JEE Main chapters · Class 11 · Physics

Gravitation: JEE Main Physics formulas and notes

Gravitation is a scoring chapter: JEE Mains usually asks 2 to 3 questions from it, most of them a single formula with a twist. The favourites are variation of g, escape speed, satellite energy, Kepler’s laws and the field of a shell. Aim for 90 seconds per question. About 9 to 10 hours in total is enough.

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Quick revision

• F = (Gm₁m₂)/(r²), [G] = M⁻¹L³T⁻²
• Field I = (GM)/(r²), potential V = −(GM)/(r), U = −(GMm)/(r)
• Shell: inside I = 0, V constant; solid sphere inside I ∝ r
• g = (GM)/(R²) = (4)/(3)πGρR
• g(h) = g(1 + (h)/(R))⁻², g(d) = g(1 − (d)/(R))
• g′ = g − ω²R cos²λ (least at the equator)
• v(e) = √(2gR) = 11.2 km/s = √2 v(o)
• v(o) = √(GM)/(r), T = 2π√(r³)/(GM), T² ∝ r³
• K = (GMm)/(2r), U = −(GMm)/(r), E = −K
• Geostationary: T = 24 h, r = 4.2 × 10⁴ km, equatorial plane
• Binding energy = (GMm)/(2r)
• Kepler: r v = constant at the ends, T² ∝ a³, E = −(GMm)/(2a)
• Zero field between M₁ and M₂: x = (d√M₁)/(√M₁ + √M₂)
• Height reached: h = (Rv²)/(2gR − v²)

Formulas with solved examples

Law of gravitation

Newton’s law of gravitation

• F = (Gm₁m₂)/(r²)
• G = 6.67 × 10⁻¹¹ N m²/kg²
• dimensions of G: M⁻¹L³T⁻²
• the force is attractive, acts along the line joining the centres and is the same on both bodies
• for spheres, r is the distance between the centres

How to apply. The force depends on the masses and the distance, not on the medium between them. Use the distance between the centres for spherical bodies, and add forces as vectors.

Example. Masses of 10 kg and 20 kg at 1 m: F = 6.67 × 10⁻¹¹ × 200 = 1.33 × 10⁻⁸ N.

Several masses and symmetry

• net force = vector sum of the forces
• equal masses at the corners of an equilateral triangle: the force on a mass at the centre is zero
• three masses M at the corners of a triangle of side a: force on each = (√3GM²)/(a²)
• ring of mass M on its axis: F = (GMmx)/((R² + x²)^3/2)

How to apply. Draw the force from each mass, resolve along two axes and add. Use symmetry to cancel components, and look for the line along which the forces add.

Example. Three equal masses M at the corners of an equilateral triangle of side a: each feels 2 × (GM²/a²) × cos30° = √3GM²/a².

Point where the field is zero

• two masses M₁ and M₂ a distance d apart: (GM₁)/(x²) = (GM₂)/((d − x)²)
• x = (d√M₁)/(√M₁ + √M₂) from M₁
• the point lies between them and nearer the smaller mass
• the potential is not zero there

How to apply. Equate the two fields in magnitude: they point in opposite directions between the masses. Take the square root to get a linear equation.

Example. M₁ = 9M and M₂ = M, d = 4 m: x = 4 × 3 ÷ (3 + 1) = 3 m from the larger mass.

Field and potential

Gravitational field and potential

• field I = (F)/(m) = (GM)/(r²) (towards the mass)
• potential V = −(GM)/(r)
• I = −(dV)/(dr)
• potential energy of two masses U = −(Gm₁m₂)/(r)
• potential energy of a system: add over all pairs
• potential is a scalar, field a vector

How to apply. The potential at a point is the work done per unit mass to bring a body from infinity to that point, so it is negative. For several masses, add the potentials as numbers and the fields as vectors.

Example. At the Earth’s surface V = −gR = −10 × 6.4 × 10⁶ = −6.4 × 10⁷ J/kg (g = 10).

Shell and solid sphere

• thin shell (mass M, radius R): outside I = (GM)/(r²), inside I = 0
• V outside = −(GM)/(r), inside V = −(GM)/(R) (constant)
• solid sphere inside: I = (GMr)/(R³), V = −(GM(3R² − r²))/(2R³)
• V at the centre of a solid sphere = −(3GM)/(2R)
• at the surface I = (GM)/(R²)

How to apply. Outside, any spherical body acts as a point mass at its centre. Inside a shell the field is zero but the potential is not. Inside a solid sphere only the mass within the radius r pulls.

Example. Inside a solid sphere at r = R/2: I = GM ÷ (2R²), which is half the surface value.

Ring on its axis

• ring of mass M and radius R: I = (GMx)/((R² + x²)^3/2)
• V = −(GM)/(√(R² + x²))
• I = 0 at the centre and is largest at x = (R)/(√2)
• V at the centre = −(GM)/(R)

How to apply. The potential of a ring is the same as for a point mass at the distance √(R² + x²), because every part of the ring is at that distance. The field is the negative slope of the potential.

Example. A ring of mass M and radius R: the potential at its centre is −GM/R, and the field there is zero.

Cavity in a uniform sphere

• sphere of density ρ with a spherical cavity: field inside the cavity is uniform, I = (4)/(3)πGρ d⃗ (d⃗ = vector from the centre of the sphere to the centre of the cavity)
• method: solid sphere + sphere of negative mass in the cavity

How to apply. Treat the cavity as a sphere of the same density but negative mass and add the two fields as vectors. Inside the solid-sphere field I = (4/3)πGρ r⃗ (towards the centre).

Example. Cavity of radius R/2 centred at R/2 from the centre of a sphere of radius R: field inside the cavity = (GM)/(2R²), since (4/3)πGρ(R/2) = GM/(2R²).

Acceleration due to gravity

Acceleration due to gravity

• g = (GM)/(R²) = (4)/(3)πGρR
• g ∝ ρR for planets of different density and size
• g(planet) ÷ g(earth) = (M)/(M(e)) × (R(e)²)/(R²)
• g at the Earth’s surface 9.8 m/s²

How to apply. Write g in terms of density and radius to compare two planets. If density is the same, g is proportional to the radius.

Example. A planet with the Earth’s density and twice its radius: g′ = 2g.

Variation of g with height and depth

• at height h: g′ = g(1 + (h)/(R))⁻² ≈ g(1 − (2h)/(R)) for h ≪ R
• at depth d: g′ = g(1 − (d)/(R))
• at the centre g = 0
• the fall with height is twice as fast as with depth for small h

How to apply. Use the exact form when h is comparable to R. Inside the Earth only the mass below the point pulls, so g falls linearly to zero at the centre (for uniform density).

Example. At a height R above the surface g′ = g/4. At a depth R/2, g′ = g/2.

Effect of the Earth’s rotation

• at latitude λ: g′ ≈ g − ω²R cos²λ
• at the equator g′ = g − ω²R
• at the poles g′ = g
• ω of the Earth = 7.3 × 10⁻⁵ rad/s
• the body would be weightless at the equator if ω = √(g)/(R), a day of about 1.4 hours

How to apply. The effective g is smaller at the equator because part of the gravitational pull is used for the circular motion. The apparent weight is least at the equator and greatest at the poles.

Example. ω²R = (7.3 × 10⁻⁵)² × 6.4 × 10⁶ = 0.034 m/s². Weightless at the equator if the day is 2π√(R/g) = 5000 s, about 1.4 h.

Weight and weightlessness

• mass is the same everywhere, weight W = mg changes
• g(moon) ≈ (g)/(6)
• apparent weight is zero in free fall and in a satellite
• inside a hollow shell the gravitational force is zero

How to apply. Weight is the force from the support, so it vanishes when body and support fall together. Weightlessness does not mean there is no gravity: g at a satellite’s height is not zero.

Example. A 60 kg astronaut weighs 600 N on the Earth and 100 N on the Moon, but her mass is 60 kg in both places.

Potential energy and escape

Gravitational potential energy

• U = −(GMm)/(r) (zero at infinity)
• near the surface the change in U is mgh
• from R to R + h: ΔU = (mgh)/(1 + h/R)
• at h = R: ΔU = (mgR)/(2)

How to apply. Use −GMm/r for large heights. Work done against gravity is the change in potential energy, and it does not depend on the path.

Example. Raising a 10 kg body to a height R (g = 10): ΔU = ½ × 10 × 10 × 6.4 × 10⁶ = 3.2 × 10⁸ J.

Escape speed

• v(e) = √(2GM)/(R) = √(2gR) = R√(8πGρ)/(3)
• for the Earth 11.2 km/s
• v(e) = √2 v(o)
• for the Moon 2.4 km/s
• independent of the mass and of the direction of the body

How to apply. Escape speed is found by setting total energy ½mv² − GMm/R to zero. It does not depend on the mass of the body, and the same speed works in any direction (ignoring air).

Example. g = 10 m/s² and R = 6.4 × 10⁶ m: v(e) = √(2 × 10 × 6.4 × 10⁶) = 11.3 km/s.

Height reached and speed at infinity

• projected with v < v(e): ½mv² − (GMm)/(R) = −(GMm)/(R + h), so h = (Rv²)/(2gR − v²)
• projected with v > v(e): speed at infinity v∞ = √(v² − v(e)²)
• if v = v(e)/2, h = (R)/(3)

How to apply. Write the total energy at the start and at the highest point (where the speed is zero). For a speed above escape, the leftover kinetic energy at infinity gives v∞.

Example. v = v(e)/2: h = R × ¼ ÷ (1 − ¼) = R/3.

Binding energy and moving between orbits

• binding energy of a satellite = (GMm)/(2r)
• energy to move a satellite from r₁ to r₂: ΔE = GMm((1)/(2r₁) − (1)/(2r₂))
• energy to launch from the surface into a circular orbit of radius r: GMm((1)/(R) − (1)/(2r))

How to apply. The binding energy is the work needed to take the satellite to infinity from its orbit. Use E = −GMm/2r for the total energy in each orbit and take the difference.

Example. A satellite of mass m just above the Earth’s surface has a binding energy of mgR/2.

Speed on falling from a height

• dropped from rest at height h: ½mv² = GMm((1)/(R) − (1)/(R + h))
• v = √(2gh)/(1 + h/R)
• for h ≪ R, v = √(2gh)
• from infinity: v = √(2gR) = v(e)

How to apply. Use energy conservation with U = −GMm/r, not v² = 2gh, when h is comparable to R.

Example. Dropped from a height R: v² = 2gR ÷ 2 = gR, so v = √(gR) = 8 km/s (g = 10, R = 6.4 × 10⁶ m).

Satellites

Orbital speed and period

• v(o) = √(GM)/(r) = R√(g)/(r)
• near the surface: v(o) = √(gR) = 7.9 km/s
• T = 2π√(r³)/(GM)
• T² ∝ r³
• near the Earth T ≈ 84 minutes
• ω = √(GM)/(r³)
• independent of the satellite’s mass

How to apply. Gravity supplies the centripetal force: GMm/r² = mv²/r. The speed and the period depend only on the orbit radius and the central mass.

Example. g = 10, R = 6.4 × 10⁶: v(o) = √(6.4 × 10⁷) = 8 km/s and T = 2π√(R/g) = 5026 s, about 84 min.

Energy of a satellite

• K = (GMm)/(2r)
• U = −(GMm)/(r)
• E = −(GMm)/(2r) = −K = (U)/(2)
• the total energy is negative: the satellite is bound
• a larger r means a smaller K and a larger (less negative) E
• air drag reduces r and increases the speed

How to apply. Remember the ratio K : U : E = 1 : −2 : −1. When a satellite loses energy to drag it moves closer, speeds up and its kinetic energy increases.

Example. A 100 kg satellite at r = 2R: E = −GMm/4R = −mgR/4 = −100 × 10 × 6.4 × 10⁶ ÷ 4 = −1.6 × 10⁹ J.

Geostationary satellite

• period T = 24 h, same ω as the Earth
• radius about 4.2 × 10⁴ km from the centre (height 36 000 km)
• orbit in the equatorial plane, west to east
• v ≈ 3.1 km/s
• a satellite with a period of n days: r ∝ n^2/3

How to apply. The satellite stays above one point of the equator, so its period equals the Earth’s. Use T² ∝ r³ to compare with it.

Example. A satellite with a period of 8 days orbits at r = r₀ × 8^2/3 = 4r₀, where r₀ is the geostationary radius.

Changing the orbit; two-body orbits

• speed suddenly raised to √2 v(o): it escapes
• suddenly halved: it falls into an ellipse with this point as apogee
• binary stars: (G(M₁ + M₂))/(d³) = ω²
• r₁ : r₂ = M₂ : M₁
• two equal masses M at distance d: v = √(GM)/(2d)

How to apply. Both stars of a binary orbit their common centre of mass with the same ω. Each is pulled by the full gravitational force, but its orbit radius is only its share of d.

Example. Two equal stars of mass M at a distance d: each moves in a circle of radius d/2 at v = √(GM/2d), with T = 2π√(d³/2GM).

Satellite near the surface and tunnel through the Earth

• satellite just above the surface: T = 2π√(R)/(g) = √(3π)/(Gρ) (depends only on the density)
• tunnel along a diameter: a = (g)/(R)x towards the centre, so SHM with T = 2π√(R)/(g) = 84 min
• speed at the centre = √(gR)
• for a chord tunnel the time is the same 84 min/2

How to apply. Inside a uniform Earth the force is F = −(GMm)/(R³)x, a linear restoring force, so the motion is simple harmonic with the same period as a surface-skimming satellite.

Example. A ball dropped into a tunnel through the Earth reaches the other end in T/2 = 42 min, and its maximum speed at the centre is √(gR) = 8 km/s (g = 10, R = 6.4 × 10⁶ m).

Kepler’s laws

Kepler’s laws

• the orbit is an ellipse with the Sun at one focus
• areal velocity (dA)/(dt) = (L)/(2m) is constant
• T² ∝ a³ (a = semi-major axis), T² = (4π²)/(GM)a³
• the planet is fastest at perihelion: r v = constant at the ends

How to apply. The second law is the conservation of angular momentum for a central force. The third law links the period with the semi-major axis of any orbit about the same central mass.

Example. Perihelion r = 1 unit and aphelion r = 4 units: v(perihelion) ÷ v(aphelion) = 4.

Elliptical orbit: energy and speed

• E = −(GMm)/(2a)
• v² = GM((2)/(r) − (1)/(a))
• a = (r₁ + r₂)/(2)
• at the ends: v₁r₁ = v₂r₂ and ½v₁² − (GM)/(r₁) = ½v₂² − (GM)/(r₂)

How to apply. For an ellipse, the total energy depends only on the semi-major axis, exactly as for a circle of radius a. Use the two conservation laws between perihelion and aphelion.

Example. An orbit with r₁ = R and r₂ = 3R: a = 2R and the speed at the nearest point is √(GM(2/R − 1/2R)) = √(1.5GM/R).

Comparing planets and orbits

• T ∝ r^3/2
• v ∝ r^−1/2
• ω ∝ r^−3/2
• L = m√(GMr) for a circular orbit
• if r becomes 4 times, T becomes 8 times and v halves

How to apply. Write the ratio of any two quantities from T² ∝ r³ and v² = GM/r. Both planets must orbit the same central body.

Example. Planet A has an orbit radius 4 times that of planet B: T(A) = 8T(B) and v(A) = v(B)/2.

Most asked question types

How do I solve “Force between masses and superposition” questions in JEE Main?

Several masses are placed in a line or at the corners of a figure. Find the net force on one of them. 1. Draw the force from each other mass. 2. Resolve along two axes and add. 3. Use symmetry to cancel components. Example: Three equal masses M at the corners of an equilateral triangle: the force on each is √3 GM²/a².

How do I solve “Point of zero field” questions in JEE Main?

Two masses are a distance d apart. Find the point where the net field or force is zero. 1. Put the point between the masses. 2. Equate GM₁/x² and GM₂/(d − x)². 3. Take the square root and solve for x. Example: 9M and M at 4 m apart: the point is 3 m from the larger mass.

How do I solve “Field and potential of a shell or a sphere” questions in JEE Main?

A shell or a solid sphere is given. Find the field or the potential at a point inside, on the surface or outside. 1. Decide whether the point is inside or outside. 2. For a shell: inside the field is 0. 3. For a solid sphere: inside I = GMr/R³. Example: Inside a solid sphere at R/2: the field is half the surface value.

How do I solve “g on another planet” questions in JEE Main?

A planet has a different mass, radius or density. Find g, the weight or the time of fall on it. 1. Write g = GM/R² = (4/3)πGρR. 2. Take the ratio with the Earth. 3. Use g for the motion. Example: Same density and twice the radius: g′ = 2g.

How do I solve “g at a height or a depth” questions in JEE Main?

Find the weight or g at a height above or a depth below the surface. 1. Height: g′ = g(1 + h/R)⁻². 2. Depth: g′ = g(1 − d/R). 3. For small h use g(1 − 2h/R). Example: At a height R: g/4. At a depth R/2: g/2.

How do I solve “Rotation of the Earth and latitude” questions in JEE Main?

The effective g changes with latitude or the Earth spins faster. Find the weight or the time of a day. 1. Write g′ = g − ω²R cos²λ. 2. For zero weight at the equator ω = √(g/R). 3. Convert ω into the length of the day. Example: Weightless at the equator when the day is about 1.4 hours.

How do I solve “Potential energy and work to lift” questions in JEE Main?

A body is raised from the surface to a height, or two masses are brought together. Find the work or the change in energy. 1. Write U = −GMm/r at each position. 2. The work is the difference. 3. Check with ΔU = mgh for small heights. Example: 10 kg raised to a height R: ΔU = mgR/2 = 3.2 × 10⁸ J.

How do I solve “Escape speed” questions in JEE Main?

Find the escape speed from a planet, or compare it between planets. 1. Use v(e) = √(2GM/R) = √(2gR). 2. Compare using √(M/R) or R√ρ. 3. Remember it does not depend on the mass of the body. Example: g = 10 and R = 6.4 × 10⁶ m: v(e) = 11.3 km/s.

How do I solve “Height reached and speed at infinity” questions in JEE Main?

A body is thrown up with a given speed. Find the greatest height or its speed far away. 1. Write the total energy at the start. 2. Equate it to the energy at the top (or at infinity). 3. Solve for h or for v∞. Example: v = v(e)/2 reaches a height R/3.

How do I solve “Orbital speed, period and energy” questions in JEE Main?

A satellite moves in a circular orbit. Find the speed, period or any form of energy. 1. Equate GMm/r² to mv²/r. 2. Use T = 2πr/v. 3. Use K = −E and U = 2E. Example: Near the Earth: v = 8 km/s, T = 84 min.

How do I solve “Geostationary and comparing satellites” questions in JEE Main?

Compare satellites with different periods or radii, or use the geostationary orbit. 1. Use T² ∝ r³. 2. Geostationary: T = 24 h, equatorial plane. 3. Express the ratio. Example: An 8-day period gives an orbit radius of 4 times the geostationary radius.

How do I solve “Binding energy and changing the orbit” questions in JEE Main?

A satellite is moved to a new orbit or taken to infinity. Find the energy needed. 1. Write E = −GMm/2r for each orbit. 2. The energy needed is the difference. 3. Binding energy = GMm/2r. Example: A satellite close to the Earth has a binding energy of mgR/2.

How do I solve “Kepler’s laws” questions in JEE Main?

A planet moves in an ellipse or two planets are compared. Find the period, the speed or the ratio of speeds. 1. Use r v = constant at the ends. 2. Use T² ∝ a³. 3. Compare through the ratio. Example: Perihelion 1 unit and aphelion 4 units: the speed ratio is 4.

How do I solve “Binary stars and elliptical orbits” questions in JEE Main?

Two stars orbit each other, or a body moves in an ellipse with given extreme distances. Find the period, the speed or the energy. 1. For binaries, use the same ω and the centre of mass. 2. For ellipses, use a = (r₁ + r₂)/2 and E = −GMm/2a. 3. Apply both conservation laws. Example: Two equal masses M at a distance d: v = √(GM/2d).

Common myths and the truth

• Myth: There is no gravity inside a satellite, so things float. Truth: Gravity still acts (nearly as g). Things float because the satellite and its contents fall together.
• Myth: The gravitational field is zero at the centre of a solid sphere only. Truth: The field is zero at the centre, but inside a thin shell it is zero everywhere.
• Myth: A heavier satellite orbits faster. Truth: The orbital speed depends on the orbit radius and the central mass, not on the satellite’s mass.
• Myth: A heavier body needs a larger speed to escape. Truth: Escape speed does not depend on the mass of the body.
• Myth: The gravitational potential energy is positive above the ground. Truth: It is negative everywhere when the zero is taken at infinity. It rises towards zero as the body goes up.
• Myth: g is the same everywhere on the Earth. Truth: It is smaller at the equator (rotation and bulge), at greater heights and inside the Earth.
• Myth: When a satellite loses energy to air drag it slows down. Truth: It moves to a lower orbit and speeds up, because K = −E.
• Myth: A planet moves at a constant speed in its orbit. Truth: Only for a circular orbit. In an ellipse it is fastest at perihelion.
• Myth: A geostationary satellite can be placed over any city. Truth: It must be over the equator, since the orbit plane has to contain the Earth’s centre and match its rotation.
• Myth: G and g are the same thing. Truth: G is a universal constant; g is the acceleration at a particular place and changes from place to place.

Mistakes to avoid

• Inside a shell the field is zero, but the potential is not zero.
• g with height falls as 1/(R + h)², not linearly except for small h.
• The orbital speed and the period do not depend on the mass of the satellite.
• Escape speed does not depend on the mass or the direction of the body.
• Gravitational potential energy is negative: U = −GMm/r, with zero at infinity.
• A geostationary satellite has a fixed period of 24 h and a fixed height.
• Weightlessness in an orbit does not mean that gravity is zero.
• Use the distance from the centre of the Earth (R + h), not from the surface, in the formulas.

Simulations for Gravitation

• Falling through a tunnel in the Earth — Drop a ball into a straight frictionless tunnel through a uniform planet. It oscillates in SHM.
• Escape from a planet — Explore escape speed for planets with different mass and radius.
• Geostationary orbit — Pick an orbital period and find the radius where a satellite keeps pace with the turning Earth.
• Satellite orbit — Move a satellite to another altitude and compare its speed and orbital period.
• Gravity above and below Earth — Move a probe from inside Earth into space and compare the two different gravity laws.
• Kepler’s elliptical orbit — Follow a planet around an elliptical orbit with the star at one focus.
• Energy of a circular orbit — Raise a satellite and split its energy into kinetic and potential parts.

More: Class 11 Physics simulations · all JEE Main chapters.