Important concept: Principle of Equivalence

Gravitational Charge Mass = Inertial Mass

Acceleration of a body in the earth's gravitational field is independent of its mass.

Kepler's Laws

The First Law: Closed Elliptical Orbits.

The Second Law: Equal Areas in Equal Durations.

The Third Law: $\frac{T^2}{R^3}=$ Constant for all planets.

Ptolemy placed the Earth at the centre of his geocentric model.

Saturn may be appearing to move in opposite direction to the earth.

The apparent motion of a planet in a direction opposite to that of other bodies within its system: Retrograde motion.

Planets are moving around Sun .

Circular Orbits with the Sun at the centre.

Orbits are nearly circular.

Kepler was able to fit all the orbits to elliptical trajectories.

Sun is on the one of the focal points.

The angle subtended is the same in equal intervals of time.

Equal areas are swept in equal intervals of time.

$A_1=A_2$

Excellent Observations

Universal patterns common to all planetary motions.

How many universal patterns do we see?

3 patterns + 1 Results

Move in elliptical orbits.

Equal area in equal time.

$\frac{T^2}{R^3}=$ constant.

All these are independent of the mass of the planet.

Combine Kinematical Results with dynamics.

In the first law, there is a force acting. If there were no force acting, all the planets would have been moving with straight line orbits.

Newton's second law: $\frac{d \vec{P}}{d t}=\vec{F}$

Third law of motion: $\overrightarrow{F}_ { A \rightarrow B}=-\vec{F}_{B \rightarrow A}$

Independent of the mass of the moving bodies (Heliocentric model).

Marry kinematics to Dynamics.

Simplification

Circular Orbits

Centripetal Force

The Third Law

$A\left(M_A\right) \rightleftarrows B\left(M_B\right)$

Use the Third Law

$F_{A \leftrightarrow B} \propto M_A M_B$

Use Centripetal Concept

$\vec{F_{A \leftrightarrow B}} \propto M_A M_B f(r) \hat{r_{A B}}$

$\vec{F_{A \rightarrow B}}=-\vec{F_{B \rightarrow A}}$

$\vec{F_{B \rightarrow A}} \propto M_A =\vec{M}_A \vec{a}_A$

Kinematics of the circular orbit: velocity is tangential and the force is radial inward.

The acceleration is always in the direction of the force.

Centrifugal force is an inertial force. It is a pseudo force.

Centrifugal force will be outward radial.

Centripetal force is inward.

The center of the orbit at the origin.

$\vec{F} \propto -\vec{r}$.

Acceleration $\vec{a} \propto -\vec{r}$.

$\vec{F}=-G M_A M_B \vec{r} f(r)$

$f(r) > 0$

Let $f(r)=r^n$

Second Law of Kepler automatically satisfied.

$\frac{d}{d t}(\vec{r} \times \vec{p})=\vec{r} \times \vec{F}$.

$m \omega^2 r=K m f(r)$

$T=\frac{2 \pi}{\omega}=\frac{2 \pi \sqrt{r}}{\sqrt{K f(r)}}$

Kepler's Third Law $\Longrightarrow$ $f(r)=\frac{1}{r^2}$

Determination of f(r): $f(r) \propto r^n$

$F =m \omega^2 r= mMG f(r)$

$\omega^2 r=K f(r)$

$\omega =\frac{2 \pi}{T} \Rightarrow \omega^2=\frac{(2 \pi)^2}{T^2}$

$\omega^2 r=k f(r)$

$(2 \pi)^2 \frac{r}{T^2}=k f(r)$

$r^2 \times \frac{r}{T^2}=K^{\prime} f(r) \times r^2$ = constant

$\frac{r^3}{T^2} \Rightarrow f(r) r^2$ = Constant

$f(r) \propto \frac{1}{r^2} \Rightarrow$

Gravitational Force $F_G=\frac{G M_A M_B}{r^2}$

There is a constant of proportionality which can be absorbed in G.

$A: \quad M_A ; \quad \vec{R}_A$

$B: \quad M_B ; \quad \vec{R}_B$

$\vec{R} _ {A, B}=\vec{R} _ A-\vec{R} _ B$

$\vec{F} _ {A \rightarrow B}$ = $-\frac{G M_A M_B}{R_{A B}^3} \vec{R} _{A B}$

Let a body be at a height $h$ above the Earth's surface.

$\frac{1}{(R+h)^2}=\frac{1}{R^2\left(1+2 h / R+h^2 / R^2\right)}$

Employ binomial expansion.

$F=\frac{G M m}{R^2}\left(1-\frac{2 h}{R}\right)$

Uniform Mass Density

All the mass is concentrated in the circle at the center of the sphere.

$F=\frac{G M m}{r^2}$

$M=\rho \int d V=\frac{4}{3} \pi R^3 \rho$

Falling body

r is the radius of the earth.

h is the height above the center.

$a=+\frac{G M_E}{(R+h)^2} ; \quad {h \ll R}$

Zeroth order: Take $h \approx 0$

$\frac{h}{R} \approx 0$

$a=\frac{G M}{R^2}=g$