Gravitation Formula Sheet — JEE Main Physics

Every key Gravitation formula, definition and theorem for JEE Main Physics in one place — with common examiner traps and worked examples. Free to read; blurt from memory, then check your gaps.

Syllabus — topics coveredNTA · 10 sub-topics

The universal law of gravitation
Acceleration due to gravity and its variation with altitude and depth
Gravitational potential energy
Gravitational potential
Escape velocity
Orbital velocity of a satellite
Geo-stationary satellites
Kepler's laws of planetary motion
Gravitational potential energy near the surface of the earth
Gravitational potential in the gravitational field

Universal Law & Gravitational Field

Newton's universal law

F = \frac{G m _{1} m _{2}}{r ^{2}}

m_{1}, m_{2} =

masses,

r =

centre-to-centre separation,

G = 6.67 \times 1 0^{- 11} N m^{2} kg^{- 2}

(universal); always attractive, along the join line

▸Vector form: $F_{12} = - \frac{G m _{1} m _{2}}{r ^{2}} \overset{r}{^}_{12}$ , and $F_{21} = - F_{12}$ (Newton's third law).
▸ $Superposition$ : net force on a mass $=$ vector sum of forces from every other mass, $F = \sum_{i} F_{i}$ .
▸Strictly for $point masses$ ; a uniform sphere acts as if all mass is at its centre (shell theorem).
▸G measured by the $Cavendish$ torsion-balance experiment ('weighing the Earth').

Gravitational field intensity: Force per unit test mass

E_{g} = \frac{F}{m}

; for a point mass M,

E_{g} = \frac{GM}{r ^{2}}

— numerically equal to the acceleration g it produces (units

N kg^{- 1} = m s^{- 2}

⚡ Shortcut ·

g

vs field

Gravitational field

E_{g}

and acceleration g are the

same number

at a point — compute whichever is easier;

E_{g}

is the cause, g the effect on a free body.

Field of standard bodies

M =

total mass,

R =

body radius,

r / x =

distance from centre/axis

Point / outside any sphere (

r \geq R

)

E_{g} = \frac{GM}{r ^{2}}

Ring, on its axis

E_{g} = \frac{GM x}{( x ^{2} + R ^{2} ) ^{3/2}}

Ring — maximum field

at x = \frac{R}{2} (E_{g} = 0 at centre)

Solid sphere, inside (

r < R

)

E_{g} = \frac{GM r}{R ^{3}} \propto r

Shell, inside (

r < R

)

E_{g} = 0

Shell theorem (NCERT):A uniform spherical shell attracts an

external

point mass as if its entire mass were at the centre; it exerts

zero

force on a point mass

inside

it.

💡 Tip · No shielding

Gravitational

shielding is impossible

— unlike a conductor for electric fields, a shell does not screen the field of outside masses on a particle inside.

🚫 Examiner Trap · Universal law & field

(1) r is the

centre-to-centre

distance, NOT the surface gap — for two spheres add both radii. (2) Force

\propto 1/ r^{2}

only for points / external spheres;

inside

a solid sphere

E_{g} \propto r

(rises from

0

). (3) G (universal constant)

\neq = g

(local, varies). (4) Inside a shell

E_{g} = 0

but a shell does

not

shield — outside fields still act.

Acceleration due to Gravity & its Variation

Surface value

g = \frac{GM}{R ^{2}} = \frac{4}{3} π GρR

M =

Earth mass,

R =

Earth radius,

ρ =

mean density;

g \approx 9.8 m s^{- 2}

, independent of the falling body's mass

Altitude

h

above surface

g_{h} = g (\frac{R}{R + h})^{2} \approx g (1 - \frac{2 h}{R})

h =

height above surface; binomial approximation valid only for

h ≪ R

Depth

d

below surface

g_{d} = g (1 - \frac{d}{R})

d =

depth; only the inner sphere of radius

R - d

contributes (shell outside cancels);

g = 0

at the centre

Latitude

λ

(Earth's rotation)

g_{λ} = g - ω^{2} R cos^{2} λ

λ =

latitude,

ω =

Earth's spin rate; max at poles (

λ = 9 0^{\circ}

), min at equator;

ω^{2} R \approx 0.034 m s^{- 2}

⚡ Shortcut · Equal fall, equal cause

For small

h ≪ R

, g falls the same amount at height h as at depth

d = 2 h

(since

- 2 h / R

- d / R

). Useful to compare 'up vs down' problems fast.

g vs distance from centre: rises linearly (g proportional to r) inside, then falls as 1/r^2 outside, peaking at the surface r=R — g is $maximum at the surface$ : inside $\propto r$ , outside $\propto 1/ r^{2}$ .

Comparative: how

g

changes

Cause	Formula	Effect on $g$	Inside variation
Altitude $h$	$g (1 - 2 h / R)$	decreases	--
Depth $d$	$g (1 - d / R)$	decreases	$\propto r$ (linear)
Latitude $λ$	$g - ω^{2} R cos^{2} λ$	min equator, max pole	--
Rotation $ω ↑$	$g - ω^{2} R cos^{2} λ$	decreases	equator only

▸Near the surface the depth fall ( $1 - d / R$ ) is $slower$ than the altitude fall ( $1 - 2 h / R$ ).
▸Earth is an oblate ellipsoid: $R_{e q} - R_{p o l e} \approx 21$ km, so $g_{p o l e} > g_{e q}$ (with rotation, $Δ g \approx 0.05 m s^{- 2}$ ).
▸At the equator gravity would vanish if $ω$ rose $\approx 17 \times$ ( $T \approx 84$ min) — bodies would float.

★ Remember

Both altitude

and

depth

decrease

g. A body weighs most at the poles and at the surface.

🚫 Examiner Trap · Variation of

g

(1)

g_{h} = g (1 - 2 h / R)

is only the

approximation

for

h ≪ R

— for large h use the exact

g (R / (R + h))^{2}

. (2) Depth uses

1 - d / R

(factor

1

), altitude uses

1 - 2 h / R

(factor

2

) — do NOT swap. (3) Rotation reduces g at the

equator

, not the poles (

cos 9 0^{\circ} = 0

). (4) g is independent of the test body's mass — heavier objects do not fall faster.

Gravitational Potential & Potential Energy

Gravitational potential

V = - \frac{GM}{r}; E_{g} = - \frac{d V}{d r}

V =

work per unit mass to bring it from

\infty

(scalar, J/kg); always negative,

V = 0

r = \infty

; field points down the potential gradient

Potential of standard bodies (

M

total,

R

radius)

Ring (centre

/

axis x)

- \frac{GM}{R} / - \frac{GM}{R ^{2} + x ^{2}}

Shell, inside (

r \leq R

)

V = - \frac{GM}{R} (constant)

Solid sphere, inside (

r < R

)

V = - \frac{GM ( 3 R ^{2} - r ^{2} )}{2 R ^{3}}

Solid sphere, centre

V_{c} = - \frac{3 GM}{2 R} = 1.5 V_{s u r f a ce}

Two panels: field g(r) and potential V(r) for a solid sphere (rising/parabolic inside) and a thin shell (zero field, constant potential inside); both behave as a point mass outside — Solid sphere (red) vs thin shell (blue) — identical outside.

🚫 Examiner Trap · Potential

(1) V is a

scalar

and always

negative

(zero only at

\infty

) — never add it like a vector. (2) Inside a shell V is

constant = - GM / R

(NOT zero) even though

E_{g} = 0

. (3) Surface potential of a solid sphere is

- GM / R

but the

centre

1.5 \times

deeper,

- 3 GM /2 R

. (4)

E_{g} = - d V / d r

— a flat V means

E_{g} = 0

, not the reverse.

Potential energy

Two point masses

U = - \frac{G m _{1} m _{2}}{r}

U =

interaction PE (J); system PE

=

sum over all pairs (superposition); for a mass in a field

U = mV

Mass

m

with Earth

U = - \frac{GM m}{r} h ≪ R Δ U = mgh

r =

centre distance; mgh is the

small-height

approximation of the true PE difference, valid only for

h ≪ R

Self-energy (assembling a body)

U_{s h e ll} = - \frac{G M ^{2}}{2 R}; U_{so l i d} = - \frac{3 G M ^{2}}{5 R}

work done by gravity in assembling the mass from

\infty

; solid sphere is more tightly bound (more negative)

🎯 Exam · Field

\leftrightarrow

potential

Field can be zero where potential is not (inside a shell:

E_{g} = 0

but

V = - GM / R \neq = 0

). The field points toward

decreasing

potential.

⚡ Shortcut ·

Δ U

to height h

Exact rise:

Δ U = \frac{m g h}{1 + h / R}

. Reduces to mgh for

h ≪ R

; use the full form whenever h is comparable to R (e.g. raising a satellite).

Escape Velocity & Projectile Paths

Escape velocity

v_{e} = \frac{2 GM}{R} = 2 g R = R \frac{8}{3} π Gρ

v_{e} =

min surface speed to reach

\infty

with

v = 0

;

\approx 11.2 km/s

for Earth,

2.3 km/s

for the Moon

▸Independent of the $mass$ of the body and of the $direction$ of projection.
▸ $v_{e} = 2 v_{o}$ — escape speed is $2 \times$ the surface orbital speed.
▸Extra speed to escape $from orbit$ : $Δ v = (2 - 1) v_{o} \approx 0.414 v_{o}$ .
▸The Moon retains no atmosphere because its low $v_{e}$ lets gas molecules escape.

Comparative: orbital vs escape velocity

	Orbital $v_{o}$	Escape $v_{e}$
Surface formula	$g R = GM / R$	$2 g R = 2 GM / R$
Earth value	$\approx 7.9$ km/s	$\approx 11.2$ km/s
Relation	$v_{o}$	$v_{e} = 2 v_{o}$
Total energy	$- GM m /2 r < 0$	$= 0$ (just escapes)

★ Remember

A bound body has

total energy E < 0

. It just escapes when

E \geq 0

(minimum at

E = 0

, reaching infinity with zero speed).

Speed of projection

\to

path

Projection speed $v$	Resulting path
$v < v_{o}$	ellipse (falls back / decays)
$v = v_{o} = GM / r$	circle
$v_{o} < v < v_{e}$	ellipse (larger orbit)
$v = v_{e} = 2 v_{o}$	parabola (just escapes)
$v > v_{e}$	hyperbola (escapes with surplus KE)

💡 Tip · Perigee vs apogee

For a horizontal projection at distance r, the launch point is the

apogee

when

v < v_{o}

(body curves inward) and the

perigee

when

v_{o} < v < v_{e}

(body moves outward).

🚫 Examiner Trap · Escape & paths

(1)

v_{e}

is independent of mass and launch

direction

— angle does not matter (no air). (2)

v_{e} = 2 v_{o}

ONLY at the same radius; do not mix surface

v_{e}

with a high orbit's

v_{o}

. (3)

v = v_{e}

gives a

parabola

= 0

), not a straight line;

v > v_{e}

is a hyperbola. (4) Escaping from a circular orbit needs only

0.414 v_{o}

extra, not a full

v_{e}

Kepler's Laws of Planetary Motion

The three laws

▸ $1 (orbits)$ : every planet moves in an ellipse with the Sun at one focus.
▸ $2 (areas)$ : the radius vector sweeps equal areas in equal times — areal velocity $\frac{d A}{d t} = \frac{L}{2 m} =$ const (fastest at perihelion).
▸ $3 (periods)$ : $T^{2} \propto a^{3}$ , i.e. $T^{2} = \frac{4 π ^{2} a ^{3}}{G M _{s}}$ , with $a =$ semi-major axis.

Areal velocity (law of areas)

\frac{d A}{d t} = \frac{∣ r \times v ∣}{2} = \frac{L}{2 m} = constant

L =

angular momentum,

m =

planet mass; constant because gravity is a central force (no torque)

Perihelion / aphelion speeds

v_{p} r_{p} = v_{a} r_{a}

r_{p}, r_{a} =

nearest/farthest distances; from angular-momentum conservation,

v_{p} > v_{a}

since

r_{p} < r_{a}

⚡ Shortcut · Period ratio

For two orbits,

\frac{T _{1}^{2}}{T _{2}^{2}} = \frac{a _{1}^{3}}{a _{2}^{3}}

— only the semi-major axis a matters, NOT eccentricity. A circle and a thin ellipse of the same a have the same T.

Elliptical orbit with the Sun at one focus; velocity tangent and maximum at perihelion, minimum at aphelion; equal areas swept in equal times — Areal velocity constant: fast near the Sun, slow far away.

Consequences

▸Gravity is a $central force$ , so angular momentum $L$ is conserved and the motion is planar.
▸Kepler's 3rd law applies to Earth satellites too: $T^{2} = \frac{4 π ^{2}}{G M _{E}} (R_{E} + h)^{3}$ .

Binary star

ω = \frac{G ( m _{1} + m _{2} )}{r ^{3}}; \frac{v _{1}}{v _{2}} = \frac{r _{1}}{r _{2}} = \frac{m _{2}}{m _{1}}

r = r_{1} + r_{2} =

separation; both stars share one period about the common centre of mass; heavier star orbits closer & slower

🚫 Examiner Trap · Kepler's laws

(1) The Sun is at a

focus

, not the centre of the ellipse. (2) Equal

areas

(not equal arcs / equal angles) in equal times — speed is highest at perihelion. (3) In

T^{2} \propto a^{3}

, a is the

semi-major axis

, not the radius or the focal distance. (4) Areal velocity is constant; linear speed v is NOT — it varies along the orbit.

Satellites — Orbital Motion & Energy

Orbital velocity

v_{o} = \frac{GM}{r} = R \frac{g}{r}

r =

orbit radius (from Earth's centre),

R =

Earth radius; near the surface

v_{o} = g R \approx 7.9 km/s

(first cosmic speed)

Time period

T = 2 π \frac{r ^{3}}{GM}

T \propto r^{3/2}

(Kepler 3); near-Earth

T = 2 π R / g \approx 84.6 min

Earth with a satellite on a circular orbit showing tangential orbital velocity v_o, and escape velocity v_e directed outward — Orbital speed $v_{o}$ (tangential) vs escape speed $v_{e}$ .

⚡ Shortcut · Higher orbit = slower

r ↑

v_{o} ↓

(as

r^{- 1/2}

T ↑

(as

r^{3/2}

), KE

↓

U ↑

(less negative), but total

E ↑

(less negative). A satellite slows as it climbs.

Energy of a satellite

K, U, E in a circular orbit

K = \frac{GM m}{2 r}; U = - \frac{GM m}{r}; E = - \frac{GM m}{2 r}

m =

satellite mass;

∣ E ∣ = K = \frac{1}{2} ∣ U ∣

; binding energy (to free it)

= \frac{GM m}{2 r}

Energy vs orbital radius: kinetic energy positive and decreasing, potential energy negative, total energy negative and half the potential energy magnitude — KE positive, total energy negative for any bound orbit.

Energy to raise orbit

Δ E = \frac{GM m}{2} (\frac{1}{r _{1}} - \frac{1}{r _{2}})

r_{1} \to r_{2}

(

r_{2} > r_{1}

): positive energy supplied to move to a higher orbit

🚫 Examiner Trap · Satellite energy

(1) Total energy is

negative

(

E = - GMm /2 r

); a positive E means the body is not bound. (2)

U = 2 E

and

K = - E = ∣ E ∣

— note

U = - 2 K

(virial). (3) Orbital

v_{o}

is independent of the satellite's

mass

, but its energy is not. (4) To raise an orbit you ADD energy, yet the satellite ends up

slower

(KE drops) — the drag/firing paradox.

Geostationary & Polar Satellites · Special Topics

Geostationary satellite

▸ $T = 24$ h, in the $equatorial$ plane, moving W $\to$ E (synchronous with Earth's spin).
▸Height $h \approx 36, 000$ km; orbital radius $r \approx 42, 000$ km.
▸Appears fixed in the sky — used for telecommunication and broadcasting (e.g. INSAT).

Geostationary radius

R_{E} + h = (\frac{T ^{2} G M _{E}}{4 π ^{2}})^{1/3}

T = 86400

s (one sidereal day); fixes a unique radius — every geostationary satellite shares the same orbit

Polar (sun-synchronous) satellite

▸Low altitude $h \approx 500$ – $800$ km; period $\approx 100$ min in a N $\to$ S orbit.
▸Scans the whole Earth strip-by-strip — used for remote sensing, weather and surveillance.

Comparative: geostationary vs polar

Feature	Geostationary	Polar
Period $T$	$24$ h	$\approx 100$ min
Height $h$	$\approx 36, 000$ km	$500$ – $800$ km
Plane	equatorial	passes both poles
Use	telecom / broadcast	remote sensing / weather

🎯 Exam · Weightlessness

In an orbiting satellite apparent weight is zero: satellite and astronaut are in

free fall

with the same acceleration g' toward Earth, so the normal force

N = 0

Tunnel through the Earth

Tunnel SHM

F = - \frac{GM m}{R ^{3}} x \Rightarrow T = 2 π \frac{R ^{3}}{GM} = 2 π \frac{R}{g}

x =

displacement from centre;

T \approx 84.6

min — equal to the near-Earth orbital period

💡 Tip · Cavity field

Inside a spherical cavity in a uniform solid sphere the field is

uniform

E_{g} = \frac{4}{3} π Gρ d

, directed along the line joining the two centres (

d =

centre-to-centre vector).

🚫 Examiner Trap · Special topics

(1) Weightlessness in orbit is

free fall

with

N = 0

, NOT zero gravity — g' is still large at that height. (2) A geostationary satellite MUST be equatorial; a 24-h inclined orbit only looks fixed from the poles, not from the equator. (3) Tunnel motion is SHM only for a

uniform

Earth and a

straight

tunnel; period

84.6

min is the same for any chord, not just a diameter. (4) Polar

\neq =

geostationary — polar satellites move relative to the ground.

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What are the most important Gravitation formulas for JEE Main?

This Gravitation formula sheet covers all the high-yield Physics formulas, definitions and theorems you need for JEE Main, across The universal law of gravitation, Acceleration due to gravity and its variation with altitude and depth, Gravitational potential energy, Gravitational potential, Escape velocity — each shown with the key result and, where useful, a worked example.

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