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Gravitation – JEE Main Physics Formula Sheet & Class 11 Notes | Newton's Law of Gravitation, g Variation, Escape Velocity, Satellites, Kepler's Laws & All Gravity Formulas
This is the complete JEE Main Physics Formula Sheet and Class 11 Formula Sheet for Gravitation — Chapter 06 from the Aakash Rapid Revision & Formula Bank. Gravitation covers: Newton's Law of Gravitation — F=Gm₁m₂/r² (universal law, vector form); Variation of g — at height h: g'=g(1–2h/Rₑ) (h<
Gravitation JEE Main Formula Sheet PDF Preview
Scroll to explore all Gravitation formulas — JEE Main Physics Formula Sheet
Introduction: Why Gravitation Is a Conceptually Rich Scoring Chapter in JEE Main Physics
Gravitation is one of the most elegant chapters in JEE Main Physics — it extends Newton's laws to a universal scale, explains both terrestrial gravity and planetary orbits within a single framework. The same formula F=Gm₁m₂/r² that governs the fall of an apple explains the motion of the Moon, the orbits of planets, the existence of escape velocity, and the physics of satellites. Every quantitative result in gravitation — g variation, escape velocity, orbital period, satellite energy — follows mathematically from this single law combined with energy conservation and Newton's 2nd law.
For JEE Main physics, Gravitation contributes 2–3 questions per session. Questions test: variation of g with height/depth/rotation (approximate formulas), gravitational field and potential for shell and sphere (inside vs outside), escape velocity vs orbital velocity (vₑ=√2 × v₀), satellite energy (KE=–E, PE=2E, TE=E), Kepler's third law T²∝a³, and binary star system orbital velocities.
Download the Free PDF for Gravitation to access all g variation formulas, gravitational field and potential complete results, escape velocity, Kepler's laws, orbital velocity, satellite KE/PE/TE, and binary star formulas in one structured JEE Main physics revision reference.
Key Concepts and Formulas in Gravitation
Newton's Law of Gravitation, g at Surface, and Variation of g
Why g Variation Formulas for Height, Depth, and Latitude Are the Most-Tested Gravitation JEE Main Results
Newton's Universal Law of Gravitation (from Aakash PDF — Gravitation):
Every pair of bodies in the universe attract each other with a force proportional to the product of their masses and inversely proportional to the square of the distance between them.
F = Gm₁m₂/r²
Vector form: F₁₂ = –(Gm₁m₂/|r₁₂|²) r̂₁₂ (force on 1 due to 2, directed from 2 toward 1).
G = 6.674×10⁻¹¹ N·m²·kg⁻² = 6.67×10⁻¹¹ in SI. Dimensional formula: [M⁻¹L³T⁻²].
G is universal (same for all pairs, all locations). Gravitational force is always attractive. It acts along the line joining the centres of the two masses (central force).
Acceleration due to Gravity at Earth's Surface (from Aakash PDF — Gravitation):
g = GMₑ/Rₑ²
g ≈ 9.8 m/s² ≈ 10 m/s². Mₑ = 6×10²⁴ kg; Rₑ = 6400 km = 6.4×10⁶ m.
Variation of g with Height (from Aakash PDF — Gravitation JEE Main):
At height h above Earth's surface (exact formula):
g' = g/(1+h/Rₑ)² = gRₑ²/(Rₑ+h)²
Approximate formula for h << Rₑ (binomial approximation):
g' ≈ g(1 – 2h/Rₑ)
Special case: at h = Rₑ (one Earth radius above surface): g' = g/(1+1)² = g/4.
At h = 2Rₑ: g' = g/9. General: g' at nRₑ above surface = g/(n+1)².
Variation of g with Depth (from Aakash PDF — Gravitation JEE Main):
At depth x below Earth's surface:
g' = g(1 – x/Rₑ)
At centre of Earth (x = Rₑ): g' = 0. Weight = 0 at the centre.
g decreases linearly with depth. g decreases with height as well, but more slowly near the surface than at large heights.
Comparison: g decreases at rate 2g/Rₑ per unit height (near surface), and decreases at rate g/Rₑ per unit depth.
Variation of g due to Earth's Rotation and Latitude (from Aakash PDF — Gravitation JEE Main):
Apparent acceleration due to gravity at latitude λ:
g' = g – Rₑω²cos²λ
where ω = angular velocity of Earth's rotation, λ = latitude angle.
At poles (λ = 90°, cosλ = 0): g' = g (maximum value — no centrifugal effect).
At equator (λ = 0°, cosλ = 1): g' = g – Rₑω² (minimum value).
Earth's rotation reduces apparent g by at most Rₑω² ≈ 0.034 m/s² at equator.
Download the Free PDF for Gravitation for all g variation examples for JEE Main.
Gravitational Field Intensity and Potential — Shell and Solid Sphere
Why Gravitational Field and Potential Inside/Outside Shell and Sphere Are the Most Graph-Tested Gravitation Topics
Gravitational Field Intensity (from Aakash PDF — Gravitation JEE Main):
Gravitational field intensity I at a point = force per unit mass at that point:
I = F/m = –GM/r² r̂ (directed toward the source mass)
SI unit: N/kg = m/s². Relation: g = I at Earth's surface.
Gravitational Potential (V): Work done per unit mass in bringing a unit test mass from infinity to that point:
V = –W/m = –GM/r (always negative; at infinity V=0)
Relation: I = –dV/dr (field = –gradient of potential).
For Spherical Shell of Mass M, Radius R (from Aakash PDF — Gravitation JEE Main):
Case I — Inside shell (r < R):
I = 0 (field is zero everywhere inside a uniform spherical shell)
V = –GM/R (potential is constant inside = value at surface)
Case II — On surface (r = R):
I = GM/R²; V = –GM/R
Case III — Outside shell (r > R):
I = GM/r² (same as point mass); V = –GM/r
For Uniform Solid Sphere of Mass M, Radius R (from Aakash PDF — Gravitation JEE Main):
Case I — Inside sphere (r < R):
I = GMr/R³ (field increases linearly from centre to surface)
V = –GM(3R²–r²)/(2R³) (parabolic variation)
At centre (r=0): I = 0; V_centre = –3GM/2R = (3/2)V_surface
Case II — On surface (r = R):
I = GM/R²; V = –GM/R
Case III — Outside (r > R):
I = GM/r²; V = –GM/r (same as point mass)
Key Comparisons (from Aakash PDF — Gravitation JEE Main):
For shell: I=0 inside (flat); jumps at surface; decreases as 1/r² outside.
For solid sphere: I increases linearly (∝r) inside; reaches max at surface GM/R²; decreases as 1/r² outside.
Potential for shell: constant (–GM/R) inside; continuous at surface; decreases as –GM/r outside (becomes less negative).
Potential for solid sphere: parabolic inside (most negative at centre: –3GM/2R); continuous at surface; decreases toward 0 as r→∞.
Gravitational Field and Potential on Axis of Ring (from Aakash PDF — Gravitation):
For a uniform ring of mass M, radius R, at axial distance x from centre:
I = GMx/(R²+x²)^(3/2) (along axis, toward centre if behind the ring)
I = 0 at centre (x=0); maximum at x = R/√2; Imax = 2GM/(3√3·R²)
V = –GM/√(R²+x²) (potential at axial point)
Download the Free PDF for Gravitation for all field and potential graphs for JEE Main.
Gravitational PE, Binding Energy, and Escape Velocity
Why Escape Velocity vₑ=√(2gRₑ) and Its Relation to Orbital Velocity Are Key Gravitation JEE Main Formulas
Gravitational Potential Energy (from Aakash PDF — Gravitation JEE Main):
PE of a mass m at distance r from Earth's centre:
U = –GMₑm/r
U = 0 at r = ∞ (reference). U is always negative (bound system). As r increases toward ∞, U increases from negative toward 0.
At Earth's surface (r = Rₑ): U_surface = –GMₑm/Rₑ = –mgRₑ
At height h: U(h) = –GMₑm/(Rₑ+h)
Change in PE when raised from surface to height h: ΔU = –GMₑm/(Rₑ+h) – (–GMₑm/Rₑ) = GMₑm×h/[Rₑ(Rₑ+h)] ≈ mgh (for h<
Binding Energy (from Aakash PDF — Gravitation JEE Main):
Energy required to remove an object completely from the gravitational field (to r=∞):
Binding Energy = –U_surface = +GMₑm/Rₑ = mgRₑ
This is the magnitude of the total mechanical energy of the mass at rest on the surface.
Escape Velocity (from Aakash PDF — Gravitation JEE Main):
Minimum initial velocity needed for a body to escape from Earth's gravitational field (reach r=∞ with zero final KE):
By energy conservation: ½mvₑ² + (–GMₑm/Rₑ) = 0 → ½mvₑ² = GMₑm/Rₑ
vₑ = √(2GMₑ/Rₑ) = √(2gRₑ)
vₑ = 11.2 km/s for Earth.
In terms of density ρ: vₑ = √(8πGρRₑ²/3) [using Mₑ = (4/3)πRₑ³ρ]
vₑ depends on the mass and radius of the planet/star, NOT on the mass of the escaping object.
For Moon: vₑ ≈ 2.4 km/s (smaller mass and radius); for Sun: vₑ ≈ 618 km/s.
Neutral Point between Two Masses (from Aakash PDF — Gravitation):
The point P between masses M₁ (at origin) and M₂ (at distance d) where the net gravitational field is zero. Let P be at distance r₁ from M₁ and r₂ = d–r₁ from M₂:
GM₁/r₁² = GM₂/r₂² → M₁/r₁² = M₂/r₂² → r₁/r₂ = √(M₁/M₂)
r₁ = d√M₁/(√M₁+√M₂); r₂ = d√M₂/(√M₁+√M₂)
Energy required to take mass m from neutral point to infinity = [V_neutral × m] (potential at neutral point is not zero — only field is zero). Download the Free PDF for Gravitation for all escape velocity and PE examples for JEE Main.
Kepler's Three Laws of Planetary Motion
Why Kepler's Laws — Especially T²∝a³ and Equal Areas — Are Direct Gravitation JEE Main Questions
Kepler's First Law (Law of Ellipses) (from Aakash PDF — Gravitation):
All planets revolve around the Sun in elliptical orbits with the Sun at one of the two foci of the ellipse.
For an ellipse with semi-major axis a and eccentricity e:
Distance at perigee (closest approach): rₚ = (1–e)a
Distance at apogee (farthest point): rₐ = (1+e)a
Sum: rₚ + rₐ = 2a (major axis length)
For circular orbit: e = 0, rₚ = rₐ = a = r (radius).
Kepler's Second Law (Law of Equal Areas) (from Aakash PDF — Gravitation JEE Main):
A line joining a planet and the Sun sweeps out equal areas in equal time intervals.
This is equivalent to: angular momentum L = constant (since no torque from gravitational force about the Sun).
Areal speed = dA/dt = L/2m = constant.
Consequence: planet moves faster near perigee and slower near apogee.
Speed at perigee vs apogee: by angular momentum conservation mvₚrₚ = mvₐrₐ:
vₚ/vₐ = rₐ/rₚ = (1+e)/(1–e)
Kepler's Third Law (Law of Periods) (from Aakash PDF — Gravitation JEE Main):
The square of the time period of revolution of a planet is proportional to the cube of the semi-major axis of its elliptical orbit:
T² ∝ a³
Exact formula: T² = 4π²a³/(GM_sun) → T = 2π√(a³/GM_sun)
For two planets 1 and 2 orbiting the same star: T₁²/T₂² = a₁³/a₂³
Ratio of angular velocities at apogee and perigee (from Aakash PDF):
ωₐ/ωₚ = (rₚ/rₐ)² = (1–e)²/(1+e)²
[Since L = mωr² = constant → ω₁r₁² = ω₂r₂² → ωₐ/ωₚ = rₚ²/rₐ²]
Download the Free PDF for Gravitation for all Kepler's law examples for JEE Main.
Orbital Velocity, Time Period, and Satellite Energy
Why Satellite KE=–E and PE=2E Are the Most Elegant Gravitation JEE Main Formulas
Orbital Velocity (from Aakash PDF — Gravitation JEE Main):
A satellite of mass m orbits Earth in a circular orbit of radius r (r = Rₑ+h). Gravitational force = centripetal force:
GMₑm/r² = mv₀²/r → v₀ = √(GMₑ/r) = √(gRₑ²/r)
At Earth's surface (r = Rₑ): v₀ = √(gRₑ) ≈ 7.9 km/s
Relation: vₑ = √2 × v₀ (at same radius). Escape velocity is √2 times orbital velocity.
v₀ decreases as r increases (farther from Earth → slower orbital speed).
Time Period of Satellite (from Aakash PDF — Gravitation JEE Main):
T = 2πr/v₀ = 2π√(r³/GMₑ)
Equivalently: T² = 4π²r³/GMₑ = 4π²r³/gRₑ² (this is Kepler's 3rd law for Earth satellites).
Minimum time period (satellite grazing Earth's surface, r = Rₑ):
T_min = 2π√(Rₑ³/GMₑ) = 2π√(Rₑ/g)
T_min = 84.6 min ≈ 1.4 hours for Earth.
All satellites have T > 84.6 min. Geostationary satellite: T = 24 h, r ≈ 42,000 km ≈ 6.6Rₑ.
Satellite Energy (from Aakash PDF — Gravitation JEE Main):
Kinetic Energy: KE = ½mv₀² = GMₑm/2r
KE = GMₑm/2r
Potential Energy: PE = –GMₑm/r
PE = –GMₑm/r = –2KE
Total Mechanical Energy: E = KE + PE = GMₑm/2r – GMₑm/r = –GMₑm/2r
E = –GMₑm/2r (total energy is negative → satellite is bound)
Key Relations (from Aakash PDF — Gravitation JEE Main):
KE = –E (kinetic energy = magnitude of total energy)
PE = 2E (potential energy = twice total energy, i.e., twice as negative as KE)
|PE| = 2×KE (always, for circular orbit)
When satellite is taken to higher orbit: r increases → v₀ decreases → KE decreases → PE increases (becomes less negative) → TE increases (becomes less negative). Paradox: as energy is added to satellite, its speed DECREASES.
Download the Free PDF for Gravitation for all satellite energy examples for JEE Main.
Binary Star System and Additional Gravitation Results
Why Binary Star System Formulas Complete the Gravitation JEE Main Formula Set
Binary Star System (from Aakash PDF — Gravitation JEE Main):
Two stars of mass M₁ and M₂ orbit their common centre of mass (CM) under mutual gravitational attraction. Distance between them = r = r₁ + r₂.
CM condition: M₁r₁ = M₂r₂ → r₁ = M₂r/(M₁+M₂); r₂ = M₁r/(M₁+M₂)
Both stars have the same angular velocity ω (both complete one orbit in the same period T):
Gravitational force on M₁ = centripetal force for M₁:
GM₁M₂/r² = M₁ω²r₁ = M₁ω²M₂r/(M₁+M₂)
ω² = G(M₁+M₂)/r³
Time period: T = 2π/ω = 2π√(r³/G(M₁+M₂))
This is Kepler's third law with total mass M₁+M₂ replacing M_sun.
Orbital speeds:
v₁ = ωr₁ = M₂√(G/((M₁+M₂)r))
v₂ = ωr₂ = M₁√(G/((M₁+M₂)r))
Note: v₁/v₂ = M₂/M₁ and v₂/v₁ = M₁/M₂ (heavier star orbits slower, closer to CM).
Special case M₁ = M₂ = M: ω = √(2GM/r³); v₁ = v₂ = √(GM/2r); r₁ = r₂ = r/2.
Geosynchronous / Geostationary Satellite (from Aakash PDF — Gravitation):
A satellite appears stationary relative to Earth when it orbits in the equatorial plane with T = 24 hours in the same direction as Earth's rotation. This is the geostationary orbit.
Radius: r_geo ≈ 42,000 km ≈ 6.6Rₑ. Height: h_geo ≈ 35,800 km ≈ 36,000 km above surface.
v_geo = √(GMₑ/r_geo) ≈ 3.07 km/s.
Weight — Apparent vs Real (Gravitation — JEE Main):
Apparent weight = Normal force N. Real weight = mg (gravitational force).
In satellite (free fall): N = 0 → apparent weight = 0 → weightlessness.
In lift accelerating up: N = m(g+a) → apparent weight increases.
In lift accelerating down: N = m(g–a). In free fall (a=g): N = 0.
Download the Free PDF for Gravitation for all binary star and satellite examples for JEE Main.
Download Free PDF — Gravitation JEE Main Physics Formula Sheet
All Gravitation formulas from the Aakash Rapid Revision PDF: Newton's law F=Gm₁m₂/r², G=6.67×10⁻¹¹ Nm²kg⁻² [M⁻¹L³T⁻²], g=GMₑ/Rₑ², g at height exact g/(1+h/Rₑ)², approximate g(1–2h/Rₑ), at h=Rₑ g/4, at nRₑ g/(n+1)², depth g(1–x/Rₑ), centre g=0, rotation g'=g–Rₑω²cos²λ (poles max, equator min=g–Rₑω²), gravitational field I=GM/r² (outside); shell: inside I=0/V=–GM/R, surface I=GM/R², outside I=GM/r²/V=–GM/r; solid sphere: inside I=GMr/R³/V=–GM(3R²–r²)/2R³, centre I=0/V=–3GM/2R=3/2×V_surface, surface and outside same as shell; ring axis I=GMx/(R²+x²)^(3/2) max at x=R/√2, V=–GM/√(R²+x²); I=–dV/dr; gravitational PE U=–GMm/r, at surface –mgRₑ, binding energy=GMm/Rₑ=mgRₑ; escape velocity vₑ=√(2GMₑ/Rₑ)=√(2gRₑ)=11.2 km/s=√8πGρRₑ²/3; neutral point M₁/r₁²=M₂/r₂²; Kepler 1st (ellipse, rₚ=(1–e)a, rₐ=(1+e)a); 2nd (equal areas ↔ L=const, vₚ/vₐ=rₐ/rₚ=(1+e)/(1–e), ωₐ/ωₚ=(rₚ/rₐ)²); 3rd (T²∝a³, T=2π√(a³/GMsun)); orbital velocity v₀=√(GMₑ/r), at surface 7.9 km/s, vₑ=√2×v₀; time period T=2π√(r³/GMₑ), T_min=2π√(Rₑ/g)=84.6 min; satellite KE=GMₑm/2r; PE=–GMₑm/r; TE=E=–GMₑm/2r; KE=–E; PE=2E; |PE|=2KE; geostationary T=24h/r≈42000km; binary star M₁r₁=M₂r₂, ω²=G(M₁+M₂)/r³, T=2π√(r³/G(M₁+M₂)), v₁=M₂√(G/(M₁+M₂)r), v₂=M₁√(G/(M₁+M₂)r).
Why Gravitation Is a Reliable Scoring Chapter in JEE Main Physics
The three g variation formulas (height, depth, rotation) are pure direct substitution in JEE Main. Height (exact): g'=g/(1+h/Rₑ)² — use this when h is not small compared to Rₑ. Height (approximate): g'≈g(1–2h/Rₑ) — use only when h<
The satellite energy relations KE=–E and PE=2E are the most frequently cited Gravitation results in JEE Main. These say: a satellite's kinetic energy equals the magnitude of its total (negative) energy; its potential energy is twice its total energy. These elegant relations mean: if you know any one of KE, PE, TE for a satellite — you know all three. Also: v₀ = √(GMₑ/r) and vₑ = √2×v₀ mean that if a satellite increases its speed to √2 times its orbital speed at the same orbit, it escapes from that orbit.
Kepler's 3rd Law T²∝a³ is directly tested in every few JEE Main sessions. The standard question: planet A has orbital period 8 years; planet B has orbital period 1 year. Find ratio of semi-major axes. Answer: (a_A/a_B)² = (T_A/T_B)² → T_A²/T_B² = a_A³/a_B³ → (8)²/(1)² = a_A³/a_B³ → a_A/a_B = (64)^(1/3) = 4. The calculation is always a simple ratio problem.
Gravitational field inside a spherical shell is ZERO — the single most conceptually important Gravitation result. Inside a uniform spherical shell, every field contribution from one patch of the shell is exactly cancelled by the contribution from the opposite patch. This means: a person inside a hollow Earth would feel zero gravity. The potential inside is constant (= –GM/R, the surface value). This discontinuity of field (zero inside, GM/R² at surface) is a standard JEE Main graph-based question. Download the Free PDF for Gravitation to have all formulas ready.
Who Should Use This Gravitation Formula Sheet?
Learning Outcomes After Completing Gravitation
After working through Gravitation using this formula sheet, a student should confidently accomplish the following for JEE Main physics. On Newton's law: state and apply F=Gm₁m₂/r²; give value and dimensions of G; compute g=GMₑ/Rₑ².
On g variation: apply g'=g/(1+h/Rₑ)² (exact) and g'≈g(1–2h/Rₑ) (approximate) for height; apply g'=g(1–x/Rₑ) for depth; state g=0 at centre; apply g'=g–Rₑω²cos²λ for rotation; identify poles (g max) and equator (g min).
On gravitational field and potential: state I and V for inside and outside a spherical shell (I=0 inside, constant V=–GM/R); state I and V for inside and outside a solid sphere (I=GMr/R³ inside, V parabolic, V_centre=–3GM/2R); state ring axis formulas; apply I=–dV/dr.
On escape velocity: derive and apply vₑ=√(2GMₑ/Rₑ)=√(2gRₑ)=11.2 km/s; state vₑ=√2×v₀; compute binding energy=GMₑm/Rₑ; find neutral point between two masses.
On Kepler's laws: state all three laws; apply T²∝a³ for ratio problems; compute vₚ/vₐ=(1+e)/(1–e) from angular momentum conservation; compute ωₐ/ωₚ=(rₚ/rₐ)².
On orbital velocity and satellite: apply v₀=√(GMₑ/r); apply T=2π√(r³/GMₑ); compute T_min=84.6 min; apply KE=GMₑm/2r; PE=–GMₑm/r; TE=–GMₑm/2r; use KE=–E and PE=2E; describe geostationary orbit; binary star: apply M₁r₁=M₂r₂, ω²=G(M₁+M₂)/r³, T formula. Download the Free PDF for Gravitation to test all outcomes before your JEE Main exam.
Get the Free PDF for Gravitation — JEE Main Quick Revision
The Aakash Rapid Revision & Formula Bank PDF for Gravitation contains all g variation formulas, complete gravitational field and potential results for shell and solid sphere, escape velocity, Kepler's three laws, orbital velocity, time period, satellite energy KE/PE/TE with key relations, and binary star system formulas in one structured JEE Main physics reference.
Conclusion — Gravitation: The Universal Force Framework in JEE Main Physics
Gravitation reveals the universality of physics — the same mathematical law that describes terrestrial gravity also explains planetary orbits, satellite motion, black holes, and the structure of the universe. For JEE Main, the chapter is compact and predictable: five key formula groups (g variation, field/potential shell/sphere, escape velocity, Kepler's laws, satellite energy) cover every question type. The most important result to internalise: satellite energy E=–GMm/2r with KE=–E and PE=2E. This single trio of equations answers every satellite energy question in JEE Main in one step.
The second most important result: field inside a uniform spherical shell = 0. This is tested in graph-based questions showing I vs r for a shell — the flat line inside (I=0) jumping to GM/R² at the surface and decreasing as 1/r² outside. For a solid sphere the graph shows linear increase inside and 1/r² decrease outside. These two graphs are the canonical Gravitation graph questions in JEE Main. Use this page and the Free PDF Download for Gravitation as your complete JEE Main revision foundation.
Frequently Asked Questions — Gravitation Formulas
What are all the formulas for variation of g in Gravitation for JEE Main?
In Gravitation, g variation formulas from the Aakash PDF: (1) At height h (exact): g'=GMₑ/(Rₑ+h)²=g/(1+h/Rₑ)². (2) At height h (h<
What is the gravitational field inside and outside a spherical shell in Gravitation?
In Gravitation, for a uniform spherical shell of mass M, radius R: Inside (r
What is the escape velocity formula in Gravitation and how is it related to orbital velocity?
In Gravitation, escape velocity: minimum initial speed for a body to escape Earth's gravitational pull (reach r=∞ with v=0). Derivation: ½mvₑ²+U_surface=0 → ½mvₑ²=GMₑm/Rₑ → vₑ=√(2GMₑ/Rₑ). Since GMₑ/Rₑ=gRₑ²/Rₑ=gRₑ: vₑ=√(2gRₑ)=√(2×10×6.4×10⁶)=√(1.28×10⁸)≈11,300 m/s=11.2 km/s. In terms of density: Mₑ=(4/3)πRₑ³ρ → GMₑ=4πGρRₑ³/3 → vₑ=Rₑ√(8πGρ/3). vₑ depends on Rₑ and Mₑ (or g and Rₑ), not on the mass of escaping body or direction. Orbital velocity at radius r: v₀=√(GMₑ/r). At surface r=Rₑ: v₀_surface=√(gRₑ)=√(10×6.4×10⁶)≈7.9 km/s. Relation: vₑ=√2×v₀(at same radius). So at surface: vₑ/v₀=√2 → vₑ=√2×7.9≈11.2 km/s ✓. Practical meaning: to transfer from circular orbit to escape trajectory, speed must be increased by factor √2. For satellite already in orbit at radius r: energy to escape = ½mvₑ²–½mv₀²=½mv₀²(2–1)=½mv₀²=KE. So to escape from orbit, satellite needs additional energy equal to its current KE.
What are Kepler's three laws of planetary motion in Gravitation and what is their mathematical form?
In Gravitation, Kepler's three laws: First Law (Law of Ellipses): planets orbit the Sun in elliptical paths with the Sun at one focus. Ellipse parameters: semi-major axis a, eccentricity e. Perigee (closest): rₚ=(1–e)a. Apogee (farthest): rₐ=(1+e)a. rₚ+rₐ=2a. e=0→circle. Second Law (Law of Equal Areas): the line joining planet to Sun sweeps equal areas in equal time intervals. Equivalent to conservation of angular momentum: L=mvr=m(ωr)r=mωr²=constant. Consequence: vₚrₚ=vₐrₐ (by L conservation) → vₚ/vₐ=rₐ/rₚ=(1+e)/(1–e). Also: ωₚrₚ²=ωₐrₐ² → ωₚ/ωₐ=(rₐ/rₚ)²=(1+e)²/(1–e)². Third Law (Law of Periods): T²∝a³. Exact form: T=2π√(a³/GMₛ) where Mₛ=Sun's mass. For two planets: T₁/T₂=(a₁/a₂)^(3/2) → T₁²/T₂²=a₁³/a₂³. JEE Main Gravitation application: Earth T=1 yr, a=1 AU; Mars a=1.52 AU → T_Mars=(1.52)^(3/2)≈1.87 years. These Gravitation Kepler results are tested in JEE Main as ratio problems using T²∝a³.
What are the satellite energy formulas in Gravitation and what do KE=–E and PE=2E mean?
In Gravitation, for a satellite of mass m in circular orbit of radius r around Earth: v₀=√(GMₑ/r). KE=½mv₀²=GMₑm/2r. PE=–GMₑm/r (gravitational PE). TE=KE+PE=GMₑm/2r–GMₑm/r=–GMₑm/2r. So: KE=GMₑm/2r>0; PE=–GMₑm/r<0; TE=–GMₑm/2r<0. Key relations: KE=–TE → TE is negative of KE → if total energy E=–GMₑm/2r, then KE=|E|=GMₑm/2r. PE=2×TE → PE equals twice the (negative) total energy → PE=–GMₑm/r=2×(–GMₑm/2r)=2E. |PE|=2KE always (potential energy magnitude = twice kinetic energy). Physical meaning of KE=–E: satellite's kinetic energy equals the magnitude of its total mechanical energy. Paradox: if satellite is given more energy (boosted to higher orbit), r increases → KE=GMₑm/2r DECREASES → satellite moves slower. This "anti-intuitive" result is because the increase in PE (less negative) exceeds the decrease in KE. Net energy increases (becomes less negative) as orbit goes higher. To de-orbit: remove energy → satellite moves to lower, faster orbit. These Gravitation satellite energy results are tested in JEE Main as conceptual MCQs and direct calculation questions.
What is the binary star system in Gravitation and what are the key formulas?
In Gravitation, binary star system: two stars M₁ and M₂ separated by distance r, orbiting their common CM. CM fixed: M₁r₁=M₂r₂; r₁+r₂=r → r₁=M₂r/(M₁+M₂); r₂=M₁r/(M₁+M₂). Both orbit with SAME angular velocity ω and SAME time period T. Finding ω: For M₁, gravity from M₂ provides centripetal force: GM₁M₂/r²=M₁ω²r₁=M₁ω²M₂r/(M₁+M₂). Simplify: GM₂/r²=ω²M₂r/(M₁+M₂) → Gω²=Gr²(M₁+M₂)/r² → ω²=G(M₁+M₂)/r³. T=2π/ω=2π√(r³/G(M₁+M₂)). Linear speeds: v₁=ωr₁=M₂ω r/(M₁+M₂)=M₂√(G/(M₁+M₂)r). v₂=M₁√(G/(M₁+M₂)r). Speed ratio: v₁/v₂=M₂/M₁ (heavier star moves slower). Equal masses M₁=M₂=M: r₁=r₂=r/2; ω=√(2GM/r³); T=2π√(r³/2GM); v₁=v₂=√(GM/2r). JEE Main Gravitation: binary star time period question → T=2π√(r³/G(M₁+M₂)). Compare with single star satellite: T=2π√(r³/GM). Replace M→M₁+M₂ for binary.
What is the orbital velocity formula and minimum time period of a satellite in Gravitation?
In Gravitation, orbital velocity at radius r: v₀=√(GMₑ/r)=√(gRₑ²/r). At Earth's surface (r=Rₑ): v₀=√(gRₑ)=√(10×6.4×10⁶)≈8 km/s. Time period: T=2πr/v₀=2π√(r³/GMₑ)=2π√(r³/gRₑ²). Minimum time period is for satellite skimming Earth's surface (r=Rₑ): T_min=2π√(Rₑ³/GMₑ)=2π√(Rₑ/g). Numerically: 2π√(6.4×10⁶/10)=2π×800=5027 s≈84.6 min≈1.4 h. Any satellite around Earth must have T>84.6 min. Geostationary satellite: T=24h=86400 s. From T²=4π²r³/GMₑ: r=(GMₑT²/4π²)^(1/3)≈42,000 km. Height above surface: h=42000–6400=35600 km≈36000 km. v_geo=√(GMₑ/r_geo)≈3.07 km/s. Recall vₑ/v₀=√2 → at any orbit radius r, escape speed=√2×orbital speed. At geostationary orbit: v_esc=√2×3.07≈4.3 km/s. These Gravitation satellite period formulas are tested in JEE Main as Kepler's 3rd law ratio problems.
What is the neutral point between two massive bodies in Gravitation?
In Gravitation, the neutral point P between two masses M₁ and M₂ is where the total gravitational field is zero. Let M₁ be at origin, M₂ at distance d. P is between them at distance r₁ from M₁ and r₂=d–r₁ from M₂. Condition for zero field: GM₁/r₁²=GM₂/r₂² → M₁/r₁²=M₂/r₂². Solving: r₁/r₂=√(M₁/M₂). With r₁+r₂=d: r₁=d√M₁/(√M₁+√M₂); r₂=d√M₂/(√M₁+√M₂). Important: at the neutral point, the FIELD is zero but the POTENTIAL is NOT zero. Potential at P: V_P=–GM₁/r₁–GM₂/r₂ (sum of both contributions, both negative, both non-zero). Energy needed to move mass m from neutral point to infinity (from Earth-Moon system perspective): W=0–(m×V_P)=–mV_P=m(GM₁/r₁+GM₂/r₂). Special case M₁=M₂: r₁=r₂=d/2 (neutral point at midpoint). If M₁>>M₂ (Earth-Moon): r₁≈d×(1/(1+√(M₂/M₁)))≈d(1–√(M₂/M₁))≈d. The neutral point is closer to the lighter mass M₂. This is how gravity assist trajectories and Lagrange points work in space mission planning.
What is the gravitational potential energy at height h and what is binding energy in Gravitation?
In Gravitation, gravitational PE of mass m at distance r from Earth's centre: U=–GMₑm/r (negative, bound system). At surface (r=Rₑ): U_surface=–GMₑm/Rₑ=–mgRₑ (using g=GMₑ/Rₑ²). At height h (r=Rₑ+h): U(h)=–GMₑm/(Rₑ+h). Change in PE when raised from surface to height h: ΔU=U(h)–U_surface=–GMₑm/(Rₑ+h)–(–GMₑm/Rₑ)=GMₑmh/(Rₑ(Rₑ+h)). For h<
What is the gravitational potential inside a solid sphere in Gravitation and how does it compare to the surface value?
In Gravitation, for a uniform solid sphere of mass M, radius R, the potential at interior point at distance r from centre: V(r)=–GM(3R²–r²)/2R³. This can be written as V(r)=–GM/2R³×(3R²–r²). At r=0 (centre): V_centre=–GM×3R²/2R³=–3GM/2R. At r=R (surface): V_surface=–GM(3R²–R²)/2R³=–GM×2R²/2R³=–GM/R. So V_centre=–3GM/2R=(3/2)×(–GM/R)=(3/2)×V_surface. Key result: V_centre=(3/2)V_surface. Since V_surface is negative, V_centre is 3/2 times more negative (lower potential). Potential is most negative at centre, increases parabolically to –GM/R at surface, then continues increasing (less negative) as –GM/r for r>R. The gravitational field inside: I=–dV/dr=–d/dr×[–GM(3R²–r²)/2R³]=GMr/R³ (linearly increases from 0 at centre to GM/R² at surface). This parabolic potential variation inside and the V_centre=(3/2)V_surface result are tested in JEE Main Gravitation as direct questions about "potential at centre" and in graph interpretation.
Related Formula Sheets — JEE Main Physics
Gravitation – JEE Main Physics Formula Sheet
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