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01 Kepler’s Second Law

Aim

To verify Kepler’s second law

Subjects

Diagram

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Figure 1:.

Equipment

Presentation

By means of the software program an elliptical orbit is projected (see Figure 2A). The speed in the orbit is visualized when in the orbit points are plotted at constant time-intervals. This orbit is also plotted when the time-interval applied is 16 times larger (Figure 2B).

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Figure 2:.

Before, we had this figure projected on a large sheet of cardboard and the elliptical time-segments were cut out (see Diagram). The similarity between the software-image and the cardboard-model is shown to the students. When the segments are placed one after the other on the balance, equal mass is observed. This also means that the areas of the segments are equal (Kepler’s second law).

Explanation

Kepler “found” his law while working on the astronomical data of Tycho Brahe. So in our demonstration the law should arise from watching the areas and “seeing” the equality. Since Newton we can use the law of conservation of angular momentum. Using this law we can explain the statement of Kepler’s second law.

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Figure 3:.

Consider the area in Figure 3 swept by the vector rr in a time Δt\Delta t.

ΔA=12∣r⃗×Δr⃗∣\Delta A = \frac{1}{2}\left| \vec{r} \times \Delta\vec{ r}\right|
(1)

ΔAΔt=12∣r⃗×Δr⃗Δt∣=12∣r⃗×Δv⃗∣\frac{\Delta A}{\Delta t} = \frac{1}{2}\left| \vec{r} \times \frac{\Delta \vec{r}}{\Delta t}\right|=\frac{1}{2}\left| \vec{r} \times \Delta \vec{v}\right|
(2)

Since L⃗=mr⃗×v⃗=const.\vec{L} = m \vec{r}\times\vec{v}=const. →\rightarrow ΔAΔt=12∣L⃗∣m=const.\frac{\Delta A}{\Delta t}=\frac{1}{2}\frac{|\vec{L}|}{m}=const.

So, when Δt\Delta t is constant (equal time-intervals), then ΔA\Delta A is constant.

Simulations

On the internet you can find many simulations that are appropriate. For instance on:

Remarks

Sources

1L10 Universal Gravitation
01 Weighing the Earth
1L20 Orbits
02 Kepler’s Third Law