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Mechanics
Gravitational Force and Central Field Mechanics
Gravitational force is a fundamental universal attractive interaction acting between any two point masses. It is a conservative central force governed by Newton's Law of Universal Gravitation.
UP TGT ScienceUP PGT PhysicsStudy Notes + Practice
Scope and prerequisites
TGT core
Newton's law of gravitation, variation of acceleration due to gravity (g) with depth and altitude, gravitational potential energy, escape velocity, and Kepler's laws of planetary motion.
PGT extension
Gravitational field and potential for continuous mass distributions (spherical shell, solid sphere), planetary motion under inverse-square central forces, Kepler's equation, and non-inertial frame transformations involving centrifugal and Coriolis forces.
Prerequisites
Vector calculus and spherical polar coordinate systems
Newton's laws of motion and kinematics
Conservation laws of linear and angular momentum
Newton's Law of Universal Gravitation
States that 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.
Expressed mathematically as $F = G (m_{1}m_{2} / r^{2}) r̂$, where G is the universal gravitational constant.
It acts along the line joining the point masses, making it a central force and hence a conservative force field.
Gravitational Field and Potential
Gravitational field intensity E is defined as the gravitational force experienced by a unit test mass placed at that point: $E = F / m$.
Gravitational potential V at a point is the work done per unit mass in bringing a test mass from infinity to that point without acceleration.
The relation between field and potential is given by the negative gradient: $E = -∇V$.
Variation of Acceleration Due to Gravity (g)
At altitude h above the Earth's surface (for h << R), g' = g(1 - 2h/R).
At depth d below the Earth's surface, g' = g(1 - d/R), assuming a uniform density Earth model.
Due to the Earth's diurnal rotation with angular velocity omega, the effective gravity at latitude lambda is given by g' = g - Rω2 cos2λ.
Kepler's Laws of Planetary Motion
First Law (Law of Orbits): Planets move in elliptical orbits with the Sun located at one of the two foci.
Second Law (Law of Areas): The position vector of a planet from the Sun sweeps out equal areas in equal intervals of time, which is a direct consequence of conservation of angular momentum.
Third Law (Law of Periods): The square of the orbital period of a planet is directly proportional to the cube of the semi-major axis of its orbit (T2 ∝ a3).
Derivations
Variation of Gravitational Acceleration with Depth
Conditions: Assuming the Earth is a homogeneous sphere of radius R and uniform density ρ.
Consider a point at depth d below the surface. The effective distance from the center is $r = R - d$.
According to Newton's shell theorem, the mass outside this radius sphere exerts no net gravitational force on a body inside.
The effective mass causing attraction is M' = (ρ)(4/3)π(R - d)3 and total mass $M = (\rho)(4/3)\piR^{3}$, giving M' = M(R - d / R)3.
The gravitational acceleration at depth d is g' = G M' / (R - d)2 = G M (R - d) / R3.
Substituting GM / R2 = g, we obtain g' = g (1 - d / R).
Result: g' = g (1 - d / R)
Orbital Velocity and Total Energy of a Satellite
Conditions: Satellite of mass m revolving in a circular orbit of radius r around a central mass M.
Equate the gravitational force to the required centripetal force: G M m / r2 = m v2 / r.
Solve for orbital velocity: $v = \sqrt{G M / r}$.
Kinetic energy $K = (1/2) m v^{2} = G M m / (2r)$.
Potential energy $U = - G M m / r$.
Total mechanical energy $E = K + U = - G M m / (2r)$.
Result: $E = - \frac{G M m}{2r}$
Formula sheet
Formula / law
Equation
Variables and units
Conditions
Newton's Law of Gravitation
$$F = - \frac{G m_1 m_2}{r^2} \hat{r}$$
G = Gravitational constant, m1, m2 = masses, r = separation distance N
Point masses or spherically symmetric bodies
Gravitational Potential Energy
$$U = - \frac{G M m}{r}$$
G = Gravitational constant, M = central mass, m = orbiting mass, r = distance J
Zero potential energy referenced at infinity
Escape Velocity
$$v_e = \sqrt{\frac{2 G M}{R}} = \sqrt{2gR}$$
G = Gravitational constant, M = mass of planet, R = radius of planet, g = surface gravity m/s
Body projected from the surface of a spherically symmetric planet
Orbital Velocity
$$v_o = \sqrt{\frac{G M}{r}}$$
G = Gravitational constant, M = central mass, r = orbital radius m/s
Circular orbit close to or at altitude h above the surface
Kepler's Third Law
$$T^2 = \left( \frac{4 \pi^2}{G M} \right) r^3$$
T = Orbital period, $r = radius of circular orbit or semi-major axis$, M = central mass s^2 / m^3 (proportionality)
Keplerian central force orbit
Gravity Variation with Altitude
$$g' = g \left(1 - \frac{2h}{R}\right)$$
g = surface gravity, h = height above surface, R = radius of Earth m/s^2
h << R
Gravity Variation with Latitude
$$g' = g - R \omega^2 \cos^2(\lambda)$$
$\omega = angular velocity of Earth$, $\lambda = latitude angle$, R = Earth radius m/s^2
Rotating reference frame of the Earth
Gravitational Field of a Uniform Spherical Shell (Internal)
$$E = 0$$
E = Gravitational field intensity N/kg
For any point situated inside the hollow spherical shell (r < R)
Graphs and experiments
Variation of g with Distance from Earth's Center
Draw: A graph with r on the x-axis and g(r) on the y-axis. For r ≤ R, a straight line passing through the origin with positive slope. For r > R, a curve falling sharply as 1/r2.
Interpret: Inside the Earth, g increases linearly with radius. At the surface, it achieves its maximum value g. Outside the Earth, it drops inversely proportional to the square of the distance.
Exam use: Frequently tested in TGT/PGT conceptual MCQs regarding turning points and continuity at r = R.
Effective Potential Energy Curve for Central Orbits
Draw: Graph of U_eff(r) vs r featuring a centrifugal barrier term (L2 / 2mr2) and attractive gravitational potential (-GMm/r), showing a minimum representing stable circular orbits.
Interpret: The minimum of the effective potential corresponds to the stable equilibrium radius where attractive gravitational force equals the centrifugal force.
Exam use: Evaluated in PGT questions to determine apsidal distances and stability of orbits.
Cavendish Torsion Balance Experiment
Draw: Schematic setup showing two small masses on a light horizontal rod suspended by a fine quartz fiber, attracted by two large lead spheres.
Interpret: Used to measure the weak gravitational force between laboratory masses, enabling the direct experimental determination of the universal gravitational constant G.
Exam use: Tested as a factual conceptual question regarding the experimental verification of G.
Solved numericals
TGT core
At what depth below the surface of the Earth does the acceleration due gravity become 75% of its value at the surface? (Assume Earth radius R = 6400 km)
Formula for gravity at depth d is g' = g (1 - d / R).
Given g' = 0.75 g.
0.75 $g = g (1 - d / R) \Rightarrow 0.75 = 1 - d / R$.
d / $R = 1 - 0.75 = 0.25 = 1 / 4$.
$d = R / 4 = 6400 km / 4 = 1600 km$.
Answer: 1600 km
PGT extension
A satellite is orbiting the Earth in a circular orbit of radius 4R (where R is the Earth's radius). What additional velocity must be imparted to the satellite so that it escapes the Earth's gravitational field?
Calculate the current orbital velocity v_o at $r = 4R: v_o = \sqrt{GM / 4R} = \frac{1}{2} \sqrt{GM / R}$.
Calculate the escape velocity v_e from the orbit radius $r = 4R: v_e = \sqrt{2GM / 4R} = \frac{1}{\sqrt{2}} \sqrt{GM / R}$.
The required additional velocity (boost Δv) is Δ$v = v_e - v_o$.
Two uniform solid spheres of mass M and radius R are touching each other. Find the gravitational potential energy of this two-body system.
The distance between the centers of the two touching spheres is r = 2R.
The gravitational potential energy of a two-point mass system is $U = - \frac{G m_1 m_2}{r}$.
Substitute $m_1 = M$, $m_2 = M$, and r = 2R into the formula.
$U = - \frac{G M M}{2R} = - \frac{G M^2}{2R}$.
Answer: - \frac{G M^2}{2R}
Assumptions and limitations
Newton's law of gravitation assumes instantaneous propagation of gravitational interaction, which is a limitation rectified by General Relativity where gravity propagates at the speed of light.
The uniform density shell theorems assume isotropic and spherically symmetric mass distributions, failing for irregularly shaped celestial bodies unless expanded in multipole moments.
Common misconceptions
Common error: Gravitational potential inside a solid sphere is zero because the field inside is zero.
Correct: The gravitational field is zero at the center, but the potential is non-zero and finite.
Potential is the integral of the field from infinity; even if the field vanishes at a point or inside a shell, integrating from the surface inward accumulates potential.
Common error: An astronaut inside an orbiting satellite experiences zero gravitational force.
Correct: The astronaut experiences free fall where gravity provides the centripetal acceleration, resulting in a sensation of apparent weightlessness.
The gravitational force at orbital heights is about 90% of surface gravity; the normal reaction from the floor is zero because both astronaut and spacecraft accelerate equally towards the Earth.
Quick revision
Gravitational force is a conservative central force yielding an inverse-square field law.
Acceleration due to gravity varies as g' = g(1 - h/R) for small heights and g' = g(1 - d/R) for depths.
Orbital energy is negative, demonstrating bound states, whereas total energy ≥ 0 corresponds to unbound parabolic or hyperbolic escape trajectories.
Must remember
Newton's Law of Universal Gravitation states that $F = G \frac{m_1 m_2}{r^2}$, and the gravitational force is always a central and conservative force.
The gravitational field intensity $E$ is related to the gravitational potential $V$ via the negative gradient relation: $E = -\nabla V$.
The variation of acceleration due to gravity with altitude ($h \ll R$) is given by $g' = g\left(1 - \frac{2h}{R}\right)$ and with depth ($d$) by $g' = g\left(1 - \frac{d}{R}\right)$.
Due to Earth's rotation with angular velocity $\omega$, the effective gravity at latitude $\lambda$ is $g' = g - R\omega^2 \cos^2\lambda$.
Escape velocity from a planet of mass $M$ and radius $R$ is $v_e = \sqrt{\frac{2GM}{R}} = \sqrt{2gR}$.
Orbital velocity for a circular orbit at radius $r$ is $v_o = \sqrt{\frac{GM}{r}}.
Kepler's Third Law relates orbital period $T$ and semi-major axis $r$ as $T^2 = \left(\frac{4\pi^2}{GM}\right)r^3$.
Inside a uniform hollow spherical shell, the gravitational field intensity is exactly zero ($E = 0$), but the gravitational potential remains constant and non-zero.
Glossary
Universal Gravitational Constant ($G$)A fundamental physical constant representing the proportionality in Newton's law of gravitation, with value approximately $6.674 \times 10^{-11} \text{ N m}^2/\text{kg}^2$.
Gravitational Field Intensity ($E$)The gravitational force experienced by a unit test mass placed at a point in space, measured in $\text{N/kg}$.
Gravitational Potential ($V$)The work done per unit mass in bringing a test mass from infinity to a point without acceleration, measured in $\text{J/kg}$.
Escape Velocity ($v_e$)The minimum initial velocity required for an object to break free from the gravitational attraction of a celestial body.
Kepler's Second LawThe law of areas stating that the position vector from the Sun to a planet sweeps out equal areas in equal intervals of time, reflecting conservation of angular momentum.
Velocity needed to maintain a stable circular orbit
Velocity needed to escape to infinity from the surface
Gravity Variation: Altitude vs Depth
Parameter
Altitude ($h \ll R$)
Depth ($d$)
Formula
$g' = g\left(1 - \frac{2h}{R}\right)$
$g' = g\left(1 - \frac{d}{R}\right)$
Trend
Gravity decreases as height increases
Gravity decreases as depth increases towards the center
Value at Center/Infinity
Approaches zero as $h \to \infty$
Becomes exactly zero at the center ($d = R$
Practice Quiz
Choose an option to reveal the explanation.
1. What is the nature of the gravitational force field?
Explanation: Gravitational force acts along the line joining the centers of two point masses, making it a central force, and since work done is path-independent, it is a conservative force field.
2. If the distance between two point masses is doubled, the gravitational force between them becomes:
Explanation: According to Newton's Law of Universal Gravitation, $F \propto \frac{1}{r^2}$. When $r$ is doubled, the force becomes $\frac{1}{2^2} = \frac{1}{4}$ of its original value.
3. The relationship between gravitational field intensity $E$ and gravitational potential $V$ is given by:
Explanation: The gravitational field intensity is the negative spatial gradient of the gravitational potential, expressed as $E = -\nabla V$.
4. At what depth $d$ below the Earth's surface does the acceleration due to gravity become half of its surface value?
5. The escape velocity of a body from the surface of the Earth is proportional to which of the following?
Explanation: Escape velocity is given by $v_e = \sqrt{\frac{2GM}{R}}$, which means $v_e \propto \sqrt{\frac{M}{R}}$.
6. Kepler's Second Law (Law of Areas) is a direct consequence of the conservation of:
Explanation: The areal velocity $\frac{dA}{dt} = \frac{L}{2m}$ is constant when torque is zero, which implies the conservation of angular momentum $L$.
7. What is the gravitational field intensity inside a uniform hollow spherical shell of radius $R$ at a distance $r < R$?
Explanation: By Gauss's law for gravity, the gravitational field inside a hollow spherical shell enclosing no mass is zero ($E = 0$).
8. Due to the Earth's rotation, the effective acceleration due to gravity is minimum at the:
Explanation: The effective gravity formula is $g' = g - R\omega^2 \cos^2\lambda$. At the equator, latitude $\lambda = 0^\circ$, making $\cos^2\lambda = 1$, which gives the maximum reduction and thus minimum effective gravity.
9. The orbital period of a satellite in a circular orbit of radius $r$ around a planet of mass $M$ is proportional to:
Explanation: From Kepler's Third Law, $T^2 \propto r^3$, taking the square root gives $T \propto r^{3/2}$.
10. What is the gravitational potential energy of a mass $m$ at a distance $r$ from a central mass $M$, taking infinity as the zero reference?
Explanation: The gravitational potential energy referenced at infinity is $U = -\frac{GMm}{r}$.
11. If the Earth suddenly shrinks to half its present radius while keeping its mass constant, the acceleration due to gravity on its surface will:
Explanation: Since $g = \frac{GM}{R^2}$, if $R$ becomes $R/2$, the new gravity $g' = \frac{GM}{(R/2)^2} = 4\left(\frac{GM}{R^2}\right) = 4g$, meaning it quadruples.
12. An astronaut inside an orbiting space station experiences weightlessness because:
Explanation: The astronaut is in free fall; both the spacecraft and astronaut accelerate towards Earth at the same rate, resulting in zero normal contact force (apparent weightlessness).
13. The gravitational potential at the center of a solid sphere of mass $M$ and radius $R$ is:
Explanation: Integrating the gravitational field from infinity through the solid sphere gives the potential at the center as $V_{center} = -\frac{3GM}{2R}$.
14. What is the ratio of escape velocity $v_e$ to orbital velocity $v_o$ for a satellite orbiting close to the Earth's surface?
Explanation: Escape velocity is $v_e = \sqrt{2gR}$ and close orbital velocity is $v_o = \sqrt{gR}$, so their ratio is $\sqrt{2}$.
15. According to Kepler's First Law, planets move in:
Explanation: Kepler's First Law (Law of Orbits) states that every planet moves in an elliptical orbit with the Sun located at one of the two foci.
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 contracts by $1\%$ with its mass remaining constant, the acceleration due to gravity on its surface will:
Explanation: Since $g = \frac{GM}{R^2}$, taking logarithms and differentiating gives $\frac{\Delta g}{g} = -2\frac{\Delta R}{R}$. A $1\%$ decrease in radius ($\Delta R / R = -0.01$) results in a $2\%$ increase in $g$.
2. A satellite is orbiting the Earth at a height $h$ equal to the radius of the Earth ($R$). Its orbital velocity is:
Explanation: The orbital radius is $r = R + h = R + R = 2R$. Orbital velocity $v_o = \sqrt{\frac{GM}{r}} = \sqrt{\frac{GM}{2R}} = \sqrt{\frac{gR^2}{2R}} = \sqrt{\frac{gR}{2}}$.
3. What is the minimum energy required to launch a satellite of mass $m$ from the Earth's surface into a circular orbit at an altitude equal to the Earth's radius $R$?
Explanation: Initial mechanical energy at surface $E_i = -\frac{GMm}{R}$. Final mechanical energy in orbit at $r = 2R$ is $E_f = -\frac{GMm}{2(2R)} = -\frac{GMm}{4R}$. The required energy $\Delta E = E_f - E_i = -\frac{GMm}{4R} - \left(-\frac{GMm}{R}\right) = \frac{3GMm}{4R}$.
4. The gravitational potential at a distance $r$ from the center of a uniform solid sphere of mass $M$ and radius $R$ (where $r < R$) is given by:
Explanation: For an internal point of a solid sphere ($r < R$), the potential is evaluated by integrating the interior field, yielding $V = -\frac{GM}{2R^3}(3R^2 - r^2)$.
5. At what latitude does the effective acceleration due to gravity become zero due to the Earth's rotation, assuming the Earth rotates fast enough?
Explanation: The condition for zero effective gravity is $g - R\omega^2 \cos^2\lambda = 0$, which requires $\cos^2\lambda = g / (R\omega^2)$. For Earth, $R\omega^2 \approx 0.034 \text{ m/s}^2$ and $g \approx 9.8 \text{ m/s}^2$, meaning $\cos^2\lambda > 1$, which is physically impossible at current rotation speeds.
6. Two bodies of masses $m_1$ and $m_2$ are initially at rest at infinite distance apart. They move towards each other under mutual gravitational attraction. What is their relative velocity of approach when their separation is $r$?
Explanation: By conservation of mechanical energy, initial energy is zero. Final energy is $\frac{1}{2}\mu v_{rel}^2 - \frac{Gm_1 m_2}{r} = 0$, where $\mu = \frac{m_1 m_2}{m_1 + m_2}$ is the reduced mass. Solving gives $v_{rel} = \sqrt{\frac{2G(m_1 + m_2)}{r}}$.
Timed Topic Test
Attempt these questions under a ten-minute limit using a separate timer.
1. The work done in bringing a test mass $m$ from infinity to a point in a gravitational field is $W$. The gravitational potential at that point is:
Explanation: Gravitational potential $V$ is defined as the work done per unit mass in bringing a test mass from infinity: $V = W / m$.
2. If the Earth were to stop rotating completely, the value of acceleration due to gravity ($g$) at the equator would:
Explanation: Effective gravity at the equator is $g' = g - R\omega^2$. If rotation stops ($\omega = 0$), $g'$ increases to $g$.
3. A planet revolves around the Sun in an elliptical orbit. The ratio of the maximum to minimum distance of the planet from the Sun is $4$. The ratio of their linear speeds at these positions is:
Explanation: By conservation of angular momentum, $mv_{max}r_{min} = mv_{min}r_{max}$, which gives $\frac{v_{max}}{v_{min}} = \frac{r_{max}}{r_{min}} = 4:1$.
4. What is the gravitational potential energy of a body of mass $m$ placed on the surface of the Earth of mass $M$ and radius $R$?
Explanation: Using the potential energy formula $U = -\frac{GMm}{r}$ with $r = R$ gives $U = -\frac{GMm}{R}$.
5. The time period of a satellite revolving very close to the Earth's surface depends primarily on:
Explanation: The close-orbit time period is $T = 2\pi \sqrt{\frac{R}{g}}$, which depends only on Earth's radius and surface gravity.
6. Inside a solid sphere of uniform density, the gravitational field intensity $E$ at a distance $r$ from the center varies as:
Explanation: For $r < R$, the gravitational field inside a uniform solid sphere is $E = \frac{GMtr}{R^3}$, which means $E \propto r$.
7. Which of the following quantities remains constant during planetary motion around the Sun?
Explanation: Since the gravitational force is conservative, the total mechanical energy of the planet-Sun system is conserved throughout the orbit.
8. The escape velocity from a planet is $v_e$. If the mass of the planet is doubled and its radius is halved, the new escape velocity will be:
Explanation: Escape velocity is $v_e = \sqrt{\frac{2GM}{R}}$. If $M' = 2M$ and $R' = R/2$, then $v_e' = \sqrt{\frac{2G(2M)}{R/2}} = \sqrt{4 \cdot \frac{2GM}{R}} = 2v_e$.
9. What is the dimensional formula of the Universal Gravitational Constant $G$?
Explanation: From $F = \frac{Gm_1 m_2}{r^2}$, $G = \frac{F r^2}{m_1 m_2}$. Dimensions are $\frac{[MLT^{-2}][L^2]}{[M^2]} = [M^{-1} L^3 T^{-2}].
10. If a tunnel is bored straight through the center of the Earth and a particle is dropped into it, the motion of the particle is:
Explanation: Inside the Earth, the restoring gravitational force is proportional to displacement from the center ($F = -\frac{mg}{R}r$), resulting in Simple Harmonic Motion.