File:Newtons proof of Keplers second law.gif

Summary

<a href="https://en.wikipedia.org/wiki/en:Isaac_Newton" class="extiw" title="w:en:Isaac Newton">Isaac Newton</a>'s proof of <a href="https://en.wikipedia.org/wiki/en:Kepler%27s_laws_of_planetary_motion#Second_law" class="extiw" title="w:en:Kepler's laws of planetary motion">Kepler's second law of planetary motion</a>.

At left, a large yellow disk represents a star, and at right, the smaller blue disk represents a planet.

The blue arrow represents the planet's instantaneous orbital velocity. Assuming no force acting upon it, it will move in a straight line at a fixed rate. After some fixed amount of time, it will have reached a new position. Then, at this position, an instantaneous centripetal force (red arrow) is acted upon the planet towards the star, altering its path as shown with the addition of the blue and red arrows (violet arrow).

The planet is shown traversing this new path, as well as a "shadow" (faint, gray disk) of the planet describing its trajectory during the same amount of time if the attractive force hadn't been acted upon it (that is, if it had remained in its original trajectory).

We have then three different positions for the planet, and an alternate position given the lack of the centripetal force. These four positions describe three triangles, which share the star as one of their vertices.

We then select the first and second triangles. We can see they both share the same length for a base (red double arrow). Tracing parallel lines, we can see they also share the same height (in orange). Since the area of a triangle is given by Failed to parse (Missing <code>texvc</code> executable. Please see math/README to configure.): <semantics><mrow class="MJX-TeXAtom-ORD"><mstyle displaystyle="true" scriptlevel="0"><mi>A</mi><mo>=</mo><mrow class="MJX-TeXAtom-ORD"><mstyle displaystyle="false" scriptlevel="0"><mfrac><mrow><mi>b</mi><mi>a</mi><mi>s</mi><mi>e</mi><mo>×<!-- &times; --></mo><mi>h</mi><mi>e</mi><mi>i</mi><mi>g</mi><mi>h</mi><mi>t</mi></mrow><mn>2</mn></mfrac></mstyle></mrow></mstyle></mrow><annotation encoding="application/x-tex">{\displaystyle A={\tfrac {base\times height}{2}}}</annotation></semantics> <img src="https://wikimedia.org/api/rest_v1/media/math/render/svg/14671645e9ae00ed095731783bb696bf64c87b4b" class="mwe-math-fallback-image-inline mw-math-element" aria-hidden="true" style="vertical-align: -1.171ex; width:14.781ex; height:4.009ex;" alt="{\displaystyle A={\tfrac {base\times height}{2}}}">, we conclude that both triangles have the same area.

A similar process is performed with the second and the third triangle, showing that the area of the triangle defined by the altered trajectory is the same as the area of the unaltered trajectory.

The time interval can be made arbitrarily small, until the instantaneous force can be considered acting continuously. Therefore, the line connecting the planet to the sun will always sweep an equal area, as described by Kepler's second law.

Licensing

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