Gravity and Gravitation
Central Force
A central force between two particles is one which is directed along the line joining the two particles and whose magnitude is a function of the distance between them.
If one particle is fixed in its position, the central force acting on the other is $\vec{F}$, then the torque acting on it is:
$$ \vec{\tau} = \vec{r} \times \vec{F} = 0 $$Since,
$$ \vec{\tau} = \frac{d\vec{J}}{dt} $$ $$ 0 = \frac{d\vec{J}}{dt} $$ $$ \Rightarrow \vec{J} = \text{constant} $$Again, we know,
$$ \vec{J} = \vec{r} \times \vec{p} $$ $$ \vec{r} \cdot \vec{J} = \vec{r} \cdot (\vec{r} \times \vec{p}) = (\vec{r} \cdot \vec{r})\, \vec{p} = 0 $$This shows that $\vec{r}$ and $\vec{J}$ are perpendicular to each other, and the motion of the particle is confined to a plane.
Central force can be represented as:
$$ \vec{F} = \frac{C}{r^2}\,\hat{r} $$For gravitational force,
$$ C = Gm_1 m_2, \qquad \text{then} \qquad \vec{F} = \frac{Gm_1 m_2}{r^2}\,\hat{r} $$For electrostatic force,
$$ C = kq_1 q_2, \qquad \text{where } k = \frac{1}{4\pi\varepsilon_0} $$ $$ \vec{F} = \frac{1}{4\pi\varepsilon_0}\,\frac{q_1 q_2}{r^2} $$Inverse Square Law
"The force between any two particles is inversely proportional to the square of the distance between them." This is called the inverse square law.
Examples:
For gravitational force,
$$ F = \frac{Gm_1 m_2}{r^2} \;\Rightarrow\; F \propto \frac{1}{r^2} $$For electrostatic force,
$$ F = \frac{1}{4\pi\varepsilon_0}\frac{q_1 q_2}{r^2} \;\Rightarrow\; F \propto \frac{1}{r^2} $$Gravitational Field and Intensity of Gravitational Field
The area around a mass, up to which its influence (gravitational attraction) can be felt, is called the gravitational field.
The force experienced by a unit mass inside the gravitational field of a given mass is called gravitational intensity. Mathematically, gravitational intensity is given by:
$$ E = \frac{F}{m} $$We know, the gravitational force between two masses is:
$$ \vec{F} = -\frac{GMm}{r^2}\,\hat{r} $$($-$ negative sign is for attractive force.)
Then, gravitational intensity will be:
$$ E = \frac{F}{m} = -\frac{GM}{r^2}\,\hat{r} $$Gravitational intensity can also be written as:
$$ E = -\frac{dV}{dr} $$where $V$ = gravitational potential.
Gravitational Potential and Gravitational Potential Energy (G.P.E.)
The amount of work done in bringing a unit mass from infinity to a given point inside the gravitational field of another mass is called gravitational potential at that point.
We know,
$$ E = -\frac{dV}{dr} \;\;\text{or,}\;\; dV = -E\,dr $$ $$ \therefore V = -\int_{\infty}^{r} dV = -\int_{\infty}^{r} E\,dr = -\int_{\infty}^{r} \left(-\frac{GM}{r^2}\right)dr = GM\int_{\infty}^{r} r^{-2}\,dr $$ $$ = GM\left[-\frac{1}{r}\right]_{\infty}^{r} $$ $$ V = -\frac{GM}{r} $$The amount of work done in bringing a mass $m$ from infinity to a given point inside the gravitational field of another mass is called gravitational potential energy at that point.
$$ \therefore \text{G.P.E.}\ (U) = m\int_{\infty}^{r} E\,dr = -m\int_{\infty}^{r} \frac{GM}{r^2}\,dr = GMm\int_{\infty}^{r} r^{-2}\,dr $$ $$ = GMm\left[-\frac{1}{r}\right]_{\infty}^{r} $$ $$ U = -\frac{GMm}{r} $$Orbital Velocity and Escape Velocity of Satellite
The velocity of a satellite in its orbit is called orbital velocity. It is denoted by $v_o$.
The velocity with which a body should be projected upward so that it will escape from earth's gravity is called escape velocity.
Gravitational potential energy on the surface of earth:
$$ \text{G.P.E.} = -\frac{GMm}{R} $$$\therefore$ Energy required to send the body to infinity:
$$ = \frac{GMm}{R} $$And, kinetic energy that should be imparted on the body to send it to infinity:
$$ = \frac{1}{2}mv_e^2 $$where $v_e$ = escape velocity. Then,
$$ \frac{GMm}{R} = \frac{1}{2}mv_e^2 $$ $$ v_e = \sqrt{\frac{2GM}{R}} $$Since $g = \frac{GM}{R^2}$ (from $F = \frac{GMm}{R^2} = mg$),
$$ v_e = \sqrt{\frac{2R\cdot GM}{R^2}} = \sqrt{2Rg} $$ $$ = \sqrt{2 \times 6400\times 10^3 \times 9.8} $$ $$ = 11200\text{ m/s} = 11.2\text{ km/s} $$Gravitational Potential and Field due to the Spherical Shell
(A) Gravitational Potential at a Point Outside the Spherical Shell
Let us consider a spherical shell of radius $R$.
Let $O$ = centre of the shell, $P$ = point at distance $r$ from the centre of the shell, $\sigma$ = surface density (mass/surface area).
Let us take a slice $CDEF$ in the form of a ring such that the two planes $CD$ and $EF$ are close to each other and perpendicular to the radius $OA$ of the shell.
Let $\angle AOF = \theta$ and $\angle COF = d\theta$.
Now,
$$ \text{Radius of the ring} = KF = R\sin\theta $$ $$ \text{Circumference of the ring} = 2\pi(KF) = 2\pi R \sin\theta $$ $$ \text{Width of the ring } (CF) = R\,d\theta $$ $$ \text{Surface area of the ring} = 2\pi R\sin\theta \times R\,d\theta \qquad (\because \text{circumference} \times \text{width}) $$ $$ \therefore \text{Mass of the ring} = \text{surface area of ring} \times \text{surface density} $$ $$ = 2\pi R\sin\theta \times R\,d\theta \times \sigma = 2\pi R^2 \sin\theta\,d\theta\,\sigma $$Every point on the ring is at a distance $x$ from point $P$.
$$ \therefore \text{Potential at } P \text{ due to the ring } (dV) = -\frac{G \times \text{mass of ring}}{x} $$ $$ = -\frac{G \times 2\pi R^2 \sin\theta\,d\theta\,\sigma}{x} \qquad \cdots (i) $$In $\triangle FOP$,
$$ \cos\theta = \frac{R^2 + r^2 - x^2}{2Rr} $$ $$ 2Rr\cos\theta = R^2 + r^2 - x^2 $$ $$ x^2 = R^2 + r^2 - 2Rr\cos\theta $$Differentiating,
$$ 2x\,dx = 0 + 0 - 2Rr(-\sin\theta)\,d\theta $$ $$ x\,dx = Rr\sin\theta\,d\theta $$ $$ R\sin\theta\,d\theta = \frac{x\,dx}{r} \qquad \cdots (ii) $$From (i) and (ii),
$$ dV = -\frac{G\cdot 2\pi R \sigma \cdot x\,dx}{x\cdot r} = -\frac{2\pi G R \sigma\,dx}{r} \qquad \cdots (iii) $$Integrating equation (iii) from $A(r-R)$ to $B(r+R)$, it gives the potential at point $P$ due to the whole spherical shell:
$$ \therefore V = \int_{r-R}^{r+R} dV = \int_{r-R}^{r+R} -\frac{2\pi G R \sigma}{r}\,dx $$ $$ = -\frac{2\pi G R \sigma}{r}\int_{r-R}^{r+R} dx = -\frac{2\pi GR\sigma}{r}\Big[x\Big]_{r-R}^{r+R} $$ $$ = -\frac{2\pi GR\sigma}{r}\big[(r+R)-(r-R)\big] = -\frac{4\pi G R\sigma}{r} = -\frac{4\pi R^2 \sigma G}{r} $$ $$ V = -\frac{GM}{r} \qquad \text{where } M = 4\pi R^2 \sigma $$(B) Gravitational Potential at the Surface of the Shell
From equation (iii), we have (note: $r \to R$ on surface):
$$ dV = -\frac{2\pi GR\sigma}{r}\,dx $$$\therefore$ Total potential due to the shell at its surface is:
$$ V = \int_{0}^{2R} dV = \int_{0}^{2R} -\frac{2\pi GR\sigma}{R}\,dx = -\frac{2\pi GR\sigma}{R}\int_0^{2R} dx = -\frac{2\pi GR\sigma}{R}\Big[x\Big]_0^{2R} $$ $$ = -\frac{2\pi GR\sigma}{R}\cdot 2R = -\frac{4\pi R^2\sigma G}{R} $$ $$ V = -\frac{GM}{R} $$(C) Gravitational Potential Inside the Shell
When point $P$ lies inside the shell, we can calculate the total potential at $P$ by integrating equation:
$$ dV = -\frac{2\pi GR\sigma}{r}\,dx \qquad \text{from } (R-r) \text{ to } (R+r) $$$\therefore$ Total potential due to the shell inside the shell is:
$$ V = \int_{R-r}^{R+r} dV = \int_{R-r}^{R+r} -\frac{2\pi GR\sigma}{r}\,dx = -\frac{2\pi GR\sigma}{r}\int_{R-r}^{R+r} dx $$ $$ = -\frac{2\pi GR\sigma}{r}\Big[x\Big]_{R-r}^{R+r} = -\frac{2\pi GR\sigma}{r}\big[(R+r)-(R-r)\big] $$ $$ = -\frac{2\pi GR\sigma}{r}\times 2r = -4\pi GR\sigma = -\frac{4\pi R^2\sigma G}{R} $$ $$ V = -\frac{GM}{R} \qquad \text{where } M = 4\pi R^2\sigma $$$\therefore$ Gravitational potential at any point inside the shell is constant, and is equal to the potential at the surface of the shell.
Gravitational Field due to the Spherical Shell
(i) At a Point Outside the Shell
Potential outside the spherical shell is:
$$ V = -\frac{GM}{r} $$$\therefore$ Gravitational field:
$$ E = -\frac{dV}{dr} = -\frac{d}{dr}\left(-\frac{GM}{r}\right) = GM\frac{dr^{-1}}{dr} $$ $$ E = -\frac{GM}{r^2} $$(ii) At the Surface of the Shell
Potential at the surface of the shell is:
$$ V = -\frac{GM}{R} $$$\therefore$ Gravitational field ($r = R$):
$$ E = -\frac{dV}{dR} = -\frac{d}{dR}\left(-\frac{GM}{R}\right) = GM\frac{dR^{-1}}{dR} $$ $$ E = -\frac{GM}{R^2} $$(iii) At a Point Inside the Spherical Shell
$$ V = -\frac{GM}{R} = \text{constant} $$ $$ E = -\frac{dV}{dR} $$ $$ E = 0 $$$\Rightarrow$ There is no gravitational field inside the spherical shell.
Gravitational Potential and Field due to the Solid Sphere
(i) At a Point Outside the Sphere
Let us consider a solid sphere and a point $P$ that lies outside the sphere. Let $OP = r$, $R$ = radius of solid sphere, $M$ = total mass of solid sphere.
A solid sphere consists of a large number of concentric shells. Let $m_1, m_2, \ldots, m_n$ be the masses of different concentric shells. Then,
Total potential = sum of potential due to each shell.
$$ V = V_1 + V_2 + \cdots + V_n $$ $$ = -\frac{Gm_1}{r} + \left(-\frac{Gm_2}{r}\right) + \cdots + \left(-\frac{Gm_n}{r}\right) $$ $$ = -\frac{G}{r}(m_1 + m_2 + \cdots + m_n) = -\frac{GM}{r} $$Now, gravitational field is given by:
$$ E = -\frac{dV}{dr} = -\frac{d}{dr}\left(-\frac{GM}{r}\right) = GM\frac{dr^{-1}}{dr} $$ $$ E = -\frac{GM}{r^2} $$(ii) At a Point on the Surface of the Sphere
Gravitational potential at the surface of the sphere is:
$$ V = -\frac{GM}{R} $$and gravitational field is:
$$ E = -\frac{dV}{dR} = -\frac{d}{dR}\left(-\frac{GM}{R}\right) = GM\frac{dR^{-1}}{dR} $$ $$ E = -\frac{GM}{R^2} $$(iii) At a Point Inside the Surface of Sphere
Let us consider a solid sphere of radius $R$. We have to find gravitational potential at point $P$ which lies at distance $r$ from the centre ($r < R$).
Potential on the surface of a solid sphere (of matter) of radius $r$:
$$ = -\frac{G\times\text{mass}}{r} = -\frac{G\times \frac{4}{3}\pi r^3 \rho}{r} \qquad (\because \rho = \tfrac{m}{v}) $$ $$ = -\frac{4}{3}\pi r^2 \rho\, G \qquad \cdots (i) $$To determine the potential at $P$ due to the outer shells:
Let $x$ = radius of one outer shell, $dx$ = thickness of that shell. Then,
$$ \text{Volume} = 4\pi x^2\,dx $$$\therefore$ Mass of shell $= \text{volume}\times\text{density} = 4\pi x^2\,dx\,\rho$
$\therefore$ Potential at $P$ due to outer shell:
$$ (dV) = -\frac{G\times\text{mass of shell}}{x} = -\frac{G\cdot 4\pi x^2\,dx\,\rho}{x} = -4\pi G\rho\, x\,dx $$Then, potential at $P$ due to all outer shells $=\int dV$:
$$ = \int_r^R -4\pi G\rho\, x\,dx = -4\pi G\rho\int_r^R x\,dx = -4\pi G\rho\left[\frac{x^2}{2}\right]_r^R $$ $$ = -\frac{4\pi G\rho}{2}(R^2 - r^2) = -2\pi G\rho (R^2 - r^2) = -\frac{4}{3}\pi\rho G\cdot\frac{3}{2}(R^2-r^2) \qquad \cdots (ii) $$Now, total potential at point $P$:
$$ = -\frac{4}{3}\pi r^2 \rho G + \left[-\frac{4}{3}\pi\rho G\cdot\frac{3}{2}(R^2-r^2)\right] $$ $$ = -\frac{4}{3}\pi\rho G\left[r^2 + \frac{3}{2}(R^2-r^2)\right] $$ $$ = -\frac{4}{3}\pi\rho G\left[\frac{2r^2+3R^2-3r^2}{2}\right] $$ $$ = -\frac{4}{3}\pi R^3\rho G\left(\frac{3R^2-r^2}{2R^3}\right) $$ $$ V = -GM\left(\frac{3R^2-r^2}{2R^3}\right) $$At the centre of the sphere, $r = 0$:
$$ V = -GM\times\frac{3R^2}{2R^3} = -\frac{3}{2}\frac{GM}{R} = \frac{3}{2}\left(-\frac{GM}{R}\right) = \frac{3}{2}\times\text{potential at surface} $$Now, gravitational field:
$$ E = -\frac{dV}{dr} = -\frac{d}{dr}\left[-GM\left(\frac{3R^2-r^2}{2R^3}\right)\right] = \frac{GM}{2R^3}\frac{d}{dr}(3R^2-r^2) $$ $$ = \frac{GM}{2R^3}\times(-2r) = -\frac{GMr}{R^3} $$ $$ E = -\frac{GMr}{R^3} $$At the centre, $r = 0$, $\therefore E = 0$.
Gravitational Self Energy
The work done by mutually attractive gravitational forces of the masses, while bringing them from an infinite distance to their present position, is called gravitational self energy.
Derivation of Gravitational Self Energy
Work done on bringing mass $dm$ from infinity to the surface of radius $x$ is given by:
$$ dW = dm\,(V_x - V_\infty) $$ $$ dW = dm\,V_x \qquad \left(\because V_\infty = 0,\ V = \frac{1}{r}\text{ at infinite distance}\right) $$ $$ = dm\left(-\frac{GM}{x}\right) = -\left[\frac{G\cdot\frac{4}{3}\pi x^3\rho}{x}\right]dm \qquad (M = \rho V = \tfrac{4}{3}\pi x^3\rho) $$ $$ = -4\pi x^2\,dx\,\rho\cdot G\cdot\frac{4}{3}\pi x^2\rho \qquad (dm = \rho\cdot A\cdot dx) $$ $$ dW = -\frac{16}{3}G\pi^2\rho^2 x^4\,dx $$$\therefore$ Total work done is obtained by integrating from $0$ to $R$:
$$ W = \int_0^R dW = \int_0^R -\frac{16}{3}G\pi^2\rho^2 x^4\,dx = -\frac{16}{3}G\pi^2\rho^2\int_0^R x^4\,dx $$ $$ = -\frac{16}{3}G\pi^2\rho^2\left[\frac{x^5}{5}\right]_0^R = -\frac{16}{15}G\pi^2\rho^2 R^5 $$ $$ = -\frac{16}{25}G\pi^2\cdot\frac{M^2}{\frac{16}{9}\pi^2 R^6}\cdot R^5 \qquad \left(\rho = \frac{M}{V} = \frac{M}{\frac{4}{3}\pi R^3}\right) $$ $$ W = -\frac{3}{5}\frac{GM^2}{R} $$which is the expression for the gravitational self energy.
Gravitational Self (Potential) Energy of Multi-Particle System
Let us consider three masses $m_1$, $m_2$ and $m_3$.
Let $r_{12}$ = separation between $m_1$ and $m_2$, $r_{23}$ = separation between $m_2$ and $m_3$, $r_{13}$ = separation between $m_1$ and $m_3$.
Then, work done in bringing $m_2$ from infinity to distance $r_{12}$ from $m_1$ is:
$$ U_{12} = \int_{\infty}^{r_{12}} \vec{F}\cdot d\vec{r} = \int_{\infty}^{r_{12}} \frac{Gm_1m_2}{r_{12}^2}\,dr_{12} = Gm_1m_2\int_{\infty}^{r_{12}} r_{12}^{-2}\,dr_{12} $$ $$ \therefore U_{12} = -\frac{Gm_1m_2}{r_{12}} $$Similarly,
$$ U_{13} = -\frac{Gm_1m_3}{r_{13}}, \qquad U_{23} = -\frac{Gm_2m_3}{r_{23}} $$$\therefore$ Total potential $(U) = U_{12}+U_{13}+U_{23}$
$$ = -\frac{Gm_1m_2}{r_{12}} - \frac{Gm_1m_3}{r_{13}} - \frac{Gm_2m_3}{r_{23}} $$ $$ = -G\left(\frac{m_1m_2}{r_{12}} + \frac{m_1m_3}{r_{13}} + \frac{m_2m_3}{r_{23}}\right) $$For $n$ particles, we can write:
$$ U = -\sum_{i,j}^{n} \frac{Gm_im_j}{r_{ij}} \qquad (i\neq j) $$Relation Between Two-Body Problem and One-Body Problem
Two particle system can be reduced into one particle system.
Let us consider two masses $m_1$ and $m_2$ at positions $r_1$ and $r_2$ respectively from the reference point. Let no external force act on them; that means it is acted upon only by the interaction force.
In this case, force experienced by $m_1$ due to $m_2$ is:
$$ F_{12} = m_1\frac{d^2r_1}{dt^2} \;\Rightarrow\; \frac{d^2r_1}{dt^2} = \frac{F_{12}}{m_1} \qquad \cdots (i) $$Similarly, force experienced by $m_2$ due to $m_1$ is:
$$ F_{21} = m_2\frac{d^2r_2}{dt^2} \;\Rightarrow\; \frac{d^2r_2}{dt^2} = \frac{F_{21}}{m_2} \qquad \cdots (ii) $$Eq.(ii) $-$ Eq.(i):
$$ \frac{d^2r_2}{dt^2} - \frac{d^2r_1}{dt^2} = \frac{F_{21}}{m_2} - \frac{F_{12}}{m_1} $$ $$ \frac{d^2}{dt^2}(\vec{r_2}-\vec{r_1}) = \frac{F_{21}}{m_2} - \frac{F_{12}}{m_1} $$Let $F_{21} = -F_{12} = F$. Then,
$$ \frac{d^2}{dt^2}(\vec{r_2}-\vec{r_1}) = F\left(\frac{1}{m_2}+\frac{1}{m_1}\right) $$ $$ \frac{d^2r}{dt^2} = F\,\frac{(m_1+m_2)}{m_1m_2} $$ $$ F = \left(\frac{m_1m_2}{m_1+m_2}\right)\frac{d^2r}{dt^2} $$ $$ F = \mu\,\frac{d^2r}{dt^2} \qquad \text{where } \mu = \frac{m_1m_2}{m_1+m_2}\ \text{(reduced mass)} $$Gauss's Law for Gravitational Field
Total flux of gravitational intensity over a closed surface in a gravitational field is $-4\pi G$ times the total mass enclosed by the surface.
$$ \text{i.e.,} \qquad \phi = -4\pi G\times \sum m $$where $\phi = \int E_n\,dS$ ($E_n$ = normal component).
Let $E$ = gravitational field at point $P$ on the surface $dS$, $E_n$ = normal component of $E$.
Then, flux over the surface $dS$ is:
$$ d\phi = E_n\,dS = E\cos\theta\,dS = -\frac{GM}{r^2}\cos\theta\,dS \qquad \left(E = -\frac{GM}{r^2}\right) $$ $$ = -\frac{GM\cos\theta}{r^2}\,dS $$ $$ d\phi = -GM\,d\omega \qquad \left(\because d\omega = \frac{\cos\theta\,dS}{r^2},\ \text{solid angle}\right) $$Now, flux over the whole closed surface is given by:
$$ \phi = \int d\phi = \int -GM\,d\omega = -G\omega\sum m $$ $$ = -G\omega M = -G\cdot 4\pi\times\sum m $$ $$ \phi = -4\pi G\times\sum m $$If no mass is enclosed inside the surface, then $\sum m = 0 \Rightarrow \phi = 0$. So, flux is $0$, i.e., there is no flux.
Poisson's Equation
The Poisson's equation for the gravitational potential obeys the relation:
$$ \nabla^2 V = -4\pi G\rho $$where,
$$ \nabla^2 = \frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}+\frac{\partial^2}{\partial z^2} $$$\rho$ = density (volume density).
Kepler's Laws of Planetary Motion
(i) First Law
Each planet revolves around the Sun in an elliptical orbit with the Sun at one of the foci.
(ii) Second Law
The radius vector joining the planet to the Sun sweeps out equal areas in equal intervals of time, i.e., areal velocity of the planet is constant.
$$ \frac{dA}{dt} = \text{constant} $$(iii) Third Law
The square of the time period of revolution of a planet around the Sun is directly proportional to the cube of the semi-major axis of the orbit. This law is also known as the harmonic law.
$$ T^2 \propto a^3 $$Proof of First Law
Let $S$ = position of Sun, $P$ = position of planet, $r$ = radius vector (distance between Sun and planet), $KL$ = directrix.
According to the definition of a conic section,
$$ \frac{SP}{PM} = e $$ $$ \Rightarrow SP = e\times PM = e\times(d - SN) = e(d - r\cos\theta) $$ $$ \Rightarrow r = e(d-r\cos\theta) $$ $$ r + er\cos\theta = ed $$ $$ r(1+e\cos\theta) = ed $$ $$ (1+e\cos\theta) = \frac{ed}{r} \qquad \cdots (i) $$Similarly,
$$ \frac{SS'}{S'O} = e \;\Rightarrow\; SS' = e\times S'O $$ $$ l = e\times d $$ $$ l = ed \qquad \cdots (ii) $$where $SS' = l$ = semi-latus rectum.
From equations (i) and (ii):
$$ 1+e\cos\theta = \frac{l}{r} $$which is the equation of an ellipse.
$\therefore$ Motion of a planet around the Sun is elliptical.
Proof of Second Law
Let $P$ = position of planet, $S$ = position of Sun, $r$ = radius vector. After time $\Delta t$, the planet reaches position $Q$ such that $PQ = \Delta r$.
Now, area swept out by the radius vector ($SP$) in time $\Delta t$ is:
$$ \Delta A = \frac{1}{2} r\times\Delta r $$Areal velocity (area swept per unit time) is given by:
$$ \frac{dA}{dt} = \lim_{\Delta t\to 0}\frac{\Delta A}{\Delta t} = \lim_{\Delta t\to 0}\frac{\frac{1}{2}r\times\Delta r}{\Delta t} = \frac{1}{2}r\lim_{\Delta t\to 0}\frac{\Delta r}{\Delta t} = \frac{1}{2}r\times\frac{dr}{dt} $$ $$ = \frac{1}{2}(\vec{r}\times\vec{v}) = \frac{1}{2m}(\vec{r}\times m\vec{v}) $$ $$ \frac{dA}{dt} = \frac{J}{2m} = \text{constant} $$Proof of Third Law
Let $T$ = time period of revolution of planet in elliptical orbit, $a$ = semi-major axis, $b$ = semi-minor axis.
Therefore, rate of change of area of ellipse, i.e., areal velocity, is given by:
$$ \frac{dA}{dt} = \frac{J}{2m} $$where $J$ = angular momentum, $m$ = mass.
Now, time period of revolution:
$$ (T) = \frac{\text{Area of ellipse}}{\text{Areal velocity}} $$ $$ T = \frac{\pi a b}{J/2m} $$ $$ T = \frac{2\pi m a b}{J} $$Squaring the above equation,
$$ T^2 = \frac{4\pi^2 m^2 a^2 b^2}{J^2} \qquad \cdots (i) $$If $l$ is the semi-latus rectum, then,
$$ l = \frac{b^2}{a} \;\Rightarrow\; b^2 = la \qquad \cdots (ii) $$Replacing $b^2$ in equation (i), we get,
$$ T^2 = \frac{4\pi^2 m^2 a^3 l}{J^2} $$ $$ \Rightarrow T^2 \propto a^3 \qquad (\text{other terms are constant}) $$This is the proof of the third law.
Derivation of Newton's Law of Gravitation from Kepler's Laws
From Kepler's law:
$$ T^2 \propto a^3 \;\Rightarrow\; T^2 = ka^3 $$Newton assumed motion of the planet in a circular orbit, so $a = r$:
$$ T^2 = kr^3 \qquad \cdots (i) $$For the circular motion of the planet, centripetal force is given by:
$$ F = \frac{mv^2}{r} = m\omega^2 r = \frac{m\,4\pi^2 r}{T^2} $$ $$ F = \frac{4m\pi^2 r}{kr^3} \qquad [\text{from eq. (i)}] $$ $$ F = \frac{4\pi^2}{k}\cdot\frac{m}{r^2} \qquad \cdots (ii) $$Thus, this force is provided by the mutually interacting gravitational force.
$$ \therefore \frac{4\pi^2}{k}\propto M \qquad \text{where } M = \text{mass of Sun} $$ $$ \frac{4\pi^2}{k} = GM \qquad \cdots (iii) $$where $G$ = universal gravitational constant.
From equations (i) and (iii), we get,
$$ F = \frac{GMm}{r^2} $$which is Newton's Law of Gravitation.
