Gravitation is the universal attractive interaction acting between all masses in the universe, described classically by Newton's law of universal gravitation and geometrically by Einstein's general theory of relativity. It governs the motion of planets, satellites, and projectiles, forming a foundational central force field in classical mechanics.
UP TGT ScienceUP PGT PhysicsStudy Notes + Practice
Scope and prerequisites
TGT core
Newton's law of universal gravitation, acceleration due to gravity (g) and its variation with altitude, depth, and latitude, gravitational potential and field, Kepler's laws of planetary motion, escape velocity, and orbital velocity of satellites.
PGT extension
Gravitational potential and field due to extended bodies (spherical shells, solid spheres), gravitational self-energy, motion under central forces, reduction of two-body problem to equivalent one-body problem, reduced mass, and effective potential energy curves.
Prerequisites
Vector calculus and resolution of forces
Newton's laws of motion and kinematics
Work, energy, and conservation laws (angular momentum and mechanical energy)
Newton's Law of Universal Gravitation
Every particle attracts every other particle with a force proportional to the product of their masses and inversely proportional to the square of the distance between them.
The vector form is given by F12 = -G(m1m2/r3) r12, where the negative sign signifies the attractive nature of the force.
It is a conservative, central force obeying the inverse-square law, making the total mechanical energy of bound systems conserved.
Variation of Acceleration Due to Gravity (g)
Due to altitude (h): g' = g(1 - 2h/R) for h << R, showing that g decreases with height above the Earth's surface.
Due to depth (d): g' = g(1 - d/R) assuming a uniform density sphere, indicating that g decreases linearly toward the Earth's center.
Due to rotation (latitude lambda): g' = g - R omega2 cos2(lambda), causing g to be minimum at the equator and maximum at the poles.
Gravitational Field and Potential
Gravitational field E is the gravitational force experienced per unit test mass placed at that point, with units N/kg or m/s2.
Gravitational potential V at a point is the work done per unit mass in bringing a test mass from infinity to that point against the gravitational field.
The relation between field and potential is given by $E = -grad(V)$, reflecting the conservative nature of the gravitational force.
Kepler's Laws of Planetary Motion
First Law (Law of Orbits): All planets move in elliptical orbits with the Sun located at one of the foci.
Second Law (Law of Areas): The line joining the planet to the Sun sweeps out equal areas in equal intervals of time, a direct consequence of conservation of angular momentum.
Third Law (Law of Periods): The square of the time period of revolution of a planet is directly proportional to the cube of the semi-major axis of its elliptical orbit.
Derivations
Variation of Gravitational Acceleration with Depth
Conditions: Earth is assumed to be a uniform solid sphere of radius R and mass M.
Consider a point at a depth d below the Earth's surface. The effective distance from the center is $r = R - d$.
The mass of the inner sphere contributing to the gravitational pull at depth d is M' = (4/3) pi r3 rho, where rh$o = M / ((4/3) \pi R^{3})$.
Substituting rho, we get M' = M (r / R)3 = M ((R - d) / R)3.
The acceleration due to gravity at depth d is g' = G M' / r2 = G (M ((R - d)/R)3) / (R - d)2.
Conditions: Body of mass m projected from the surface of Earth of mass M and radius R to infinity.
Total mechanical energy on the surface of Earth is $E = Kinetic Energy + Potential Energy = (1/2)m v_{e}^{2} - (G M m) / R$.
At infinity, the minimum velocity required makes both final kinetic energy and potential energy zero, so E = 0.
Applying conservation of mechanical energy: (1/2)m ve2 - (G M m) / R = 0.
Solving for ve gives ve2 = 2 G M / R.
Using $g = G M / R^{2}$, substitute G $M = g R^{2}$ to obtain ve = sqrt(2 g R).
Result: $v_e = \sqrt(2gR) \approx 11.2 km/s$
Formula sheet
Formula / law
Equation
Variables and units
Conditions
Newton's Law of Gravitation
$$F = \frac{G m_1 m_2}{r^2}$$
F = gravitational force, G = universal gravitational constant, m_1, $m_2 = masses$, r = separation distance N
Point masses or spherically symmetric bodies
Gravitational Acceleration
$$g = \frac{G M}{R^2}$$
g = acceleration due to gravity, G = gravitational constant, M = mass of planet, R = radius of planet m/s^2
Surface of a spherical planet
Gravitational Potential
$$V = -\frac{G M}{r}$$
V = gravitational potential, M = mass, r = distance from center J/kg
Outside a spherically symmetric mass distribution, reference at infinity
Orbital Velocity
$$v_o = \sqrt{\frac{G M}{R + h}}$$
$v_o = orbital velocity$, h = height above surface, R = radius, M = mass m/s
Circular orbit around a central body
Escape Velocity
$$v_e = \sqrt{2gR}$$
$v_e = escape velocity$, g = surface gravity, R = radius m/s
From the surface of a planet
Kepler's Third Law
$$T^2 = \frac{4\pi^2}{G M} r^3$$
T = time period, $r = semi-major axis$, M = central mass s^2 / m^3
Elliptical or circular orbit
Gravitational Potential Energy
$$U = -\frac{G M m}{r}$$
U = potential energy, M, m = masses, r = separation J
Zero potential energy at infinite separation
Variation of g with Latitude
$$g' = g - R \omega^2 \cos^2(\lambda)$$
g' = effective gravity, omega = angular velocity of Earth, lambda = latitude m/s^2
Rotating reference frame of Earth
Graphs and experiments
Variation of g with Distance from Center (r)
Draw: A graph with r on the x-axis and g on the y-axis. For r < R, a straight line passing through the origin (g proportional to r). For r > R, a curve falling off as 1/r2.
Interpret: Shows that gravity increases linearly from the center to the surface, reaches a maximum at R, and then decreases inversely with the square of the distance outside.
Exam use: Frequently tested in conceptual MCQs comparing g at internal points versus external points.
Effective Potential Energy Curve for Central Forces
Draw: A plot of $U_eff(r) = L^2 / (2 m r^2) - G M m / r against radial distance r$, showing a potential well with a minimum.
Interpret: The minimum of the effective potential corresponds to a stable circular orbit where the centrifugal repulsive term balances the gravitational attractive term.
Exam use: Used in PGT level problems to determine turning points, circular orbit radii, and stability.
Cavendish Experiment Setup
Draw: A torsion balance with two small lead spheres at the ends of a light rod suspended by a fine fiber, with two large lead spheres brought near them.
Interpret: Demonstrates the weak gravitational attraction between laboratory-scale masses and is used to measure the universal gravitational constant G.
Exam use: Tested as a conceptual question regarding the experimental determination of G.
Solved numericals
TGT core
Calculate the height above the Earth's surface at which the acceleration due to gravity drops to 36% of its value on the surface. (Radius of Earth R = 6400 km)
Given condition: g' = 0.36 g.
Formula for gravity at height h: g' = g (R / (R + h))^2.
Substitute g': 0.36 $g = g (R / (R + h))^2$.
Take square root on both sides: $0.6 = R / (R + h)$.
Solve for h: 0.6(R + h) = $R => 0.6 R + 0.6 h = R => 0.6 h = 0.4 R => h = (4/6) R = (2/3) R$.
A particle of mass m is placed at the center of a thin spherical shell of mass M and radius R. What is the net gravitational force on the particle, and what is the work done in moving the particle from the center to a point at distance 2R?
By shell theorem, the gravitational field inside a uniform spherical shell is zero everywhere. Thus, the net force on the particle at the center is zero.
Gravitational potential at the center of the shell: $V_center = -G M / R$.
Gravitational potential at a distance $r = 2R outside the shell: V_outside = -G M / (2R)$.
Work done by an external agent in moving the mass m from center to 2R is $W = m (V_final - V_initial)$.
$W = m [(-G M / (2R)) - (-G M / R)] = m (G M / 2R)$.
Answer: Force = 0, Work don$e = (G M m) / (2 R)$
TGT core
Find the orbital velocity and time period of an artificial satellite revolving in a circular orbit at a height of 600 km above the Earth's surface. (Take R = 6400 km, $g = 9.8 m/s^2)$
Radius of orbit $r = R + h = 6400 + 600 = 7000 km = 7 * 10^6 m$.
Orbital velocity $v_o = \sqrt(G M / r) = \sqrt(g R^2 / r)$.
Newton's law assumes instantaneous action at a distance, which fails for high-speed gravitational interactions governed by general relativity (finite speed of gravity equal to light speed c).
Earth is treated as a spherically symmetric body of uniform or spherically stratified density in elementary problems, whereas actual Earth is an oblate spheroid (geoid).
Common misconceptions
Common error: Gravity is zero at the center of the Earth because you are equidistant from all sides.
Correct: Gravity is zero at the center because the mass shells enclosing spheres of zero radius exert no net force, and symmetrical mass cancellation results in zero vector sum.
Vector addition of forces from all directions inside a symmetric hollow or solid distribution results in total cancellation.
Common error: Escape velocity depends on the mass of the escaping body.
Correct: Escape velocity is independent of the mass of the projectile.
Mass m cancels out in the energy conservation equation equating kinetic energy to gravitational potential energy.
Quick revision
Gravitational force is always attractive, conservative, and central, obeying the inverse-square law.
Newton's Law of Gravitation: $F = \frac{G m_1 m_2}{r^2}$
Gravitational Acceleration: $g = \frac{G M}{R^2}$
Must remember
Newton's Law of Universal Gravitation states that the force between two point masses is $F = \frac{G m_1 m_2}{r^2}$ and represents a conservative, central force.
Acceleration due to gravity varies with altitude as $g' \approx g(1 - \frac{2h}{R})$, with depth as $g' = g(1 - \frac{d}{R})$, and with latitude as $g' = g - R \omega^2 \cos^2(\lambda)$.
Escape velocity from a planet of mass $M$ and radius $R$ is given by $v_e = \sqrt{2gR}$ and is independent of the mass of the escaping body.
Kepler's Second Law (Law of Areas) is a direct consequence of the conservation of angular momentum for planetary motion.
Gravitational potential $V$ and gravitational field $E$ are related by $E = -\text{grad}(V)$.
Glossary
Universal Gravitational Constant ($G$)A proportionality constant in Newton's law of gravitation, equal to $6.674 \times 10^{-11} \text{ N m}^2/\text{kg}^2$.
Gravitational Field ($E$)The gravitational force experienced per unit test mass placed at a point in space, measured in $\text{N/kg}$ or $\text{m/s}^2$.
Gravitational Potential ($V$)The work done per unit mass in bringing a test mass from infinity to a point against the gravitational field, measured in $\text{J/kg}$.
Escape Velocity ($v_e$)The minimum velocity required for an object to escape the gravitational field of a celestial body to infinity, $v_e = \sqrt{2gR}$.
Reduced Mass ($mu$)In a two-body problem, the effective inertial mass $\mu = \frac{m_1 m_2}{m_1 + m_2}$ used to reduce it to a single-body equivalent motion.
Orbital Velocity vs Escape Velocity
Parameter
Orbital Velocity ($v_o$)
Escape Velocity ($v_e$)
Formula (near surface)
$v_o = \sqrt{gR}$
$v_e = \sqrt{2gR}$
Ratio ($v_e / v_o$)
$\sqrt{2} \approx 1.414$
$1$
Trajectory
Stable circular orbit around the central body
Open parabolic or hyperbolic trajectory to infinity
Variation of $g$ with Altitude and Depth
Condition
Formula
Behavior as distance changes
Altitude ($h \ll R$)
$g' = g(1 - \frac{2h}{R})$
Decreases linearly with height $h$
Depth ($d$) for uniform sphere
$g' = g(1 - \frac{d}{R})$
Decreases linearly to zero at the center
Center of Earth ($d = R$)
$g' = 0$
Zero effective gravitational acceleration
Practice Quiz
Choose an option to reveal the explanation.
1. What is the nature of the gravitational force given by $F_{12} = -G\frac{m_1 m_2}{r^3} \mathbf{r}_{12}$?
Explanation: Newton's law of gravitation represents a conservative, central force that obeys the inverse-square law, allowing total mechanical energy to be conserved.
2. How does the acceleration due to gravity $g'$ vary at a small height $h$ above the Earth's surface compared to the surface value $g$?
Explanation: For $h \ll R$, expanding the exact formula yields $g' = g(1 - \frac{2h}{R})$, showing that gravity decreases with altitude.
3. At what location on the Earth's surface is the effective acceleration due to gravity maximum due to the Earth's rotation?
Explanation: The effective gravity is given by $g' = g - R\omega^2 \cos^2(\lambda)$. At the poles, latitude $\lambda = 90^{\circ}$, so $\cos(90^{\circ}) = 0$, making $g' = g$ (maximum).
4. What is the relation between the gravitational field $E$ and gravitational potential $V$?
Explanation: The gravitational field is the negative gradient of the gravitational potential, $E = -\text{grad}(V)$, reflecting the conservative nature of the force.
5. Which of Kepler's laws is a direct consequence of the law of conservation of angular momentum?
Explanation: Kepler's Second Law states that the areal velocity $\frac{dA}{dt} = \frac{L}{2m}$ is constant, which directly stems from the conservation of angular momentum $L$ under central forces.
6. What is the escape velocity $v_e$ from the surface of a planet of mass $M$ and radius $R$?
Explanation: By equating total mechanical energy to zero for an object escaping to infinity, $\frac{1}{2}m v_e^2 - \frac{GMm}{R} = 0$, giving $v_e = \sqrt{\frac{2GM}{R}} = \sqrt{2gR}$.
7. What is the gravitational potential energy of a two-mass system with masses $M$ and $m$ separated by distance $r$?
Explanation: Gravitational potential energy is defined with zero reference at infinite separation, yielding $U = -\frac{GMm}{r}$.
8. If the Earth suddenly shrinks to half its present radius while keeping its mass constant, what will happen to the acceleration due to gravity $g$ on its surface?
Explanation: Since $g = \frac{GM}{R^2}$, if $R$ becomes $\frac{R}{2}$, then $g$ becomes $\frac{GM}{(R/2)^2} = 4 \frac{GM}{R^2} = 4g$.
9. What is the orbital velocity $v_o$ of a satellite revolving in a circular orbit at a height $h$ above a planet of mass $M$ and radius $R$?
Explanation: Equating the gravitational force to the centripetal force: $\frac{GMm}{(R+h)^2} = \frac{m v_o^2}{R+h}$, which gives $v_o = \sqrt{\frac{GM}{R+h}}$.
10. According to Kepler's Third Law, how does the time period $T$ of a planet relate to the semi-major axis $r$ of its orbit?
Explanation: Kepler's Third Law states that the square of the time period of revolution is directly proportional to the cube of the semi-major axis, $T^2 = \frac{4\pi^2}{GM}r^3$.
11. (PGT Extension) How is the reduced mass $\mu$ defined for a two-body system with masses $m_1$ and $m_2$?
Explanation: The two-body problem under central forces is reduced to an equivalent one-body problem using the reduced mass $\mu = \frac{m_1 m_2}{m_1 + m_2}$.
12. (PGT Extension) What is the gravitational self-energy of a uniform solid sphere of mass $M$ and radius $R$?
Explanation: Integrating the potential energy contributions of spherical shell increments yields the self-energy of a uniform solid sphere as $U = -\frac{3}{5} \frac{GM^2}{R}$.
13. What is the value of the acceleration due to gravity $g'$ at a depth $d$ inside the Earth assuming uniform density?
Explanation: Assuming uniform density, only the mass contained within a sphere of radius $(R-d)$ contributes to the effective gravitational force, yielding $g' = g(1 - \frac{d}{R})$.
14. Why does a body weigh more at the poles than at the equator?
Explanation: The effective gravity formula $g' = g - R\omega^2\cos^2(\lambda)$ shows that the centrifugal term is zero at the poles ($\lambda=90^{\circ}$) and maximum at the equator ($\lambda=0^{\circ}$), reducing effective weight at the equator.
15. (PGT Extension) For a bound system in an elliptical orbit under an inverse-square central force, what is the nature of its total mechanical energy $E$?
Explanation: Bound systems (such as planets or satellites in closed orbits) always possess negative total mechanical energy ($E < 0$), with kinetic energy being less than the absolute magnitude of potential energy.
PYQ Patterns
These are pattern-based practice questions. Authentic year labels are only used when verified from a source.
1. If the radius of the Earth were to shrink by $1\%$ while its mass remained constant, what would be the approximate percentage change in the acceleration due to gravity $g$ on its surface?
Explanation: Since $g = \frac{GM}{R^2}$, taking natural logarithms gives $\frac{\Delta g}{g} = -2 \frac{\Delta R}{R}$. If $\frac{\Delta R}{R} = -1\%$, then $\frac{\Delta g}{g} = -2(-1\%) = +2\%$, meaning $g$ increases by $2\%$.
2. A satellite is orbiting the Earth in a circular orbit of radius $r$ with velocity $v_o$. If the orbital radius is increased to $4r$, what will be its new orbital velocity?
Explanation: Orbital velocity is given by $v_o = \sqrt{\frac{GM}{r}}$, so $v_o \propto \frac{1}{\sqrt{r}}$. When the radius becomes $4r$, the new velocity is $\frac{v_o}{\sqrt{4}} = \frac{v_o}{2}$.
3. At what depth $d$ below the Earth's surface is the value of acceleration due to gravity equal to half of its value at the surface?
Explanation: Using the depth variation formula $g' = g(1 - \frac{d}{R})$, setting $g' = \frac{g}{2}$ gives $\frac{g}{2} = g(1 - \frac{d}{R}) \implies \frac{d}{R} = \frac{1}{2} \implies d = \frac{R}{2}$.
4. The escape velocity of a body from the Earth's surface is $11.2 \text{ km/s}$. If a body is projected with twice this escape velocity, what will be its speed far away in interstellar space?
5. (PGT Extension) What is the gravitational potential at a distance $r$ from the center inside a uniform solid sphere of mass $M$ and radius $R$ ($r < R$)?
Explanation: Integrating the interior field $E = \frac{GMr}{R^3}$ from $r$ to $R$ and matching potential at the surface gives $V = -\frac{GM}{2R^3}(3R^2 - r^2)$ for $r < R$.
6. Two uniform spheres of masses $m_1$ and $m_2$ are placed with their centers separated by distance $r$. If the separation distance $r$ is doubled, how does the gravitational force between them change?
Explanation: According to Newton's law of gravitation, $F \propto \frac{1}{r^2}$. If $r$ is doubled ($2r$), the force becomes $\frac{1}{(2r)^2} = \frac{1}{4}$ of the original value.
Timed Topic Test
Attempt these questions under a ten-minute limit using a separate timer.
1. What is the gravitational field intensity inside a thin uniform spherical shell of mass $M$ and radius $R$?
Explanation: By Gauss's Law for gravitation, the net gravitational field inside a closed thin uniform spherical shell is zero due to symmetrical mass cancellation from all directions.
2. If the Earth's angular velocity of rotation about its axis increases such that the effective acceleration due to gravity at the equator becomes zero, what would be the length of the day?
Explanation: Setting $g' = g - R\omega^2 = 0$ gives $\omega = \sqrt{\frac{g}{R}} = \sqrt{\frac{9.8}{6.4 \times 10^6}} \approx 1.24 \times 10^{-3} \text{ rad/s}$. The time period $T = \frac{2\pi}{\omega} \approx 5060 \text{ s} \approx 84.6 \text{ minutes}$.
3. (PGT Extension) What is the effective potential energy curve term representing the centrifugal barrier for a particle moving under a central gravitational force with angular momentum $L$?
Explanation: The effective potential energy includes the true gravitational potential and the centrifugal barrier term, $U_{\text{eff}}(r) = -\frac{GMm}{r} + \frac{L^2}{2\mu r^2}$.
4. What is the ratio of escape velocity $v_e$ to orbital velocity $v_o$ for a satellite orbiting just above the Earth's surface?
Explanation: Since $v_e = \sqrt{2gR}$ and $v_o = \sqrt{gR}$, their ratio is $\frac{v_e}{v_o} = \sqrt{2}$.
5. A planet revolves in an elliptical orbit around the Sun. At which position is its speed maximum?
Explanation: By conservation of angular momentum ($L = mvr = \text{constant}$), speed is inversely proportional to distance from the Sun. Thus, speed is maximum at the perihelion (closest approach).
6. What happens to the total mechanical energy of an orbiting satellite if small frictional drag forces due to upper atmosphere air resistance act on it?
Explanation: Air resistance is a non-conservative dissipative force that removes mechanical energy from the satellite, causing its total mechanical energy to decrease and its orbit to spiral inward.
7. If two masses $m_1$ and $m_2$ are separated by distance $r$, the work done by the gravitational force when the distance is reduced from $r_1$ to $r_2$ (where $r_2 < r_1$) is:
Explanation: Work done by a conservative force is negative of change in potential energy: $W = -(\Delta U) = -(-\frac{GMm}{r_2} - (-\frac{GMm}{r_1})) = GMm_1m_2(\frac{1}{r_2} - \frac{1}{r_1})$.
8. What is the dimensional formula of the universal gravitational constant $G$?
9. (PGT Extension) For a spherical shell of mass $M$ and radius $R$, what is the gravitational potential at any point inside the shell?
Explanation: The gravitational field inside a thin spherical shell is zero, meaning the potential is constant throughout the interior and equal to its value at the surface, which is $-\frac{GM}{R}$.
10. What is the time period of a simple pendulum inside a satellite orbiting around the Earth?
Explanation: Inside an orbiting satellite, effective gravity $g_{\text{eff}} = 0$ (weightlessness). Since $T = 2\pi \sqrt{\frac{L}{g_{\text{eff}}}}$, the time period becomes infinite (the pendulum does not oscillate).