The rate at which area is swept is then $$\frac{\mathrm{d}\vec A}{\mathrm{d}t} = \frac{1}{2}\vec{r}\times\dot{\vec{r}} = \frac{\vec L}{2},$$ where $\vec L$ is the orbital angular momentum (per mass) about $\vec r=0$. Kepler's second law amounts to saying that $\mathrm{d}^2\vec A/\mathrm{d}t^2=0$, which is thus the same as saying that angular momentum is conserved.

There are fundamental reasons angular momentum conservation is connected to spherical symmetry (see Noether's theorem). But we can also just explicitly note that $$\frac{\mathrm{d}^2\vec A}{\mathrm{d}t^2}=\frac{1}{2}\vec r\times\ddot{\vec r}$$ (since $\dot{\vec r}\times\dot{\vec r}=0$), so Kepler's second law means $\vec r\times\ddot{\vec r}=0$, i.e., that the force $\ddot{\vec r}$ is parallel to the position vector $\vec r$. If the force is the gradient of a potential $\phi$, then this means $\phi$ can only vary with respect to $|\vec r|$, i.e. it must be spherically symmetric.

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  • $\begingroup$ just curious; is angular momentum not conserved for none/some/all non-spherically symmetric potentials? $\endgroup$
    – uhoh
    Commented Feb 21 at 5:26
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    $\begingroup$ @uhoh Symmetry of the potential about an axis is associated with conservation of the angular momentum component along that axis. Spherical symmetry implies symmetry about all axes, but for example, we could consider the potential of a galactic disk instead. Angular momentum about the disk axis is conserved, but other angular momentum components are not. For example, objects inside a disk undergo vertical oscillations with a higher frequency than the azimuthal orbit frequency. $\endgroup$
    – Sten
    Commented Feb 21 at 6:57
  • $\begingroup$ So for the trajectory of a test particle about an irregularly shaped object (e.g. a hydrogen atom around comet 67P), it's angular momentum varies due to torques from the "lumps"? That reminds me of my old and I think incompletely answered question Which mass distributions guarantee two bodies have non-Keplerian orbits? Which non-spherical distributions still allow noncircular Keplerian orbits? $\endgroup$
    – uhoh
    Commented Feb 21 at 7:07
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    $\begingroup$ @uhoh Yeah, the atom's orbital angular momentum would not be conserved, and neither would its orbital energy if the comet is rotating (so the potential is time dependent). $\endgroup$
    – Sten
    Commented Feb 21 at 18:18
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    $\begingroup$ @ZeroTheHero The rate at which area is swept is precisely half the orbital angular momentum (per mass). Why do you think this should fail for non-closed orbits? $\endgroup$
    – Sten
    Commented Feb 22 at 16:27
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I think your procedure is not completely correct... When deriving the Kepler law you have to derive the product as: \begin{eqnarray} \frac{dA}{dt} &=& \frac{1}{2}\vec{r}\wedge\dot{\vec{r}} = c \\ \frac{d^2A}{dt^2} &=& \frac{1}{2}(\dot{\vec{r}} \wedge \dot{\vec{r}} + \vec{r} \wedge \ddot{\vec{r}}) = 0. \end{eqnarray}

The first term is null due to the properties of vectorial product and then, the second Kepler's law says that the movement is governed by a central force perpendicular to the movement, i.e. $\vec{r} \perp \ddot{\vec{r}}$, which is other manner to say that the total angular momentum is conserved, and the movement is contained on a plane. By this way, we do not have any information of the potential.

Nevertheless, I appreciate your physical intuition because it points out to a very interesting equation in physics: the Binet equation. This equation relates the potentials with the trajectory of the orbits. One can feed the equation with a polar-parametrized curve and compute what potential can produce the orbit, or one can put a given potential and solve it for, obtaining the family of curves the potential can handle. The derivation on Wikipedia is easy to understand for your purpose: https://en.wikipedia.org/wiki/Binet_equation

When solving the Binet equation for the Newtonian potential, one obtains the 3 family of curves we observe in the solar system: elliptical, parabolic and hyperbolic, which corresponds with the conic sections. But when varying slightly the Newton potential $V(r) = G\frac{M}{r^{1+\epsilon}}$ with $\epsilon << 1$ the orbits are open instead of closed, in contradiction with Kepler 1st Law. Furthermore, the Bertrand's theorem states that the only two potentials which orbits are closed are the Newtonian and the harmonic potential. (see https://en.wikipedia.org/wiki/Bertrand%27s_theorem)

In conclusion, Kepler's Second Law is applicable to any two-body gravitational system where total angular momentum is conserved but is not linked to a specific gravitational potential. Instead, there exists a connection between trajectories and potential: between the Kepler's First Law and the Newtonian potential.

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starting with
\begin{align*} &\frac{d\,\vec{p}}{dt}=\vec{F}\\ &\vec{L}=\vec{r}\times\vec{p} \end{align*} where $~\vec{L}~$ is the angular momentum and $~\vec{p}=m\vec{\dot{r}}~$ is the linear momentum

obtain the time derivative of the angular momentum \begin{align*} &\frac{d\vec L}{dt}=\underbrace{\vec{\dot{r}}\times\vec{p}}_{=\vec{0}}+\vec{r}\times\vec{\dot{p}}=\vec{r}\times\vec{F} \end{align*}

hence, the angular momentum is conserved $~\left(\frac{d\vec L}{dt}=0\right)~$ only if the force $~\vec{F}~$ is central force: \begin{align*} &\vec{F}=f(r)\,\frac{\vec{r}}{|\vec{r}|}\quad\Rightarrow\\ &\frac{d\vec{L}}{dt}=\vec{r}\times\vec{\dot{r}}=\frac{1}{m}\left(\vec{r}\times f(r)\,\frac{\vec{r}}{|\vec{r}|}\right)=\vec{0} \end{align*}

with \begin{align*} &\frac{d\vec A}{dt}=\frac{1}{2}\left(\vec{r}\times\vec{\dot{r}}\right)=\frac{1}{2}\frac{d\vec{L}}{dt}\quad,\Rightarrow \vec{A}=\frac{1}{2}\vec{L} \end{align*}

Which type of potentials lead to Kepler's second law "same area in same time"?

the potential is \begin{align*} U(r)=-\int f(r)\,dr=-\int \frac{m\,M_s\,G}{r^2}\,dr=\frac{m\,M_s\,G}{r} \end{align*} with $~f(r)~$ the gravitation force between the planet and the sun

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