Kepler's Law of Planetary Motion

Kepler's Law of Planetary Motion:

Kepler's First Law:

The path of a planet is an elliptical orbit around the sun with the sun at one of the foci. This is also known as the law of the elliptical orbits and given the shape of the path of a planet around the sun.

Proof:

Consider a planet if mass m moving around the sun in a closed path. Let \(\vec{r}\) be the position vector of the planet at any instant from the sun.

The force acting on the planet due to sun is

\begin{align*}\vec{F}=-\frac{GMm}{r^{2}}\hat{r}\rightarrow 1 \end{align*}The accleration of the planet is\begin{align*}\vec{a}=-\frac{GM}{r^2}\hat{r}\rightarrow 2 \end{align*}The general equation of acceleration in polar co-ordinate is\begin{align*}\vec{a}=[\frac{d^2r}{dt^2}-r(\frac{d\theta}{dt})^2]\hat{r}+\frac{1}{r}\frac{d}{dt}(r^2\frac{d\theta }{dt})]\hat{\theta }\rightarrow 3 \end{align*}Comparing Equation 2 and 3, we get\begin{align*}\frac1r\frac{d}{dt}(r^2\frac{d\theta }{dt})=0\rightarrow 4\space and \end{align*}\begin{align*}\frac{d^2r}{dt^2}-r(\frac{d\theta }{dt})^2=-\frac{GM}{r^2}\rightarrow\space 5 \end{align*}Integrating equation 4, we get,\begin{align*}r^2\frac{d\theta }{dt}=constant \end{align*}\begin{align*}or,r^2\frac{d\theta }{dt}=h(say)=\frac{J}{m}\rightarrow 6 \end{align*}where h is angular momentum per unit mass.

Let r=\(\frac1u\)\begin{align*}\frac{dr}{dt}=\frac{d}{du}(\frac1u)\frac{du}{dv}.\frac{dv}{dt} \end{align*}\begin{align*}=-\frac{1}{u^2} \frac{h}{r^2}\frac{du}{d\theta }\end{align*}\begin{align*}=-\frac{1}{u^2}.hu^2\frac{du}{d\theta } \end{align*}\begin{align*}-h\frac{du}{d\theta } \end{align*}\begin{align*}\frac{dr}{dt}=-h\frac{du}{d\theta }\rightarrow 7 \end{align*}\begin{align*}Also,\space \frac{d^2r}{dt^2}=-h\frac{d}{d\theta }(\frac{du}{d\theta}).\frac{dv}{dt} \end{align*}\begin{align*}=-h\frac{d^2u}{d\theta ^2}\rightarrow 8 \end{align*}Substituting values from eqn 6 and 8,\begin{align*}-\frac{GM}{r^2}=-h^2u^2\frac{d^2u}{d\theta ^2}-\frac{1}{u}(hu^2)^2\end{align*}\begin{align*}\frac{d^2u}{d\theta ^2}+u=\frac{GM}{h^2u^2r^2} \rightarrow 9\end{align*}The general solution of equation 9 is\begin{align*}u=\frac{GM}{h^2}+Acos\theta \rightarrow 10 \end{align*}where A is constant of integration

Rearranging eqn 10, we get,\begin{align*}\frac{\frac{h^2}{GM}}{r}=1+\frac{Acos\theta }{GM}h^2\rightarrow 11 \end{align*}This is equation us in the form\begin{align*}\frac{l}{r}=1+ecos\theta \rightarrow 12\end{align*}wheree=\(\frac{Ah^2}{GM}\) is the eccentricity of the conic section and l=\(\frac{h^2}{GM}\) is semi-latus rectum.

We have, the kinetic energy of planet is \begin{align*}T=E-U\rightarrow 13 \end{align*}where, E is the total energy which must be -ve and U is the potential energy given by\begin{align*}U=-\frac{GMm}{r}=-GMmu\rightarrow 14 \end{align*}Then from 13 and 14\begin{align*}\frac12mv^2=E-GMmu\rightarrow 15 \end{align*}In polar form, we have\begin{align*}V^{2}=(\frac{dr}{dt^2}+(r\frac{d\theta}{dt})^2\rightarrow 16\end{align*}Now, from equation 16,6 and 7,\begin{align*}v^2=h^2(\frac{du}{d\theta })^2+r^2(hu^2)^2 \end{align*}\begin{align*}v^2=h^2[(\frac{du}{d\theta })^2+u^2]\rightarrow 17 \end{align*}from eqn 15 and 17\begin{align*}h^2[(\frac{du}{d\theta })^2+u^2]=\frac{2(E-GMmu)}{m}\end{align*}\begin{align*}(\frac{du}{d\theta })^2+u^2=\frac{2(E-GMmu)}{h^2m}\rightarrow 18 \end{align*}Substituting value of u from 10 to 18 and solving we get,\begin{align*}E=-\frac{mG^2M}{2h^2}(1-e^2)\rightarrow 19 \end{align*}where e=\(\frac{Ah^2}{GM}\)

As indicated earlier, E<0, for this e<1.

This suggests that the path of the planet is an ellipse around the sun.

Kepler's Second Law:

statement: The radius vector drawn from the sun to a planet sweeps out equal areas in equal interval of time i.e. the areal velocity(area swept out by a planet per unit time) of planet is constant. This is also known as the law of velocity and relates the orbital speed of the planet with its distance from the sun.

Proof:

Consider a planet of mass 'm' moving around the sun in an elliptical path.

Let at any instant the planet be at A. Let after a small time interval at the planet reaches to B subtending small angle d\(\theta \) at the sun. Let \(\vec{r}\) be the radius vector joining the planet from the sun.

The area swept out by the radius vector in moving the planet from A to B is\begin{align*}dA=Area\space OAB \end{align*}\begin{align*}=\frac12 OA.AB \end{align*}\begin{align*}=\frac12r.rd\theta (where,\space AB=rd\theta )\end{align*}\begin{align*}\frac12r^2d\theta \end{align*}Now, the areal velocity of the planet is\begin{align*}Areal\space velocity=\frac{dA}{dt}=\frac12r^2d\theta =\frac12r^2\dot{\theta } \end{align*}\begin{align*}=\frac12\frac{mr^2\dot{\theta }}{m} \end{align*}\begin{align*}=\frac{J}{2m} \end{align*}where J=\(mr^2\dot{\theta }\) is the angular momentum of the planet.

Since, the planet goes round the sun under the effect of central force, the angular momentum of the planet remains conserved.\begin{align*}\therefore Areal\space velocity=\frac{J}{2m}=constant\end{align*}

Kepler's Third Law:

statement: The square of the time period of a planet is proportional to the cube of the semi-major axis of its orbit. This law relates time period with the size of the orbit.

Proof:

Let T be the time period of a planet around the sun in an elliptical orbit having semi-major axis 'a' and seni-minor axis 'b'. Then the period of revolution of the planet going round the sun is\begin{align*}T=\frac{Area\space of\space the\space ellipse}{Areal\space velocity\space of\space the\space planet} \end{align*}\begin{align*}=\frac{\pi ab}{\frac{dA}{dt}} \end{align*}\begin{align*}=\frac{\pi ab}{\frac{J}{2m}} \end{align*}\begin{align*}T=\frac{2\pi mab}{J} \end{align*}Squaring on both sides, we get,\begin{align*}T^2=\frac{4\pi ^2m^2a^2b^2}{J^2}\rightarrow 1 \end{align*}The semi-latus rectum of an ellipse is\begin{align*}l=\frac{b^2}{a} \end{align*}\begin{align*}b^2=al \end{align*}Substituting value of \(b^2\) in eqn I, we get\begin{align*} T^2=\frac{4\pi ^2m^2a^2al}{J^2}\end{align*}\begin{align*}=\frac{4\pi ^2m^2l}{J^2}a^3 \end{align*}\begin{align*}T^2=ka^3 \end{align*}where, K=\(\frac{4\pi ^2m^2l}{J^2}\), \(=\frac{h^2}{GM}\) and \(h=r^2\frac{d\theta }{dt}=r^2\dot{\theta }=r^2\omega =\frac{J}{m}\)\begin{align*}\therefore T^2\prec a^3 \end{align*}

Deduction of Newton's Law of Gravitation from Kepler's Laws:

The general expression of acceleration in polar form is \begin{align*}\vec{a}=[\frac{d^2r}{dt^2}-r(\frac{d\theta }{dt})^2]\hat{r}+[\frac{1}{r}\frac{d}{dt}(r^2\frac{d\theta }{dt})]\hat{\theta }\rightarrow 1 \end{align*}From Kepler's second law,\begin{align*}\frac12r^2\frac{d\theta }{dt}=constant \end{align*}\begin{align*}or,\space \frac12r^2\dot{\theta }=constant \end{align*}\begin{align*}or,\space \frac12r^2\dot{\theta }=\frac{J}{2m}=\frac{h}{2} \end{align*}\begin{align*}or,\space r^2\dot{\theta }=h \rightarrow 2\end{align*}From equation 1, we get\begin{align*}\vec{a}=[\frac{d^2r}{dt^2}-r(\frac{d\theta }{dt})^2]\hat{r}+[\frac{1}{r}\frac{dh}{dt}]\hat{\theta } \end{align*}\begin{align*}\vec{a}=[\frac{d^2r}{dt^2}-r(\frac{d\theta }{dt})^2]\hat{r}\rightarrow 3(\because h\space is\space constant) \end{align*}\begin{align*}Here,\space \frac{dr}{dt}=\frac{d}{du}(\frac{1}{u}).\frac{du}{d\theta }.\frac{d\theta }{dt} \end{align*}\begin{align*}=-\frac{1}{u^2}\frac{h}{r^2}.\frac{du}{d\theta } \end{align*}\begin{align*}=-h\frac{du}{d\theta } \end{align*}\begin{align*}Again,\space \frac{d^2r}{dt^2}=-h\frac{d}{du}(\frac{du}{d\theta }).\frac{du}{dt} \end{align*}\begin{align*}=-h^2u^2\frac{d^2u}{d\theta }^2\rightarrow 4 \end{align*}Thus from equation 2,3 and 4 \begin{align*}\vec{a}=-h^2u^2\frac{d^2u}{dt^2}-\frac{(hu^2)^2}{u} \end{align*}\begin{align*}or,\space \vec{a}=-h^2u^2\frac{d^2u}{d\theta ^2}-h^2u^3 \end{align*}\begin{align*}or,\space \vec{a}=-h^2u^2[{\frac{\mathrm{d}^2u }{\mathrm{d} \theta ^2}}+ u]\hat{r}\rightarrow 5 \end{align*}From Kepler's first law, the equation of orbit of planet is\begin{align*}\frac{l}{r}=1+ecos\theta \end{align*}\begin{align*}u=\frac{1+ecos\theta }{l}\rightarrow 6 \end{align*}\begin{align*}\frac{du}{d\theta }=-\frac{esin\theta }{l} \end{align*}\begin{align*}\frac{d^2u}{d\theta ^2}=-\frac{e}{l}cos\theta \rightarrow 7\end{align*}Substituting values from 6 and 7 in equation 5, we get\begin{align*}\vec{a}=-h^2u^2[{-\frac{e}{l}cos\theta +\frac{1+ecos\theta }{l}}]\hat{r} \end{align*}\begin{align*}\vec{a}=-\frac{h^2u^2}{l}\hat{r}\rightarrow 8 \end{align*}The time period of planet is given by\begin{align*}T=\frac{\pi ab}{\frac{h}{2}}=\frac{2\pi ab}{h} \end{align*}\begin{align*}T^2=\frac{4\pi ^2a^2b^2}{h^2} \end{align*}\begin{align*}T^2=\frac{4\pi ^2a^2la}{h^2}[\because \frac{b^2}{a}=l] \end{align*}\begin{align*}=\frac{4\pi ^2l}{h^2}a^3 \end{align*}\begin{align*}Since,\space T^2\prec a^3 \end{align*}\begin{align*}\frac{4\pi ^2l}{h^2}=constant \end{align*}\begin{align*}or,\space \frac{l}{h^2}=constant=\frac{1}{k}(say)\rightarrow 9 \end{align*}from equation 8 and 9, we get\begin{align*}\vec{a}=-\frac{k}{r^2}\hat{r}\rightarrow 10 \end{align*}Thus the force on the planet due to the Sun is given by Newton's second law is\begin{align*}\vec{f_{sp}}=-\frac{mk}{r^2}\hat{r}\rightarrow 10(\because f=ma) \end{align*}Similarly force on Sun due to the planet is \begin{align*}\vec{f_{ps}}=\frac{Mk'}{r^2}\hat{r}\rightarrow 12 \end{align*}from Newton's third law,\begin{align*}\vec{f_{sp}}=-\vec{f_{ps}} \end{align*}\begin{align*}mk=-Mk' \end{align*}\begin{align*}\frac{k}{M}=-\frac{k'}{m}=constant=G(say) \end{align*}\begin{align*}k=GM \end{align*}Substituting the value of k in the expression of force, we get\begin{align*}\vec{f_{sp}}=-\frac{GMm}{r^2}\hat{r}\rightarrow 13 \end{align*}\begin{align*}similarly,\space\vec{f_{ps}}=\frac{GMm}{r^2}\hat{r}\rightarrow 14 \end{align*}Equation 13 and 14 indicate that the gravitational force between the sun and the planet is

a) directly proportional to the product of their masses

b)inversely proportional to the square of distance between them.

References

Adhikari, Pitri Bhakta. A Textbook of Physics Volume-I. Kathmandu: Sukunda Pustak Bhawan, 2015.

Feynman, Richard P. The Feynman Lectures on Physics Volume 1. Noida: Dorling Kindersley (India) Pvt. Ltd., 2014.

Mathur, D S. Mechanics. New Delhi: S. Chand & Company Pvt. Ltd., 2015.

Young, Hugh D, Roger A Freedman and A Lewis Ford. University Physics. Noida: Dorling Kindersley (India) Pvt. Ltd., 2014.