GRAVITATION: Universal Law of Gravitation

$F \propto \frac{m_{1} m_{2}}{r^{2}} \text { or } F=G \frac{m_{1} m_{2}}{r^{2}}$

where $\mathrm{G}=6.67 \times 10^{-11} \mathrm{Nm}^{2} \mathrm{~kg}^{-2}$ is the universal gravitational constant.

Newton’s Law of Gravitation in vector form :

$\vec{F_{12}} = \frac{Gm_1m_2}{r^2} \hat{r_{12}}$

$ \vec{F_{21}} = \frac{Gm_1m_2}{r^2} \hat{r_{21}}$

Now $\hat{r_{12}} = -\hat{r_{21}}$

Thus, $\vec{F_{21}}=\frac{-G m_1 m_2}{r^2} \hat{r}_{12}$

Comparing above, we get $\vec{F_{12}}=-\vec{F_{21}}$

Gravitational Field:

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

Gravitational Potential:

$V=-\frac{G M}{r} . \quad E=-\frac{d V}{d r}$

• Ring

  $V=\frac{-G M}{x \text { or }\left(a^{2}+r^{2}\right)^{1 / 2}} \quad \quad E=\frac{-G M r}{\left(a^{2}+r^{2}\right)^{3 / 2}} \hat{r}$

  $\text { or } E=-\frac{G M \cos \theta}{x^{2}}$

  Gravitational field is maximum at a distance, $r= \pm a / \sqrt{2} \text { and it is }-2 G M / 3 \sqrt{3} a^{2}$

• Thin Circular Disc

  $V=\frac{-2 G M}{a^{2}}\left[\left[a^{2}+r^{2}\right]^{\frac{1}{2}}-r\right] E=-\frac{2 G M}{a^{2}}\left[1-\frac{r}{\left[r^{2}+a^{2}\right]^{\frac{1}{2}}}\right]=-\frac{2 G M}{a^{2}}[1-\cos \theta]$

• Non conducting solid sphere

  (a) Point P inside the sphere $r \leq a$, then

  $V=-\frac{G M}{2 a^{3}}\left(3 a^{2}-r^{2}\right) E=-\frac{G M r}{a^{3}}$

  At the centre $V=-\frac{3 G M}{2 a} \quad \text{and} \quad E=0$

  (b) Point $P$ outside the sphere $r \geq a$, then

  $ V=-\frac{G M}{r} \quad E=-\frac{G M}{r^{2}}$

• Uniform Thin Spherical Shell/ Conducting solid sphere

  (a) Point $P$ Inside the shell

  $r \leq a$, then $V=\frac{-G M}{a} \quad E=0$

  (b) Point $P$ outside shell

  $r \geq a$, then $V=\frac{-G M}{r} \quad E=-\frac{G M}{r^{2}}$

Variation Of Acceleration Due To Gravity :

• Effect of Altitude

  $g_{h}=\frac{G M_{e}}{\left(R_{e}+h\right)^{2}}=g\left(1+\frac{h}{R_{e}}\right)^{-2}$

  $\simeq g\left(1-\frac{2 h}{R_{e}}\right) \text { when } h < <R \text {. }$

• Effect of depth $g_{d}=g\left(1-\frac{d}{R_{e}}\right)$

• Effect of the surface of Earth

  The equatorial radius is about $21 \mathrm{~km}$ longer than its polar radius. We know, $g=\frac{GM_{e}}{R_{e}^{2}}$

  Hence $g_{\text {pole }}>g_{\text {equator }}.$

Satellite Velocity (Or Orbital Velocity)

$v_{0}=\left[\frac{G M_{e}}{\left(R_{e}+h\right)}\right]^{\frac{1}{2}}=\left[\frac{g R_{e}^{2}}{\left(R_{e}+h\right)}\right]^{\frac{1}{2}}$

When $h«R_e$ then $v_0=\sqrt{gR}$

$\therefore \quad v_{0}=\sqrt{9.8 \times 6.4 \times 10^{6}}$

$=7.92 \times 10^{3} \mathrm{ms}^{-1}=7.92 \mathrm{km} \mathrm{s}^{1}$

Time period of Satellite:

$T=\frac{2 \pi\left(R_{e}+h\right)}{\left[\frac{g R_{e}^{2}}{\left(R_{e}+h\right)}\right]^{\frac{1}{2}}}=\frac{2 \pi}{R_{e}}\left[\frac{\left(R_{e}+h\right)^{3}}{g}\right]^{\frac{1}{2}}$

Energy of a Satellite:

$U=\frac{-G M_{e} m}{r} \quad \text{K.E.} =\frac{G M_{e} m}{2 r} ;\quad \text{then total energy:} \quad E=-\frac{G M_{e} m}{2 R_{e}}$

Kepler’s Laws

Law of area :

• The line joining the sun and a planet sweeps out equal areas in equal intervals of time.

• Areal velocity: $ v_A=\frac{\text { area swept }}{\text { time }}=\frac{\frac{1}{2} r(r d \theta)}{d t}= \frac{1}{2} r^{2} \frac{d \theta}{d t}= \text{constant .}$

• Hence $\frac{1}{2} r^{2} \omega=$ constant.

Law of periods :

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

Effective Gravitational Acceleration

Considering both gravitational and centrifugal effects, the effective gravitational acceleration at any latitude is:

$g_{\text{effective}} = g - \omega^2 r$

Where: $\omega$ is the angular velocity of the Earth’s rotation.

At the poles, this simplifies to:

$g_{\text{effective, poles}} = g, \quad \text{because} \quad a_c = 0 \quad \text{at the poles.}$

Escape velocity

The escape speed from the surface of the earth is $ v = \sqrt{2G \frac{M}{r}}$