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Is there a simple way to explain how Kepler's third law follows from the inverse square law that of gravity (and laws of motion)

For example for Kepler's second law we can say it's because Gravity is a central force

What is it about the law of gravity which ultimately gives rise to Kepler's third law?

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  • $\begingroup$ Possible duplicates: physics.stackexchange.com/q/57527/2451 and links therein. $\endgroup$ – Qmechanic Mar 21 '16 at 10:27
  • $\begingroup$ Look at the answer by ABC at the link given by Qmechanic. If that doesn't satisfy you, reword your question to be more specific. $\endgroup$ – Bill N Nov 30 '17 at 17:50
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To show that the inverse square law of gravity leads to Kepler's 3rd law for an elliptical orbit which obeys the first and second laws, we make use of the conservation of mechanical energy: $$E_{orbit}=\frac{1}{2}mv^2-\frac{GMm}{r},$$ where $E_{orbit}$ is the constant energy, $r$ is the instantaneous distance of the planet from the sun, $v$ is the instantaneous speed, $G$ is the universal gravitational constant, $M$ is the sun mass, and $m$ is the planet mass.

Let us choose two convenient points along the ellipse:

  • the perihelion ($r_1=a(1-e)$), where $e$ is the eccentricity and the $a$ is the semi-major axis. At this point the angular momentum $L=mv_1a(1-e)$.
  • the aphelion ($r_2=a(1+e)$). At this point $L=mv_2a(1+e)$.

We equate the energies at these two point, and because of the 2nd Law, we know that the angular momentum magnitudes, $L$, are equal: $$\frac{L^2}{2ma^2(1+e)^2}-\frac{GMm}{a(1+e)}=\frac{L^2}{2ma^2(1-e)^2}-\frac{GMm}{a(1-e)}.$$

After some algebra we arrive at $$L^2=GMm^2a(1-e^2).$$

Next, we examine the geometry of the ellipse and we find the relationship between the semi-major axis and the semi-minor axis lengths are $$b^2=a^2(1-e^2).$$

Using this to eliminate the eccentricity gives us $$aL^2=GMm^2b^2.$$

Finally, according to Kepler's 2nd Law (and conservation of angular momentum) we know that the ratio of the area of the ellipse to the period of revolution, $T$, is a constant, proportional to the specific angular momentum, $L/m$.: $$\frac{\pi a b}{T}=\frac{k_0L}{m}.$$ Solving for the square of the period we get $$T^2={\pi^2a^2b^2}{k_0^2} \to T^2=\frac{\pi^2a^2\left(\frac{aL^2}{GMm^2}\right)}{\frac{k_0^2L^2}{m^2}}.$$ Resolving the algebra we get $$T^2=\frac{\pi^2}{k_0^2GM}a^3.$$

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Kepler's third law states that the orbital period $P$ and the mean planet-star distance $a$ obey the relation

$$ P^2\propto a^3$$

Take the square root of both sides to get $P\propto a\sqrt{a}$. Now, the linear $a$ term is how the length of the orbit varies with $a$, and the $\sqrt{a}$ term is how the inverse of the orbital velocity varies with $a$.

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Newton's universal law of gravity is: $$F=G\frac{Mm}{r^2},$$ where $G$ is a constant, $M$ and $m$ are the masses of the two bodies and $r$ is the distance between the two bodies (for example the sun and the earth).

Kepler's third law states that the orbit time of a planet is related to the distance to the sun so that $$\frac{T^2}{a^3}=\text{constant}$$ where $T$ is the period (the time it takes to complete a full orbit) and $a$ is the average distance to the sun (the orbits of planets are ellipses. That means the distance to the sun is not constant. $a$ is the average distance of the planet to the sun). You can think of this as a circle with radius $a$ with the sun at its center. The average distance is of course the same for the whole orbit.

The formula thus states that the cube of the average distance is equal to the period squared times a constant ($a^3\cdot C=T^2$, where $C$ is the constant).

Gravity is a centripetal force (it pulls you towards the center of the sun). The formula for a centripetal force is given by: $$F=\frac{mv^2}{r}$$ where $m$ is mass, $v$ is velocity and $r$ the distance to the center (in this case the sun).

Since gravity is a centripetal force you can conclude the following: $$G\frac{Mm}{r^2}=\frac{mv^2}{r}$$ Multiplying by $r$ and dividing by $m$ gives: $$G\frac{M}{r}=v^2$$ The orbital period (the time it takes to complete an orbit) is: $$T=\frac{2\pi r}{v}$$ Solving for $v$ gives $v=2\pi r/T$. Squaring gives $$v^2=\frac{4\pi^2 r^2}{T^2}$$ Above we saw that $v^2=GM/r$ as well. We can thus set these two equations equal: $$\frac{4\pi^2 r^2}{T^2}=G\frac{M}{r}$$ solving for $T^2$ gives $$T^2=\frac{4\pi^2 r^3}{GM}=r^3\frac{4\pi^2}{GM}$$ We now have Kepler's third law! Perhaps it is not directly clear but 4 is of course constant, so is $\pi$ (and $\pi^2$). $G$ is a constant and $M$ as well (the mass of the sun is constant for anyone in the solar system). This can be rewritten as $T^2=C\cdot r^3$ where $r$ is the distance to the sun and equals $a$. Replacing $r$ by $a$ gives us: $$T^2=C\cdot a^3$$ and this is the same as Kepler's third law. Note that $4\pi^2/GM$ is constant, as we saw above, and can be replaced by $C$.

Kepler's law can be derived of Newton's law of gravitation!

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    $\begingroup$ This is rather misleading, at least to some degree. Kepler's third law is much stronger than its restriction to circular orbits - it also states that the period of an elliptical orbit only depends on its major axis and is independent of the ellipticity, but this proof does not work for elliptical orbits. The circular-orbit argument you provide is a valuable piece, but it's not the whole story. $\endgroup$ – Emilio Pisanty Dec 1 '17 at 22:33

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