1.Newton's Laws of Motion $2^{\text {nd }}$ and $3^{\text {rd }}$ Laws

2.Centripetal Force

Example: The center of the orbit from the sun to that from the earth to that of the sun.

$F_c=\frac{m v^2}{r}$

3.The Galilean Law of Freely falling bodies.

4.The second law describes how the acceleration takes place.

5.The Newton's third law describes the symmetry.

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

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

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

$\vec{R}_{A. B}=\vec{R}_A-\vec{R}_B$

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

Acceleration independent of mass.

Acceleration independent of the hight.

Two bodies start with a separation d.

The distance between them should keep on decreasing as they fall towards because both of them fall towards the center.

Acceleration should increase.

$m {{a}}=\frac{G m M_E}{\left(R_E+{{h}}\right)^2}$

$\frac{h}{R_E} \ll 1$

$a=\frac{G M_E}{h^2}\left[\frac{1}{\left(1+\frac{h}{R_E}\right)^2}\right]$

$x=\frac{h}{R_E} \ll 1$

So, $\frac{1}{\left(R_E+h\right)^2}$

$=\frac{1}{R_E^2\left(1+\left(\frac{h}{R_E}\right)\right)^2}=\frac{1}{R_E^2(1+x)^2}$

$\frac{1}{(1+x)^2}=\frac{1}{\left(1+2 x+x^2\right)}$

$x^2 \ll x \ll 1$

$\simeq \frac{1}{1+2 x}=(1-2 x)$

$m a=\frac{F_{G}}{m}=\frac{G M_E m}{R_E^2}\left(1-2 \frac{h}{R_E}\right)$

$h=100 \mathrm{~m}$

$R_E=6.4 \times 10^6$

$x$ is really small

Reconciled all aspects of law of freely falling body with the Netonian law

$R \sim 6400 \mathrm{Km} ; \quad h \sim 100 \mathrm{~m}$

$\frac{2 h}{R} \sim 3 \times 10^{-4}$

Distances are known

Masses not known

G Value not determined.

Galilean law determines $M_EG$.

The Cavendish Experiment (1797-98):Directly determine G.

Measure the bodies acceleration

Arrest the motion by known force.

The whole Apparatus was placed in a thick wooden box.

The wooden box was enclosed in a shed.

Observations were made through two small holes fitted with telescopes.

The experiment was not null. A Large period of $20 \mathrm{~min}$.

The vernier scale had a least count of $0.1 \mathrm{~mm}$.

Large Lead balls: ${158.04 \mathrm{~kg}}$

Small Lead balls: $0.73 \mathrm{~kg}$

Mass of the rod: $0.03 \mathrm{~kg}$

The distance between small rod and the big rod = 0.225 millimeter.

We want a very large gravitational force. Therefore, the distance must be as small as possible.

Measure the angle.

Make an analysis and find out how determine the value of G.

$\mathscr{T}=K \theta=L F=L \frac{G M m}{d^2}$

$\therefore G=\frac{K \theta d^2}{L M m}$

How to determine K?

Natural Oscillations.

$T=2 \pi \sqrt{\frac{I}{K}}=2 \pi \sqrt{\frac{m L^2}{2 K}}$

$T=L F=L\frac{G Mm}{d^2}$

$={k \theta}$

$\theta$= angle of deflection

$G =\frac{K \theta d^2}{M m L}$

How To Determine K?

Look at the Natural Oscillation of the system.

$T=2 \pi \sqrt{\frac{I}{K}}$

$I$: moment of Inertia

$I = 2 \times m\left(\frac{L}{2}\right)^2$

$G=\frac{2 \pi^2 L d^2}{M T^2}$

$G($ Cavendish $)=6.74 \times 10^{-11}$

$G($ Modern $)=6.67408(31) \times 10^{-11}$

Relative error is about $1\%$.

Absolute error $7\%$.

$m g=G \frac{m M_E}{R_E^2} \Longrightarrow$

$M_E=\frac{g R_e^2}{G}=\frac{4 \pi}{R_e^3} \rho_E$

$\frac{\rho_E}{\rho_w}=5.448 \pm 0.033$

$mg=\frac{G M_E m }{R_E^2}$

$M_E=\frac{g \bar{R}_E^2}{G}$

$=\frac{4 \pi}{3} \bar{R}_E^3 \bar{p}$