Gravitation revision notes

Historical Background & Kepler's Laws

The understanding of planetary motion evolved significantly over centuries. Ptolemy proposed a 'geocentric' model roughly 2000 years ago, placing Earth at the center with celestial objects moving in complicated circles. Later, Aryabhatta (5th century A.D.) and Nicolaus Copernicus (1473–1543) independently proposed the 'heliocentric' model, where planets revolve around a central Sun. Galileo Galilei recognized that all bodies accelerate towards the earth at a constant rate, regardless of mass, validating this through inclined plane experiments. Tycho Brahe compiled extensive naked-eye planetary observations, which his assistant Johannes Kepler later analyzed to formulate three fundamental laws.

Kepler’s Three Laws of Planetary Motion:

1. Law of Orbits: All planets move in elliptical orbits with the Sun situated at one of the focal points. The closest point to the Sun is the perihelion, and the farthest point is the aphelion.
2. Law of Areas: The line joining any planet to the Sun sweeps out equal areas in equal intervals of time. The areal velocity ($\Delta A / \Delta t$) is constant and is mathematically given by $\Delta A / \Delta t = L / 2m$, where $L$ is the angular momentum and $m$ is the mass of the planet.JEE TIPThis law is a direct consequence of the conservation of angular momentum, which holds true for any central force (a force directed along the position vector). [JEE TIP] Since angular momentum $L = mvr$ is constant, a planet moves faster when it is closer to the Sun ($v_p r_p = v_A r_A$), meaning $v_{perihelion} > v_{aphelion}$.
3. Law of Periods: The square of the time period of revolution ($T$) of a planet is proportional to the cube of the semi-major axis ($a$) of its elliptical orbit ($T^2 \propto a^3$). For circular orbits, this can be written as $T^2 = (4\pi^2 / GM_s) R^3$, where $M_s$ is the central mass.

Constant and Variable Quantities in Elliptical Orbits (Comet Motion) For a planet or comet in a highly elliptical orbit:

• Conserved quantities: Angular momentum (because gravity is a central force) and Total Mechanical Energy. Linear momentum is explicitly not conserved.
• Variable quantities: Linear speed, angular speed, kinetic energy, and potential energy all continuously change as the distance to the Sun varies.

Newton's Universal Law of Gravitation

Isaac Newton deduced that the same force causing apples to fall also keeps the moon in orbit, noting that the moon's centripetal acceleration is much smaller than $g$ due to its vast distance, scaling as $1/r^2$.

Universal Law: Every body in the universe attracts every other body with a force directly proportional to the product of their masses and inversely proportional to the square of the distance between them. Vector Form: The force $\vec{F}_{12}$ on point mass $m_1$ due to $m_2$ is mathematically expressed as: $\vec{F}_{12} = \frac{G m_1 m_2}{r^2} \hat{r}_{21} = -\frac{G m_1 m_2}{r^3} \vec{r}_{12}$. Here, $G$ is the universal gravitational constant, $\vec{r}_{12} = \vec{r}_2 - \vec{r}_1$ is the position vector from $m_1$ to $m_2$, and the negative sign indicates the force is attractive.

Superposition Principle: For a collection of point masses, the net gravitational force on any one mass is the vector sum of the individual gravitational forces exerted by all other masses: $\vec{F}_R = \sum \vec{F}_i$.

Extended Objects (Shell Theorem): Newton's law directly applies only to point masses, but integration yields critical rules for spherically symmetric bodies:

1. Hollow Spherical Shell (Outside): The force between a uniform hollow spherical shell and a point mass outside it acts precisely as if the shell's entire mass were concentrated at its geometric center.
2. Hollow Spherical Shell (Inside): The gravitational force exerted by a uniform hollow spherical shell on a point mass situated inside it is exactly zero.
3. Solid Homogeneous Sphere: For a point mass situated inside a uniform solid sphere, the gravitational force acts toward the center and its magnitude is directly proportional to the distance $r$ from the center ($F \propto r$). This force is exclusively exerted by the spherical mass interior to the particle's position; the outer shell contributes zero net force.

The Gravitational Constant ($G$)

The value of $G$ was first accurately measured experimentally by Henry Cavendish in 1798 using a torsional balance. In his setup, large lead spheres were brought close to small spheres suspended on a bar via a fine wire, creating a measurable gravitational torque. The gravitational torque ($F \times L$) was equated to the wire's restoring torque ($\tau \theta$), yielding the relationship: $\tau \theta = G \frac{Mm}{d^2} L$. The universally accepted value is $G = 6.67 \times 10^{-11} \text{ N m}^2/\text{kg}^2$. By calculating $G$ and knowing $g$ and the Earth's radius, Cavendish effectively "weighed the earth" for the first time.

Acceleration Due to Gravity ($g$) & Its Variation

Assuming the Earth is a spherically symmetric solid, a point outside the Earth experiences gravity as if the Earth's entire mass ($M_E$) is concentrated at its center. On the surface of the Earth ($r = R_E$), the acceleration due to gravity is: $g = \frac{GM_E}{R_E^2}$.

1. Variation with Height ($h$): For a point at height $h$ above the surface, the distance from the center is $(R_E + h)$. $g(h) = \frac{GM_E}{(R_E + h)^2}$. [JEE TIP] If the height is very small compared to the Earth's radius ($h \ll R_E$), use the binomial approximation: $g(h) \approx g(1 - \frac{2h}{R_E})$. Use the exact formula for large heights (e.g., $h = R_E$).

2. Variation with Depth ($d$): For a point at depth $d$, the distance from the center is $(R_E - d)$. The outer spherical shell of thickness $d$ exerts zero net force. Only the smaller inner sphere of radius $(R_E - d)$ contributes. Assuming uniform density, the mass of this inner sphere is $M_s = M_E (R_E - d)^3 / R_E^3$. This yields the exact formula for any depth $d$: $g(d) = g(1 - \frac{d}{R_E})$. [JEE TIP] Acceleration due to Earth's gravity is strictly maximum at the surface; it decreases linearly as you go down to the center, and falls off as $1/r^2$ as you go up into space.

Gravitational Potential and Potential Energy

Gravitational force is a conservative force, meaning the work done is independent of the path taken. For points very close to the Earth's surface, the force is roughly constant ($mg$), and potential energy is approximated as $W(h) = mgh + W_o$, where $W_o$ is the potential energy at the surface.

However, for arbitrary distances where $g$ is not constant, the generalized potential energy difference $W_{12}$ between distances $r_1$ and $r_2$ is calculated via integration: $W_{12} = GM_E m \left( \frac{1}{r_1} - \frac{1}{r_2} \right)$.

By convention, potential energy is defined as zero at infinity ($r \to \infty$). Therefore, the gravitational potential energy ($V$ or $U$) of a two-particle system separated by distance $r$ is: $V(r) = - \frac{G m_1 m_2}{r}$.

Gravitational Potential: Defined as the gravitational potential energy per unit mass at a point. Superposition: For a multi-particle system, the total potential energy is the algebraic sum of the potential energies of all possible pairs of constituent particles. [JEE TIP] For a system of $N$ particles, there are exactly $N(N-1)/2$ pairs to compute.

Escape Speed & Satellites

Escape Speed: By the principle of conservation of energy, the minimum initial speed ($v_i$) required for a projectile to escape Earth's gravity and reach infinity (where its kinetic and potential energies approach zero) is the escape speed. Equating surface energy to energy at infinity: $\frac{1}{2} m v_e^2 - \frac{GM_E m}{R_E} = 0$. $v_e = \sqrt{\frac{2GM_E}{R_E}} = \sqrt{2gR_E}$. On Earth, $v_e \approx 11.2 \text{ km/s}$. On the Moon, gravity is weaker, making $v_e \approx 2.3 \text{ km/s}$, which explains the absence of a lunar atmosphere (gas molecules exceed this thermal speed and escape).

Earth Satellites: Satellites obey Kepler's laws. For a circular orbit at height $h$, the centripetal force is provided entirely by gravitational attraction: $\frac{mV^2}{R_E + h} = \frac{GM_E m}{(R_E + h)^2}$. Orbital speed: $V = \sqrt{\frac{GM_E}{R_E + h}}$. For a satellite very close to Earth ($h \approx 0$), $V \approx \sqrt{gR_E}$.

Time period of satellite: $T = \frac{2\pi(R_E + h)}{V} = \frac{2\pi (R_E + h)^{3/2}}{\sqrt{GM_E}}$. Squaring gives Kepler's Law of Periods: $T^2 = (\frac{4\pi^2}{GM_E}) (R_E + h)^3$. For a satellite close to Earth's surface, $T_0 = 2\pi \sqrt{\frac{R_E}{g}} \approx 85 \text{ minutes}$.

Energy of an Orbiting Satellite: In an orbit of radius $r = R_E + h$:

• Kinetic Energy (KE): Positive, $KE = \frac{GM_E m}{2r}$.
• Potential Energy (PE): Negative, $PE = - \frac{GM_E m}{r}$.
• Total Energy (TE): $TE = KE + PE = - \frac{GM_E m}{2r}$.JEE TIPNotice the strict magnitude relationship for circular orbits: $TE = -KE = \frac{PE}{2}$. The total energy is negative, which is the defining characteristic of a "bound system" (closed orbit). If $TE \ge 0$, the object escapes to infinity.

Key Concepts & Definitions

Central Force:

A force directed radially along the vector joining the interacting bodies. Angular momentum is strictly conserved under central forces.

Conservative Force:

A force for which the work done is independent of the path taken between final and initial points (e.g., gravity).

Bound System:

A state where the total mechanical energy is negative, forcing the object to remain in a closed (elliptical/circular) orbit at a finite distance.

Weightlessness:

A state experienced by astronauts in orbit. It occurs not because gravity is absent, but because both the satellite and the astronaut are in continuous "free fall" towards the Earth at the same rate.

Neutral Point:

The position in space where the net gravitational pull from multiple massive bodies completely cancels out (net force is zero).

Tidal Effect:

Even though the Sun exerts a much larger absolute gravitational pull on the Earth than the Moon does, the Moon is responsible for stronger tidal effects. Tides arise from the gradient or difference in gravitational force across the Earth's diameter, which depends on 1/r31/r^31/r3. The Moon's proximity makes its differential pull greater than the Sun's.

Conditions & Limitations

• Newton's Law Formula ($1/r^2$): Strictly applicable only for point masses or outside spherically symmetric uniform bodies. For asymmetric extended objects, integration must be used.
• $mgh$ Potential Energy: The formula $\Delta U = mgh$ is a localized approximation valid only when $h$ is trivially small compared to Earth's radius, assuming $g$ remains constant. Do not use for satellite launches.
• $g(h) \approx g(1 - 2h/R)$ Approximation: Valid only when $h \ll R_E$ (typically less than 5% of Earth's radius). If $h = R_E / 2$, you must use the exact inverse square relation.
• Escape Velocity Limits: $v_e = \sqrt{2gR_E}$ assumes projection from the surface of a non-rotating Earth, neglecting atmospheric drag and the presence of other planetary bodies (like the Sun).

Previous Year JEE Topics

• Calculating variation of $g$ above and below the surface, equating $g(h)$ to $g(d)$ to find depth/height relationships.
• Applying Conservation of Mechanical Energy combined with Angular Momentum conservation to find the speed of a satellite shifting from perihelion to aphelion.
• Work required to shift a satellite from orbit $R_1$ to orbit $R_2$ ($W = E_{final} - E_{initial}$).
• Locating the Neutral Point between two massive bodies and finding the minimum initial velocity to overcome the gravitational barrier between them.
• Energy-speed relationships for projectiles launched with speeds less than, equal to, or greater than escape velocity.

Standard Derivations & Step-by-Step Problem Solving

Step-by-Step: Solving Neutral Point Projection Problems Scenario: Projecting a mass $m$ from the surface of Planet $M_1$ toward Planet $M_2$ (distance $D$ apart).

1. Find Neutral Point ($N$): Equate the gravitational fields: $\frac{G M_1}{r^2} = \frac{G M_2}{(D - r)^2}$ to find the critical distance $r$ from $M_1$ where forces cancel.
2. Setup Energy Conservation: To just reach $M_2$, the projectile only needs enough initial kinetic energy to reach point $N$. Past $N$, $M_2$'s gravity pulls it in automatically.
3. Calculate: $E_{surface} = E_{neutral\_point}$ $\frac{1}{2} m v^2 - \frac{G M_1 m}{R_1} - \frac{G M_2 m}{(D - R_1)} = 0 - \frac{G M_1 m}{r} - \frac{G M_2 m}{(D - r)}$. Solve for $v$. Note that at the neutral point, kinetic energy is effectively zero for the minimum required speed.

Step-by-Step: Shifting Satellite Orbits Scenario: A satellite of mass $m$ is moved from circular orbit $R_i$ to $R_f$.

1. Calculate Initial Total Energy: $E_i = - \frac{GM_E m}{2 R_i}$.
2. Calculate Final Total Energy: $E_f = - \frac{GM_E m}{2 R_f}$.
3. Change in Energy required (Work done by rockets): $\Delta E = E_f - E_i = \frac{GM_E m}{2} \left( \frac{1}{R_i} - \frac{1}{R_f} \right)$.JEE TIPThe change in Potential Energy is strictly twice the change in Total Energy: $\Delta V = 2\Delta E$, while the change in Kinetic Energy is $\Delta K = - \Delta E$.