What is the explanation for the second Kepler's law? Why is the law valid?

Is it that the total energy of a planet equals to the kinetic energy plus the potential energy?

  • $\begingroup$ In case you're looking for an intuitive understanding of what it means, it's just a more formal way of saying "it goes slow when far away, and fast when close to the star" $\endgroup$ Nov 8, 2021 at 5:39
  • $\begingroup$ Energy conservation explains something else. $\endgroup$
    – J.G.
    Nov 8, 2021 at 8:55

3 Answers 3


Kepler's second law (a line segment joining a planet and the Sun sweeps out equal areas during equal intervals of time) is a consequence of conservation of angular momentum. It applies not just to gravity, but also to motion under any central force i.e. a force that is always direct towards a fixed point.

Of course, Kepler himself derived his three laws empirically, by observing the motions of the known planets. It was Newton who proved that the Kepler's first and third laws are a consequence of the inverse-square nature of gravity, and that Kepler's second law applies more generally to any central force.


The area of a sector is $A = \frac{1}{2}r^2 \theta$ and for small changes in time (where the distance of the planet to the sun is assumed constant), the rate that the area is swept out is $\frac{dA}{dt} = \frac{1}{2}r^2 \omega$

For orbits the angular momentum is constant, so $mr^2\omega$ is constant and that means that the rate that the area is swept out is constant too.

So we could compare the area swept out by the earth around the Sun for a day in January (for example), when it's closest and a day in July when it's furthest away and the areas would be the same.

  • $\begingroup$ Please explain computing derivative of A, presented in a form dA/dt. In your formula for A there is no variable t... $\endgroup$
    – Jane N.
    Nov 7, 2021 at 14:07
  • $\begingroup$ @ Jan N. The angle $\theta$ is variable as it's for an orbit $\frac{d\theta}{d t} = \omega $ $\endgroup$ Nov 7, 2021 at 14:58
  • $\begingroup$ Thank you for your answer. However I still do not understand. You have computed a derivative according to a variable t in a way of replacing a Greek letter by a different one, while there was no t at all. Maybe a physicist would understand, but I am a mathematician.. Could you please clarify? Thank you. $\endgroup$
    – Jane N.
    Nov 7, 2021 at 15:26
  • $\begingroup$ @ Jan N. When a planet is in orbit, the angle to an initial reference position changes and varies with time, perhaps you would prefer $\theta(t)$, $\frac{d \theta(t)}{dt}$ is commonly called $\omega$, it is the angular velocity. That's how much the angle in radians changes with time. All the best. $\endgroup$ Nov 7, 2021 at 15:47
  • $\begingroup$ Thank you for your explanation. $\endgroup$
    – Jane N.
    Nov 7, 2021 at 15:56

Kepler's second law

according to Newton's second law

$$m\,\mathbf{\ddot{r}}=F(r)\,\frac{\mathbf r}{r}$$

where $~F(r)~$ is central force

from here

$$\mathbf r\times m\,\mathbf{\ddot{r}}=\frac{F(r)}{r}\left(\mathbf r\times \mathbf r\right)=\mathbf 0$$

with $$\frac{d}{dt}\left(\mathbf r\times m\,\mathbf{\dot{r}}\right)= \mathbf{\dot{r}}\times m\,\mathbf{\dot{r}}+\mathbf r\times m\,\mathbf{\ddot{r}}= \mathbf 0$$


$$ \mathbf r\times m\,\mathbf{\dot{r}}=\mathbf L=\textbf{const.}\tag 1$$

this is the conservation of the angular momentum

additional from equation (1) you obtain that $\mathbf r\cdot \mathbf L=\mathbf 0~$ this means that $~\mathbf r\perp\mathbf L~$ so the mass point m has a planner motion

from equation (1) $~\mathbf r\times \mathbf{\dot{r}}=\frac{\textbf{const.}}{m}$ with polar coordinate you obtain

$$\mathbf r=r\,\begin{bmatrix} \cos(\varphi) \\ \sin(\varphi) \\ 0 \\ \end{bmatrix}\\ \mathbf{\dot{r}}=\left[ \begin {array}{c} \cos \left( \varphi \right) {\dot r}-r\sin \left( \varphi \right) \dot\varphi \\ \sin \left( \varphi \right) {\dot r}+r\cos \left( \varphi \right) \dot\varphi \\ 0\end {array} \right]\quad \Rightarrow\\ |\mathbf r\times \mathbf{\dot{r}}|=r^2\,\dot\varphi $$

or $$\frac{d A}{dt}=\frac 12 r^2\,\dot\varphi\\ \boxed{~A=\frac 12 r^2\,\varphi}$$

Kepler's second law

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