Physics Gravitation

Gravitation Formula Sheet

All key Gravitation formulas for JEE Physics: Newton's law, g variation, field, potential, escape & orbital velocity, satellites, Kepler's laws. Quick revision for JEE Main & Advanced.

7 min read Updated Jun 2026 #formula sheet#quick revision#jee-main

Every formula, key relation, and must-know result from the Gravitation chapter, grouped by sub-topic for last-minute revision. Use it to scan before the exam, not to learn for the first time.

Five things JEE asks again and again

$g = \frac{GM}{R^2}$ and its variation with height/depth/latitude

Escape velocity $v_e = \sqrt{2gR} = \sqrt{2}\,v_o$

Satellite energy ratio $KE : |PE| : |E| = 1 : 2 : 1$

Kepler’s third law $T^2 \propto a^3$

Potential/PE are scalars and always negative (reference at infinity)

Universal Law of Gravitation

$$\boxed{F = G\frac{m_1 m_2}{r^2}}$$

Vector form (attractive, along line of centres):

$$\vec{F}_{12} = -G\frac{m_1 m_2}{r^2}\hat{r}_{12}, \qquad \vec{F}_{12} = -\vec{F}_{21}$$

Superposition for several masses:

$$\vec{F}_{net} = \vec{F}_1 + \vec{F}_2 + \vec{F}_3 + \dots$$
QuantityFormula / ValueNotes
Force between two masses$F = G\dfrac{m_1 m_2}{r^2}$$r$ = centre-to-centre distance
Dimensions of $G$$[G] = M^{-1}L^3T^{-2}$i.e. $\text{m}^3\,\text{kg}^{-1}\,\text{s}^{-2}$
Acceleration of a mass$a = \dfrac{F}{m} = G\dfrac{M}{r^2}$independent of $m$
Distance doubled$F \to F/4$inverse-square law
Both masses doubled$F \to 4F$
High-yield reminder

$G = 6.67 \times 10^{-11}$ N·m²/kg² is universal (same everywhere). $g$ is local and varies. Never mix them. Always use $r = R + h$, not $h$, for distance from Earth’s centre.

Acceleration Due to Gravity (g) and Its Variation

Surface value (from $mg = GMm/R^2$):

$$\boxed{g = \frac{GM}{R^2}} \qquad (GM = gR^2)$$
SituationFormulaNotes
At height $h$ (general)$g_h = g\left(\dfrac{R}{R+h}\right)^2$inverse-square
At height $h$ ($h \ll R$)$g_h \approx g\left(1 - \dfrac{2h}{R}\right)$note factor of 2; valid $h < 0.1R$
At depth $d$$g_d = g\left(1 - \dfrac{d}{R}\right)$linear; assumes uniform density
At Earth’s centre$g = 0$all forces cancel
Due to rotation, latitude $\lambda$$g' = g - \omega^2 R\cos^2\lambda$$\omega = 2\pi/T$
At poles ($\lambda = 90^\circ$)$g' = g$maximum
At equator ($\lambda = 0^\circ$)$g' = g - \omega^2 R$minimum

Headline forms:

$$\boxed{g_h = g\left(\frac{R}{R+h}\right)^2 \approx g\left(1 - \frac{2h}{R}\right)} \qquad \boxed{g_d = g\left(1 - \frac{d}{R}\right)}$$

Combined height + latitude (small $h$):

$$g' \approx g\left(1 - \frac{2h}{R}\right) - \omega^2 R\cos^2\lambda$$
Same distance, height vs depth

For equal $h = d$, height reduces $g$ more than depth: $g_h < g_d$ (decrease is $2h/R$ vs $d/R$). Weight reduction always equals $g$ reduction since $W = mg$ with $m$ constant.

Gravitational Field, Potential, and Potential Energy

$$\boxed{\vec{E} = \frac{\vec{F}}{m} = -\frac{GM}{r^2}\hat{r}} \qquad \boxed{V = -\frac{GM}{r}} \qquad \boxed{U = -\frac{GMm}{r}}$$
QuantityFormulaUnitsScalar / Vector
Force$F = \dfrac{GMm}{r^2}$NVector (attractive)
Field intensity$E = \dfrac{GM}{r^2}$N/kg = m/s²Vector (toward $M$)
Potential$V = -\dfrac{GM}{r}$J/kgScalar (negative)
Potential energy$U = -\dfrac{GMm}{r}$JScalar (negative)

Key relations:

$$E = \frac{F}{m}, \qquad V = \frac{U}{m}, \qquad E = -\frac{dV}{dr}, \qquad F = -\frac{dU}{dr}$$

Superposition (potential is scalar — add algebraically):

$$V_{total} = -G\left(\frac{M_1}{r_1} + \frac{M_2}{r_2} + \frac{M_3}{r_3}\right)$$

Values for Earth

QuantityAt surface ($r = R$)At height $h$ ($r = R+h$)
Potential$V = -\dfrac{GM}{R} = -gR$$V_h = -\dfrac{GM}{R+h}$
Potential energy$U = -\dfrac{GMm}{R} = -mgR$$U_h = -\dfrac{GMm}{R+h}$

Change in PE moving from surface to height $h$:

$$\boxed{\Delta U = \frac{GMmh}{R(R+h)}} \xrightarrow{h \ll R} \Delta U \approx mgh$$

Work done against gravity (move $m$ from $r_1$ to $r_2$):

$$\boxed{W = m(V_2 - V_1) = GMm\left(\frac{1}{r_1} - \frac{1}{r_2}\right)}$$

Surface to infinity (binding energy of mass on surface):

$$W = \frac{GMm}{R}$$

Self-energy of a uniform solid sphere:

$$\boxed{U = -\frac{3GM^2}{5R}}$$
Power-of-r and sign pattern

$F, E \propto r^{-2}$; $V, U \propto r^{-1}$. Both $V$ and $U$ are negative (reference at infinity). Differentiating $-1/r$ gives $+1/r^2$, which is how $E = -dV/dr$ and $F = -dU/dr$ work out.

Escape Velocity

$$\boxed{v_e = \sqrt{\frac{2GM}{R}} = \sqrt{2gR}}$$

Derived from total mechanical energy $= 0$ at the surface (just reaches infinity with $v = 0$).

QuantityFormulaValue for Earth
Escape velocity$v_e = \sqrt{2gR}$$\approx 11.2$ km/s
From height $h$$v_e(h) = \sqrt{\dfrac{2GM}{R+h}} = v_e\sqrt{\dfrac{R}{R+h}}$decreases with height
Relation to orbital$v_e = \sqrt{2}\,v_o \approx 1.414\,v_o$
Binding energy (surface)$E = \dfrac{GMm}{R} = mgR = \tfrac{1}{2}mv_e^2$$= mgR$
Proportionality$v_e \propto \sqrt{\dfrac{M}{R}} \propto \sqrt{gR}$mass of object independent

Energy classification of trajectories:

  • $E < 0$ → bound orbit
  • $E = 0$ → parabolic escape (minimum escape, $v_\infty = 0$)
  • $E > 0$ → hyperbolic escape ($v_\infty > 0$)

Speed at infinity when projected at $v = k\,v_e$ from the surface:

$$\boxed{v_\infty = v_e\sqrt{k^2 - 1}} \qquad (\text{generally } v_\infty = \sqrt{v^2 - v_e^2})$$

Atmospheric retention condition (gas RMS speed $v_{rms} = \sqrt{3kT/m}$):

$$v_{rms} < \frac{v_e}{6}$$
Escape velocity essentials

Independent of the object’s mass and direction of projection — depends only on the planet’s $M$ and $R$. The energy needed ($\tfrac{1}{2}mv_e^2$) does depend on mass.

Body$v_e$ (km/s)
Earth11.2
Moon2.4
Mars5.0
Jupiter60
Sun618

Satellite Motion

Gravitational force provides centripetal force: $\dfrac{GMm}{r^2} = \dfrac{mv^2}{r}$.

$$\boxed{v_o = \sqrt{\frac{GM}{r}} = \sqrt{gR}\cdot\sqrt{\frac{R}{r}}} \qquad \boxed{T = 2\pi\sqrt{\frac{r^3}{GM}} = 2\pi\sqrt{\frac{r^3}{gR^2}}}$$
ParameterFormulaVariation with $r$Value (LEO, $h \approx 0$)
Orbital velocity$v_o = \sqrt{\dfrac{GM}{r}}$$v \propto r^{-1/2}$$\approx 7.9$ km/s
Time period$T = 2\pi\sqrt{\dfrac{r^3}{GM}}$$T \propto r^{3/2}$$\approx 84$ min
Kinetic energy$KE = \dfrac{GMm}{2r}$$\propto r^{-1}$$\tfrac{1}{2}mgR$
Potential energy$PE = -\dfrac{GMm}{r}$$\propto r^{-1}$$-mgR$
Total energy$E = -\dfrac{GMm}{2r}$$\propto r^{-1}$$-\tfrac{1}{2}mgR$

Energy relations (circular orbit):

$$\boxed{E = -\frac{GMm}{2r} = \frac{PE}{2} = -KE} \qquad |PE| = 2\,KE, \quad |E| = KE$$$$\boxed{KE : |PE| : |E| = 1 : 2 : 1}$$

Period at surface ($r = R$): $T_0 = 2\pi\sqrt{\dfrac{R}{g}} \approx 84$ minutes (minimum possible period; same as a body dropped through a tunnel along Earth’s diameter).

Energy to launch from surface to circular orbit at height $h$:

$$\boxed{\Delta E = GMm\left(\frac{1}{R} - \frac{1}{2(R+h)}\right) = mgR\left(1 - \frac{R}{2(R+h)}\right)}$$

For low orbit ($h \ll R$): $\Delta E \approx \dfrac{mgR}{2}$ (half the escape energy).

Counterintuitive but examined

Higher orbit → slower speed, longer period, but larger total energy (less negative). Weightlessness in orbit is free fall, not zero gravity — at ISS height ($\approx 400$ km), $g$ is still about 90% of the surface value.

Types of Satellites

TypeHeightPeriodNotes
Low Earth Orbit (LEO)200–2000 km~90–120 minISS, Hubble; atmospheric drag
Medium Earth Orbit (MEO)2,000–35,786 km2–12 hGPS at ~20,200 km, 12 h
Geostationary (GEO)$\approx 35{,}786$ kmexactly 24 hequatorial, west→east, appears fixed
Polar~500–800 km~100 mincovers whole Earth, mapping/weather
Geostationary conditions

Must satisfy all three: period = 24 h, equatorial plane (inclination 0°), and west-to-east motion. Geosynchronous needs only the 24 h period — all geostationary are geosynchronous, not vice-versa.

Kepler’s Laws of Planetary Motion

graph LR
    K[Kepler's Laws] --> L1["1st: Ellipse — Sun at a focus"]
    K --> L2["2nd: Equal areas in equal times → L conserved"]
    K --> L3["3rd: T² ∝ a³"]

First Law (Orbits): planets move in ellipses with the Sun at one focus.

$$r = \frac{a(1-e^2)}{1 + e\cos\theta}, \qquad e = \sqrt{1 - \frac{b^2}{a^2}}, \qquad r_1 + r_2 = 2a$$
PointDistance from Sun
Perihelion (closest)$a(1-e)$
Aphelion (farthest)$a(1+e)$

Eccentricity cases: circle $e = 0$; ellipse $0 < e < 1$; parabola $e = 1$; hyperbola $e > 1$ (Earth’s orbit $e = 0.017$).

Second Law (Areas): equal areas swept in equal times.

$$\boxed{\frac{dA}{dt} = \frac{1}{2}r^2\omega = \frac{L}{2m} = \text{constant}}$$

This is conservation of angular momentum (gravity is central → zero torque), giving:

$$\boxed{r_p v_p = r_a v_a} \quad\Rightarrow\quad v_p > v_a \ (\text{faster at perihelion})$$

Third Law (Periods):

$$\boxed{T^2 \propto a^3} \qquad \boxed{\frac{T^2}{a^3} = \frac{4\pi^2}{GM} = \text{constant}}$$

For circular orbits $a = r$, so $T^2 = \dfrac{4\pi^2}{GM}r^3$. Mass of central body from an orbiting object:

$$\boxed{M = \frac{4\pi^2 a^3}{GT^2}}$$
QuantityProportionality
Period$T \propto r^{3/2}$
Orbital speed$v \propto r^{-1/2}$
Angular velocity$\omega \propto r^{-3/2}$
Kepler quick recall

“Two squared, three cubed”: $T^2 \propto a^3$. Doubling $r$ → $T \to 2^{3/2}T \approx 2.83T$. Doubling $T$ → $r \to 2^{2/3}r \approx 1.59r$. The constant depends only on the central mass — different for Sun, Earth, Mars.

Standard Constants and Reference Values

QuantityValue
Universal gravitational constant $G$$6.67 \times 10^{-11}$ N·m²/kg²
$g$ at Earth’s surface$9.8$ m/s² (use 10 for quick work)
Earth’s mass $M_E$$\approx 6 \times 10^{24}$ kg
Earth’s radius $R_E$$\approx 6.4 \times 10^6$ m
$\omega^2 R$ (rotation term)$\approx 0.034$ m/s²
Escape velocity (Earth)$11.2$ km/s
Orbital velocity (surface)$7.9$ km/s
Surface-orbit / tunnel period$\approx 84$ minutes
Geostationary height$\approx 35{,}786$ km

One-Line Connectors

  • $GM = gR^2$ — the bridge between $G$-form and $g$-form of every formula.
  • $v_e = \sqrt{2}\,v_o$ — escape vs orbital velocity at the same point.
  • $E = -KE = PE/2$ — satellite energy shortcut ($1:2:1$).
  • $r_p v_p = r_a v_a$ — angular-momentum bridge for elliptical orbits.
  • $T^2 = \dfrac{4\pi^2}{GM}a^3$ — Kepler’s third law, same form for planets and satellites.