Gravitation

Between 1609 and 1619, Johannes Kepler formulated three empirical laws based on Tycho Brahe's meticulous observational data of planetary positions:

1. First Law (Law of Orbits): Every planet moves in an elliptical orbit around the Sun, with the Sun situated at one of the two focal points of the ellipse.

   An ellipse is characterized by its semi-major axis $a$, semi-minor axis $b$, and eccentricity $e = \sqrt{1 - b^2/a^2}$ ($0 \le e < 1$). The closest approach to the Sun is the perihelion at distance $r_p = a(1 - e)$, and the farthest point is the aphelion at distance $r_a = a(1 + e)$.

2. Second Law (Law of Areas): The radius vector drawn from the Sun to a planet sweeps out equal areas in equal intervals of time. That is, the areal velocity of a planet is strictly constant:$$\frac{dA}{dt} = \text{constant}$$
   Proof via Conservation of Angular Momentum: Consider a planet of mass $m$ moving with velocity $\vec{v}$ at position $\vec{r}$ relative to the Sun. In an infinitesimal time $dt$, its displacement is $d\vec{r} = \vec{v}\,dt$. The area swept out is the triangular area $dA = \frac{1}{2}|\vec{r} \times d\vec{r}| = \frac{1}{2}|\vec{r} \times \vec{v}|\,dt$. Thus:$$\frac{dA}{dt} = \frac{1}{2}|\vec{r} \times \vec{v}| = \frac{|\vec{r} \times m\vec{v}|}{2m} = \frac{L}{2m}$$Because the gravitational force exerted by the Sun is purely central (directed along $-\hat{r}$), the net torque is zero: $\vec{\tau} = \vec{r} \times \vec{F}_g = 0$. Therefore, total angular momentum $\vec{L}$ is conserved, proving that $\frac{dA}{dt} = \frac{L}{2m} = \text{constant}$.

   A direct consequence is that the planet moves fastest at perihelion and slowest at aphelion: $v_p r_p = v_a r_a$.

3. Third Law (Law of Periods): The square of the orbital period of revolution ($T$) of a planet is directly proportional to the cube of the semi-major axis ($a$) of its elliptical orbit:$$T^2 \propto a^3 \implies \frac{T^2}{a^3} = \text{constant}$$

Assuming a planet of mass $m$ moves in a circular orbit of radius $r$ around the Sun of mass $M_s$ with constant speed $v$ and period $T$:

The required centripetal force is $F = \frac{m v^2}{r} = m \left(\frac{2\pi r}{T}\right)^2 \frac{1}{r} = \frac{4\pi^2 m r}{T^2}$.

From Kepler's Third Law, $T^2 = k r^3$. Substituting into the force equation:

Newton recognized that by his Third Law, the force must also be directly proportional to the mass of the Sun $M_s$, leading directly to the Universal Law of Gravitation: $F = G\frac{M_s m}{r^2}$.

Every particle of matter in the universe attracts every other particle with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers:

Vector Formulation: If $\vec{r}_{12}$ is the position vector of particle 1 with respect to particle 2, the gravitational force $\vec{F}_{12}$ exerted on particle 1 by particle 2 is:

Similarly, $\vec{F}_{21} = -\vec{F}_{12}$. Gravitational interaction satisfies Newton's Third Law identically, forming an action-reaction pair.

Essential Characteristics of Gravitational Force:

• It is always purely attractive.
• It is a central force (acts along the line joining the centers of mass of the two bodies).
• It is a conservative force (work done in any closed path is zero: $\oint \vec{F}\cdot d\vec{r} = 0$).
• It is completely independent of the intervening medium between the masses.
• It obeys the Principle of Superposition: The total gravitational force on a particle due to multiple surrounding masses is the vector sum of individual pairwise forces: $\vec{F}_{\text{net}} = \sum_{i=2}^N \vec{F}_{1i}$.

The constant $G$ represents the magnitude of the gravitational attraction between two point masses of $1\,\text{kg}$ each separated by a distance of $1\,\text{m}$.

• SI Value: $G = 6.6743 \times 10^{-11}\,\text{N}\cdot\text{m}^2/\text{kg}^2 = 6.6743 \times 10^{-11}\,\text{m}^3/(\text{kg}\cdot\text{s}^2)$
• CGS Value: $6.6743 \times 10^{-8}\,\text{dyn}\cdot\text{cm}^2/\text{g}^2$
• Dimensional Formula: $[G] = \frac{[F][r^2]}{[m^2]} = \frac{[MLT^{-2}][L^2]}{[M^2]} = [M^{-1}L^3T^{-2}]$

Cavendish Experiment (1798): Henry Cavendish determined $G$ experimentally using a sensitive quartz-fiber torsion balance, measuring the deflection caused by large lead spheres on small suspended gold spheres.

Inertial mass ($m_i$) is the measure of a body's resistance to acceleration when subjected to any external force ($F = m_i a$). Gravitational mass ($m_g$) is the measure of the strength of a body's gravitational interaction ($F_g = G\frac{M m_g}{r^2}$).

Experiments by Eötvös, Dicke, and Braginsky have verified to a precision of 1 part in $10^{15}$ that $m_i = m_g$. This fundamental identity is known as Einstein's Principle of Equivalence, establishing that gravitational acceleration is independent of the mass or composition of the falling body.

For a body of mass $m$ resting on the surface of Earth (mass $M$, radius $R$), its weight equals the gravitational force exerted by Earth:

Expressing mass in terms of uniform mean density $\rho$: $M = \frac{4}{3}\pi R^3 \rho$, which yields:

On Earth's surface, the standard mean value is $g \approx 9.8\,\text{m/s}^2 = 980\,\text{cm/s}^2$.

At an altitude $h$ above the Earth's surface, the distance from the center is $r = R + h$:

When $h \ll R$ (for heights up to a few hundred kilometers), expanding using the Binomial Theorem and neglecting higher-order terms:

Fractional decrease: $\frac{\Delta g}{g} = \frac{g - g_h}{g} = \frac{2h}{R}$. Percentage decrease: $\frac{2h}{R} \times 100\%$.

At a depth $d$ below the Earth's surface, the distance from the center is $(R - d)$. By Gauss's Law of gravitation, the spherical shell of thickness $d$ exerts zero net force at internal points. Only the inner solid sphere of radius $(R - d)$ attracts the mass:

Important Conclusions:

• At the center of the Earth ($d = R$): $g_{\text{center}} = g(1 - R/R) = 0$. A body at the center of the Earth is weightless.
• Comparing height and depth for small distances ($h = d$): $\Delta g_h = 2\Delta g_d$. The decrease in $g$ at height $h$ is twice the decrease at the same depth $d$.

As the Earth rotates about its polar axis with angular speed $\omega = \frac{2\pi}{86400} \approx 7.27 \times 10^{-5}\,\text{rad/s}$, a body at latitude $\lambda$ describes a circular path of radius $r = R\cos\lambda$. It experiences an outward centrifugal pseudo-force $m\omega^2 r = m\omega^2 R\cos\lambda$.

Resolving along the local vertical, the apparent acceleration due to gravity is:

• At the Equator ($\lambda = 0^\circ$): $\cos 0^\circ = 1 \implies g_{\text{eq}} = g - \omega^2 R$ (Minimum value).
• At the Poles ($\lambda = 90^\circ$): $\cos 90^\circ = 0 \implies g_{\text{pole}} = g$ (Maximum value, completely unaffected by rotation).
• Difference due to rotation: $g_{\text{pole}} - g_{\text{eq}} = \omega^2 R \approx (7.27 \times 10^{-5})^2 (6.4 \times 10^6) \approx 0.034\,\text{m/s}^2$.

Earth is an oblate spheroid, flattened at the poles and bulging at the equator: $R_{\text{eq}} - R_{\text{pole}} \approx 21\,\text{km}$. Since $g \propto 1/R^2$, the polar radius being smaller makes $g$ larger at the poles than at the equator by about $0.018\,\text{m/s}^2$. Combined with rotation, total difference is $\approx 0.052\,\text{m/s}^2$ ($g_{\text{pole}} \approx 9.83\,\text{m/s}^2$, $g_{\text{eq}} \approx 9.78\,\text{m/s}^2$).

The Gravitational Field Intensity at any point in space is defined as the gravitational force experienced per unit test mass placed at that point:

It is a vector quantity pointing radially toward the attracting mass $M$. Units: $\text{N/kg}$ or $\text{m/s}^2$, dimensions: $[LT^{-2}]$. Near Earth's surface, $\vec{E}_g = \vec{g}$.

The Gravitational Potential at a point is defined as the amount of work done by an external agent in bringing a unit mass from infinity to that point without acceleration:

Properties of Gravitational Potential:

• It is a scalar quantity.
• It is always negative throughout space, taking its maximum value of zero at infinity ($r \to \infty$). The negative sign signifies that the gravitational interaction is attractive and work is done by the field itself.
• Units: $\text{J/kg}$, dimensional formula: $[L^2T^{-2}]$.
• Relation to field intensity: $\vec{E}_g = -\nabla V = -\frac{dV}{dr}\hat{r}$.

The Gravitational Potential Energy of a system of two masses $M$ and $m$ separated by distance $r$ is the work done in bringing mass $m$ from infinity to distance $r$ in the field of $M$:

On Earth's surface ($r_1 = R$), potential energy is $U_1 = -\frac{GMm}{R}$.

At height $h$ ($r_2 = R + h$), potential energy is $U_2 = -\frac{GMm}{R + h}$.

The change in potential energy (work required by external agent) is:

Substituting $GM = g R^2$:

When $h \ll R$, $\frac{h}{R} \to 0$, giving the familiar elementary formula $\Delta U \approx mgh$. However, when $h$ is comparable to $R$ (e.g. $h = R$), $\Delta U = \frac{mgh}{1 + 1} = \frac{1}{2}mgh$.

Escape Velocity is defined as the minimum initial projection speed required for an object launched from the surface of a celestial body to completely overcome its gravitational attraction and reach infinity with zero or positive residual kinetic energy.

Let an object of mass $m$ be projected with speed $v_e$ from the surface of a planet of mass $M$ and radius $R$.

Total initial mechanical energy on the planet's surface:

At infinity ($r \to \infty$), the gravitational potential energy is zero ($U_f = 0$), and for minimum escape speed, the final kinetic energy is zero ($K_f = 0$). By Conservation of Mechanical Energy:

Using $GM = g R^2$ and $M = \frac{4}{3}\pi R^3 \rho$:

For Earth ($R = 6.4 \times 10^6\,\text{m}$, $g = 9.8\,\text{m/s}^2$):

Critical Characteristics of Escape Velocity:

• Independent of projectile mass ($m$): A pebble and a Saturn V rocket have identical escape speeds ($11.2\,\text{km/s}$).
• Independent of projection angle ($\theta$): Whether launched vertically or at $30^\circ$ to the horizontal, the escape speed is identical (neglecting air resistance).
• Depends solely on the mass $M$ and radius $R$ (or density $\rho$ and radius $R$) of the celestial body.

For the Moon ($M_m \approx \frac{M_e}{81}, R_m \approx \frac{R_e}{4}$):

The root-mean-square thermal speed of common atmospheric gas molecules ($O_2, N_2, CO_2, H_2O$) at lunar daytime temperature ($T \approx 400\,\text{K}$) is given by $v_{\text{rms}} = \sqrt{\frac{3k_B T}{m_{\text{molecule}}}}$. Because the thermal speeds exceed or are comparable to $v_{e,\text{moon}}$, the gas molecules escaped into space over cosmic timescales, leaving the Moon without any atmosphere.

If a dying star collapses such that its escape velocity equals or exceeds the speed of light $c = 3 \times 10^8\,\text{m/s}$ ($v_e \ge c$):

$R_s$ is called the Schwarzschild radius. Nothing, not even light, can escape from inside this event horizon, defining a black hole.

Consider a satellite of mass $m$ revolving in a circular orbit of radius $r = R + h$ around Earth (mass $M$, radius $R$).

The gravitational force provides the necessary centripetal force:

For a satellite orbiting close to Earth's surface ($h \ll R$):

Period of revolution ($T$):

For near-Earth orbit ($h \approx 0$): $T_0 = 2\pi\sqrt{\frac{R}{g}} = 2\pi\sqrt{\frac{6.4 \times 10^6}{9.8}} \approx 5078\,\text{s} \approx 84.6\,\text{mins}$.

Satellite Energy Relations:

• Kinetic Energy: $K = \frac{1}{2}m v_o^2 = \frac{GMm}{2r} = \frac{|U|}{2}$
• Potential Energy: $U = -\frac{GMm}{r}$
• Total Mechanical Energy: $E = K + U = \frac{GMm}{2r} - \frac{GMm}{r} = -\frac{GMm}{2r} = -K = \frac{U}{2}$
• Binding Energy: $\text{BE} = -E = \frac{GMm}{2r}$ (energy required to free satellite to infinity)

A common misconception is that astronauts in orbit experience weightlessness because "gravity is zero in space." In reality, at an ISS altitude of $400\,\text{km}$, $g_h = g(1 + h/R)^{-2} \approx 8.7\,\text{m/s}^2$, which is nearly $89\%$ of surface gravity!

The true reason for weightlessness is that the spacecraft and the astronaut are in a state of continuous free fall toward Earth with identical acceleration $a = g_h$. Because both accelerate together, the normal reaction force exerted by the spaceship floor on the astronaut is zero: $N = m(g_h - a) = 0$. Since human sensation of weight is provided by the contact reaction force $N$, the absence of $N$ produces the sensation of absolute weightlessness.