Gravitation — NEET notes

Newton's Law of Gravitation

Every two masses attract each other with force F = GMm r^2 along the line joining them. G = 6.674 10^ -11 N m ^2 kg ^ -2 is universal. This inverse-square law governs everything from falling apples to orbiting planets. NEET loves superposition of gravitational forces and comparison with Coulomb's law.

F = GMm r^2

The Universal Law

Newton's law of gravitation gives the force between any two point masses. The force is proportional to the product of their masses and inversely proportional to the square of the distance between them.

Newton's Law of Universal Gravitation: Every particle 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 them. The force acts along the line joining the two particles.

F = GMm r^2

Here M and m are the two masses, r is the distance between their centres, and G is the universal gravitational constant = 6.674 10^ -11 N m ^2 kg ^ -2 . The vector form: F = - GMm r^2 r (the negative sign means the force is attractive — directed toward the other mass).

• Gravitational force is always attractive — there's no gravitational repulsion
• G is universal — same value everywhere in the universe, independent of medium, temperature, or material
• The force obeys Newton's third law: F_ Earth on you = F_ you on Earth (same magnitude, opposite direction)
• Applies to point masses. For spheres, r is the distance between centres (shell theorem)
• Dimensions of G : [M^ -1 L^3T^ -2 ]

Superposition Principle

The net gravitational force on a mass is the vector sum of individual gravitational forces from all other masses. No force cancels or modifies another — they simply superpose.

If a mass m is near several other masses m_1, m_2, m_3, , the total gravitational force on m is: F _ net = F _1 + F _2 + F _3 + . Each F _i is calculated independently using Newton's law. The presence of other masses does NOT alter any individual force.

F _ net = _ i=1 ^ n F _i = -Gm _ i=1 ^ n m_i r_i^2 r _i

• Each pair interaction is calculated using F = GMm/r^2 independently
• Then all force vectors are added using vector addition (components or geometry)
• For symmetric arrangements, many forces cancel by symmetry — always check before computing
• For a uniform sphere, treat the entire mass as concentrated at its centre (shell theorem)

Quick recall

Three equal masses m are placed at the vertices of an equilateral triangle of side a . What is the net force on any one mass?

Each mass feels two forces of magnitude F = Gm^2/a^2 at 60° to each other. Resultant = F 2(1 + 60°) = F 3 = 3 Gm^2 a^2 , directed toward the centroid.

Gravity vs Coulomb's Law

Both are inverse-square laws, but gravity is always attractive and far weaker than the electrostatic force. NEET sometimes asks direct comparisons.

Gravitational Field & Potential

Gravitational field intensity g = - GM r^2 r tells you the force per unit mass at any point. Gravitational potential V = - GM r tells you the work done per unit mass to bring a mass from infinity. The relation g = - dV dr connects the two. Reference is at infinity where V = 0 .

E _g = - GM r^2 r | V = - GM r

Gravitational Field Intensity

The gravitational field at a point is the force a unit test mass would experience there. It's a vector field pointing toward the source mass.

The gravitational field intensity (or gravitational field) at a point is defined as the force experienced by a unit test mass placed at that point: E _g = F m = - GM r^2 r . At Earth's surface, E_g = g 9.8 m/s ^2 — this is just the familiar acceleration due to gravity.

E _g = F m = - GM r^2 r (unit: N/kg = m/s^2)

• Gravitational field is a vector — it has magnitude AND direction (toward the source mass)
• SI unit: N/kg or equivalently m/s ^2 (same as acceleration)
• At Earth's surface: E_g = g = GM_E R_E^2 9.8 m/s ^2
• Field due to a uniform sphere (outside): same as if all mass is at the centre
• Field inside a uniform spherical shell: zero everywhere (shell theorem)

Gravitational Potential

Gravitational potential V at a point is the work done by an external agent to bring a unit mass from infinity to that point without acceleration. It's always negative because gravity does positive work during the approach.

Gravitational potential at distance r from a mass M : V = - GM r . This is the work done per unit mass to bring a test mass from infinity (where V = 0 ) to the point. The negative sign means you need to do negative work (gravity assists the motion) — or equivalently, you'd need to do positive work to remove the mass back to infinity.

V = - GM r (unit: J/kg)

• Gravitational potential is a scalar — no direction, just a number at each point
• Always negative (for finite r ) because the reference ( V = 0 ) is at infinity
• At Earth's surface: V = - GM_E R_E = -gR_E
• Potential due to multiple masses: V = V_1 + V_2 + (simple addition, no vector hassle)
• SI unit: J/kg (joule per kilogram)

Relation Between Field and Potential

The gravitational field is the negative gradient of potential. Field points in the direction of decreasing potential. This relation lets you find one from the other.

The gravitational field and potential are related by: E_g = - dV dr . The field is the negative rate of change of potential with distance. This means the field points in the direction of steepest decrease of potential.

E_g = - dV dr V_B - V_A = - _A^B E _g d r

• If V is constant in some region → E_g = 0 there (no force in a region of constant potential)
• Closely spaced equipotential surfaces → strong field; widely spaced → weak field
• Equipotential surfaces are always perpendicular to field lines
• For a point mass: E_g = - d dr (- GM r ) = - GM r^2 (recovers the field formula)

Quick recall

If the gravitational potential near a planet varies as V = - k r^2 (hypothetical), what is the gravitational field?

E_g = - dV dr = - d dr (- k r^2 ) = - 2k r^3 . The negative sign indicates the field points toward the planet (attractive).

Variation of g

The value of g = 9.8 m/s ^2 is only approximate and varies with altitude ( g' = g(1 - 2h/R) ), depth ( g' = g(1 - d/R) ), and latitude (due to Earth's rotation). Understanding these variations is one of the highest-yield NEET topics in gravitation — expect 1-2 questions every year.

g' = g (1 - 2h R ) | g' = g (1 - d R )

Variation with Altitude

At height h above Earth's surface, g decreases because you're farther from the centre. For small h , the approximation g' g(1 - 2h/R) works beautifully.

At height h above Earth's surface, the distance from Earth's centre is (R + h) . Using Newton's law: g' = GM (R+h)^2 = g R^2 (R+h)^2 . This is the exact formula. For h R (small heights), expand using binomial approximation:

g' = g ( R R+h )^2 = g (1 + h R )^ -2 g (1 - 2h R ) (h R)

• Exact formula: g' = gR^2 (R+h)^2 — use this when h is comparable to R
• Approximate formula: g' g(1 - 2h/R) — use only when h R
• At h = R : g' = g/4 (exact). The approximate formula gives g' = -g which is nonsense — never use approximation for large h
• At h = R/2 : g' = 4g/9 (exact)
• g decreases continuously from the surface going upward — approaches zero at infinity

Variation with Depth

Inside the Earth, only the mass below you contributes to gravity (shell theorem). As you go deeper, g decreases linearly and reaches zero at the centre.

At depth d below the surface, only the sphere of radius (R - d) below you contributes to gravity (by the shell theorem, all the mass above you — the surrounding shell — exerts zero net force). Assuming uniform density:

g' = g (1 - d R ) = g ( R - d R )

• This formula is exact (for uniform density), not an approximation — no binomial expansion needed
• At the surface ( d = 0 ): g' = g (as expected)
• At the centre ( d = R ): g' = 0 — you'd be weightless at Earth's centre!
• g varies linearly with depth — graph is a straight line from g at surface to 0 at centre
• The formula assumes uniform density — in reality, Earth's core is denser, so g actually increases slightly before decreasing

Quick recall

Where is g maximum — at the surface, at height h , or at depth d ?

At the surface (for a uniform-density Earth). Going up decreases g (farther from centre). Going down also decreases g (less mass below you). The surface is the sweet spot where g is maximum.

Effect of Earth's Rotation

Earth's rotation creates a centrifugal effect that reduces the apparent value of g . The reduction is maximum at the equator and zero at the poles. This is why you weigh slightly less at the equator.

A body on Earth's surface participates in Earth's rotation. Part of the gravitational pull provides the centripetal force for circular motion. The effective (apparent) value of g at latitude :

g' = g - R ^2 ^2

• At the equator ( = 0° ): g' = g - R ^2 — maximum reduction. R ^2 0.034 m/s ^2
• At the poles ( = 90° ): g' = g — no reduction (rotation axis passes through poles, no circular motion there)
• g_ pole - g_ equator 0.034 m/s ^2 (due to rotation alone)
• If Earth stopped rotating, g at the equator would increase by R ^2
• Earth is also oblate (flattened at poles), so the pole is closer to the centre — this adds to the difference

Quick recall

If the Earth rotates 17 times faster, what happens at the equator?

g' = g - R '^2 = g - R(17 )^2 = g - 289 R ^2 . Since R ^2 0.034 m/s ^2 : 289 0.034 9.8 m/s ^2 = g . So g' = 0 — objects at the equator become weightless and fly off! (This is why NEET asks 'by what factor should angular velocity increase for weightlessness at equator.')

Gravitational Potential Energy

Gravitational PE of a two-body system is U = - GMm r with reference at infinity. The negative sign means the system is bound — you must supply energy to separate the masses. Near Earth's surface, this simplifies to the familiar U = mgh . For systems of multiple particles, add up PE of every pair.

U = - GMm r

Gravitational PE

The gravitational potential energy of a system of two masses separated by distance r is U = -GMm/r . The reference ( U = 0 ) is at infinite separation. Negative PE means the system is bound.

The gravitational PE of a system of two masses M and m separated by distance r is: U = - GMm r . This is defined as the work done by an external agent to assemble the system by bringing the masses from infinity (where U = 0 ) to separation r , without any change in kinetic energy.

U = - GMm r (reference: U = 0 at r = )

• PE is negative because gravity is attractive — the system naturally falls to lower r , releasing energy
• At r = : U = 0 (masses are free, no interaction)
• As r decreases: U becomes more negative (system becomes more tightly bound)
• At r = R (surface): U = - GMm R — this is the binding energy of the object to Earth
• PE belongs to the system, not to either mass individually

PE of a System of Particles

For a system of n particles, the total gravitational PE is the sum of PE contributions from every unique pair. With n particles, there are n(n-1)/2 pairs.

For n particles, the total gravitational PE is the sum of PE of all unique pairs: U_ total = _ all pairs - Gm_i m_j r_ ij . The number of pairs = n(n-1) 2 . For 3 particles: 3 pairs. For 4 particles: 6 pairs.

U_ total = _ i<j (- Gm_i m_j r_ ij ) = -G _ i<j m_i m_j r_ ij

• 3 equal masses m at corners of equilateral triangle (side a ): U = 3 (- Gm^2 a ) = - 3Gm^2 a
• 4 equal masses at corners of a square (side a ): 4 side-pairs + 2 diagonal-pairs = - Gm^2 a (4 + 2 2 ) = - Gm^2 a (4 + 2 )
• Don't double-count: each pair contributes ONE term, not two

Quick recall

What is the work done to completely disassemble a system of 3 equal masses m at the corners of an equilateral triangle of side a ?

Total PE = -3Gm^2/a . To disassemble (move all to infinity where U = 0 ), work done by external agent = 0 - (-3Gm^2/a) = +3Gm^2/a . Positive work because you're pulling bound masses apart.

Connection to mgh

Near Earth's surface, the change in gravitational PE simplifies to U = mgh . This is what you used in kinematics and work-energy chapters — it's a special case of the general formula.

When an object moves from Earth's surface (radius R ) to a height h above it (radius R + h ), the change in PE is: U = - GMm R+h - (- GMm R ) = GMm ( 1 R - 1 R+h ) = GMmh R(R+h) .

U = GMmh R(R+h) h R GMm R^2 h = mgh

• For h R : (R + h) R , so U GMmh/R^2 = mgh — recovers the familiar formula
• For large h (comparable to R ): use the exact formula U = GMmh/[R(R+h)]
• mgh is NOT the PE — it's the change in PE when you go up by h from the surface
• The choice of reference doesn't affect U — that's why mgh works regardless of where you set U = 0

Escape Velocity

Escape velocity v_e = 2gR = 2GM R is the minimum speed needed to escape a planet's gravitational pull forever. It doesn't depend on the mass or direction of the projectile — only on the planet's mass and radius. For Earth, v_e 11.2 km/s.

v_e = 2gR = 2GM R

Escape Velocity Derivation

Escape velocity is found by setting total energy at the surface equal to zero (just barely reaching infinity). It comes from energy conservation, not force analysis.

A body launched from the surface with speed v has KE = 1 2 mv^2 and PE = - GMm R . To just escape (reach infinity with zero speed), the total energy must be zero (KE at infinity = 0, PE at infinity = 0). Setting E_ total = 0 :

1 2 mv_e^2 - GMm R = 0 v_e = 2GM R = 2gR

• For Earth: v_e = 2 9.8 6.4 10^6 11.2 km/s
• Independent of the mass of the body — a marble and a rocket need the same escape speed
• Independent of direction of projection — whether fired straight up, sideways, or at an angle
• Depends only on M and R of the planet (or equivalently, g and R )
• Relationship: v_e = 2 v_o where v_o is orbital velocity at the surface

Escape Velocity from Different Bodies

Each celestial body has its own escape velocity. The Moon's is 2.3 km/s (no atmosphere), Jupiter's is 60 km/s (thick atmosphere). A body with v_e > c (speed of light) is a black hole.

Since v_e = 2GM R , heavier and denser bodies have higher escape velocities. This has real consequences: the Moon can't hold an atmosphere ( v_e = 2.3 km/s, gas molecules easily exceed this), while Jupiter's massive v_e = 60 km/s traps even hydrogen.

• Moon: v_e 2.3 km/s — too low to retain atmosphere (gas molecules escape)
• Earth: v_e 11.2 km/s — holds N_2 , O_2 but not H_2 , He (light gases escape)
• Jupiter: v_e 60 km/s — retains everything, including hydrogen and helium
• Sun: v_e 618 km/s
• If v_e c (speed of light) → even light can't escape → black hole

v_ e,1 v_ e,2 = M_1 M_2 R_2 R_1 = g_1 R_1 g_2 R_2

Quick recall

If a planet has twice Earth's mass and half Earth's radius, what is its escape velocity?

v_e = 2G(2M) R/2 = 4 2GM R = 2v_ e,Earth = 2 11.2 = 22.4 km/s.

Orbital Velocity & Satellites

Orbital velocity v_o = gR^2/(R+h) gR (near surface) is the speed needed for a satellite to maintain a circular orbit. Geostationary satellites orbit at 36 , 000 km with T = 24 hours. The relation v_e = 2 v_o connects escape and orbital velocities. NEET frequently tests time period, orbital radius, and energy of satellites.

v_o = GM R+h = R g R+h | T = 2 (R+h)^3 GM

Orbital Velocity

For a satellite in circular orbit, the gravitational force provides exactly the centripetal force needed. This condition gives the orbital velocity v_o = GM/(R+h) .

For a satellite of mass m in a circular orbit of radius r = R + h around Earth (mass M ), gravity provides the centripetal force: GMm r^2 = mv_o^2 r . Cancel m and solve:

v_o = GM r = GM R+h = R g R+h

Near Earth's surface ( h 0 ): v_o gR = 9.8 6.4 10^6 7.9 km/s.

• Orbital velocity is independent of satellite's mass — a 1 kg satellite and a 1000 kg satellite orbit at the same speed at the same height
• Higher orbit → lower orbital velocity (farther from Earth, weaker gravity, slower orbit)
• Near surface: v_o 7.9 km/s. Relation to escape velocity: v_e = 2 v_o
• If speed > v_o but < v_e : elliptical orbit. If speed = v_e : parabolic escape. If speed > v_e : hyperbolic escape
• Orbital velocity doesn't depend on the satellite's mass — a feather and a space station orbit together

Time Period of a Satellite

The time period T = 2 r/v_o = 2 r^3/(GM) depends only on the orbital radius and the planet's mass. Near Earth's surface, T 84.6 minutes.

T = 2 r v_o = 2 r 1 GM/r = 2 r^3 GM . Substituting r = R + h :

T = 2 (R+h)^3 GM = 2 R (R+h)^3 g

• Near surface ( h 0 ): T = 2 R/g 2 6.4 10^6 / 9.8 84.6 min ≈ 5076 s
• This is the minimum possible orbital period (any lower orbit would be inside the Earth)
• T increases with orbital radius — higher satellites take longer to complete one orbit
• T^2 r^3 — this is Kepler's third law (next concept)

Quick recall

A satellite orbits at a height equal to Earth's radius ( h = R ). What is its period in terms of the near-surface period T_0 ?

T = 2 (2R)^3/(GM) = 2 8R^3/(GM) = T_0 8 = 2 2 T_0 2.83 T_0 .

Geostationary & Polar Orbits

A geostationary satellite has T = 24 hours and orbits in the equatorial plane at 36 , 000 km altitude — it appears stationary from Earth. Polar orbit satellites pass over the poles and cover the entire Earth surface.

A geostationary satellite has three conditions: (1) T = 24 hours, (2) orbits in the equatorial plane (zero inclination), (3) revolves in the same direction as Earth's rotation (west to east). It appears stationary above a fixed point on the equator.

r_ geo = ( GMT^2 4 ^2 )^ 1/3 42 , 164 km from Earth's centre 35 , 786 km above surface

• Geostationary orbit is also called GEO (Geostationary Earth Orbit) or Clarke orbit (proposed by Arthur C. Clarke in 1945)
• There is only ONE geostationary orbit — a unique radius where T = 24 hrs
• INSAT series (India's communication satellites) are geostationary
• IRS (Indian Remote Sensing) satellites use polar/sun-synchronous orbits

Kepler's Laws of Planetary Motion

Kepler's three empirical laws describe planetary orbits: (1) orbits are ellipses with the Sun at one focus, (2) the line joining planet and Sun sweeps equal areas in equal times (conservation of angular momentum), (3) T^2 a^3 where a is the semi-major axis. Newton later derived all three from his law of gravitation. Kepler's 3rd law is a NEET favourite.

T^2 = 4 ^2 GM a^3 T^2 a^3

Kepler's First Law (Law of Orbits)

Every planet moves in an elliptical orbit with the Sun at one of the two foci. A circle is just a special case of an ellipse with zero eccentricity.

First Law: Every planet moves in an elliptical orbit with the Sun at one of the two foci. The closest point to the Sun is called perihelion (distance r_ min ), the farthest point is aphelion (distance r_ max ).

• An ellipse has two foci — the Sun sits at ONE focus, not at the centre
• Semi-major axis a : half the longest diameter. Semi-minor axis b : half the shortest diameter
• Eccentricity e = c/a where c is the distance from centre to focus. Circle: e = 0 . Parabola: e = 1
• Most planetary orbits are nearly circular ( e 0 ). Earth: e = 0.017 . Mercury: e = 0.206 (most eccentric planet)
• r_ min = a(1-e) (perihelion), r_ max = a(1+e) (aphelion)

Kepler's Second Law (Law of Areas)

The line joining a planet and the Sun sweeps out equal areas in equal time intervals. This means the planet speeds up near perihelion and slows down near aphelion. It's a direct consequence of conservation of angular momentum.

Second Law: The radius vector from the Sun to a planet sweeps out equal areas in equal time intervals. Mathematically: dA dt = constant . Since gravitational force is central (always directed toward the Sun), there is no torque about the Sun. So angular momentum L = mvr is conserved.

dA dt = L 2m = constant v_ perihelion r_ min = v_ aphelion r_ max

• Closer to Sun → faster speed (smaller r → larger v to keep L = mvr constant)
• Farther from Sun → slower speed
• This law is equivalent to conservation of angular momentum (since gravity exerts zero torque about the Sun)
• For circular orbits, the speed is constant (equal areas automatically because r is constant)

Quick recall

A planet's maximum distance from the Sun is 4 times its minimum distance. What is the ratio of its maximum to minimum orbital speed?

By conservation of angular momentum: v_ min r_ max = v_ max r_ min . So v_ max /v_ min = r_ max /r_ min = 4 . The planet is 4 times faster at perihelion.

Kepler's Third Law (Law of Periods)

The square of a planet's orbital period is proportional to the cube of the semi-major axis of its orbit. Newton proved this from his law of gravitation. This is the most frequently tested Kepler's law in NEET.

Third Law: T^2 a^3 where T is the orbital period and a is the semi-major axis. For circular orbits ( a = r ): equating gravitational force to centripetal force gives GMm r^2 = 4 ^2 mr T^2 , yielding:

T^2 = 4 ^2 GM r^3 T_1^2 T_2^2 = r_1^3 r_2^3

• The constant 4 ^2 GM depends only on the central body's mass M (not on the orbiting body)
• All satellites of Earth obey the same T^2/r^3 = constant
• All planets orbiting the Sun obey the same T^2/a^3 = constant
• For elliptical orbits: replace r with semi-major axis a
• Can be used to find the mass of the central body: M = 4 ^2 r^3 GT^2

Quick recall

If the orbital radius of a satellite is doubled, by what factor does its time period change?

T^2 r^3 → T r^ 3/2 . If r doubles: T' = T 2^ 3/2 = 2 2 T 2.83 T . The period increases by a factor of 2 2 .

Weightlessness & Satellite Energy

Astronauts in orbit are weightless not because gravity is zero, but because both they and the satellite are in free fall. A satellite has KE = GMm 2r , PE = - GMm r , and total energy E = - GMm 2r (negative = bound). Total energy is half the PE — a clean relation NEET loves.

E = - GMm 2r | KE = -E = GMm 2r | PE = 2E

Weightlessness

Apparent weight becomes zero when the normal force (or support force) is zero. This happens in free fall — both you and your surroundings accelerate identically under gravity, so there's no contact force between you.

Weightlessness occurs when the apparent weight (the force you feel from the floor/scale) is zero. In a satellite, both the astronaut and the satellite are falling toward Earth with the same acceleration ( g' ). There's no relative acceleration between them, so the floor doesn't push up on the astronaut. No normal force → no sensation of weight → weightlessness.

• True weight ( mg ): gravitational pull on you — never zero near a planet
• Apparent weight ( N ): the normal force from the floor/scale — this is what you 'feel'
• In free fall: a = g → N = m(g - a) = 0 → apparent weight = 0 → weightless
• In a satellite (circular orbit): the satellite is in free fall → everything inside is weightless
• A person in a freely falling lift is also weightless — same principle as a satellite

Energy of an Orbiting Satellite

A satellite in circular orbit has KE = GMm/(2r) , PE = -GMm/r , and total energy E = -GMm/(2r) . The total energy is negative (bound orbit) and equals half the PE. These clean ratios are heavily tested.

For a satellite of mass m in a circular orbit of radius r around a body of mass M :

KE = 1 2 mv_o^2 = GMm 2r | PE = - GMm r | E = KE + PE = - GMm 2r

• KE is always positive: GMm 2r
• PE is always negative: - GMm r
• Total energy is negative: - GMm 2r (negative total energy = bound orbit)
• Beautiful relations: E = -KE = PE 2 or equivalently KE = -E and PE = 2E
• Magnitude: |PE| = 2 KE — PE is twice the KE in magnitude

Quick recall

A satellite has total energy E . What is the minimum energy needed to make it escape?

To escape, total energy must become 0 . Currently E = - GMm 2r . Energy to supply = 0 - E = GMm 2r = |E| . This equals the satellite's KE — you need to double its kinetic energy to escape.

Binding Energy

Binding energy is the minimum energy required to remove a satellite from its orbit to infinity. It equals the negative of the total orbital energy, which equals the kinetic energy of the satellite.

Binding energy (BE) of a satellite is the energy needed to move it from its orbit to infinity (where E = 0 ): BE = 0 - E = -E = GMm 2r . Since E = -KE , the binding energy equals the satellite's kinetic energy.

BE = -E = GMm 2r = KE of satellite = 1 2 mv_o^2

• BE for a satellite on the surface (not orbiting): BE = GMm R = 1 2 mv_e^2 (uses escape velocity)
• BE for a satellite in orbit at radius r : BE = GMm 2r
• Higher orbit → smaller BE (easier to escape — already farther from Earth)
• If energy is added to a satellite, its orbit radius increases and its speed decreases (counterintuitive!)