Physics — Complete Formula Sheet

$F$: Gravitational force (N) · $G$: Universal gravitational constant = 6.67 × 10⁻¹¹ N·m²/kg² · $m_1,m_2$: Masses of the two bodies (kg) · $r$: Distance between their centres (m)

Every mass attracts every other mass along the line joining their centres. Attractive only — no repulsion. Inverse-square law.

$g$: Acceleration due to gravity (m/s²) · $M$: Mass of Earth ≈ 6 × 10²⁴ kg · $R$: Radius of Earth ≈ 6.4 × 10⁶ m

Acceleration experienced by a freely falling body near Earth's surface. On Earth, g ≈ 9.8 m/s².

$W$: Weight (N) · $m$: Mass (kg) — scalar, constant everywhere · $g$: Local acceleration due to gravity (m/s²)

Weight = gravitational force on the body. Unit: newton (N). Weight varies with location (g varies); mass does not.

g decreases with altitude. At h = R (one Earth radius up), g becomes g/4.

$v_{o}rb$: Orbital velocity (m/s) · $r$: Orbital radius from Earth's centre (m)

Minimum tangential speed to keep a satellite in a circular orbit of radius r from Earth's centre.

$v_e$: Escape velocity (m/s) · $R$: Radius of Earth (m)

Minimum speed to escape Earth's gravity forever. For Earth: v_e ≈ 11.2 km/s. Note: v_e = √2 × v_orb (at same r).

Law of orbits. Orbits are ellipses, not circles.

Line joining planet to Sun sweeps equal AREAS in equal times ⇒ planet moves faster when nearer the Sun. Consequence of conservation of angular momentum.

Square of orbital period is proportional to the cube of the mean orbital radius. Used by Newton to guess the inverse-square law.

$F$: Net force (N) · $m$: Mass of the body (kg) · $a$: Acceleration (m/s²)

Net force = mass × acceleration. Unit of F: newton (N) = kg·m/s².

Vector quantity. Direction same as velocity. Unit: kg·m/s.

Impulse = force × time = change in momentum. Same units as momentum.

Total momentum of an isolated system is conserved (no external force).

Weight is the gravitational force. Direction: towards centre of Earth.

Gun of mass M fires bullet of mass m with velocity v; minus sign shows recoil is opposite.

+ when lift accelerates up, − when it accelerates down. Free fall ⇒ R = 0 (weightless).

Scalar quantity. Direction not considered.

Vector quantity. For uniform acceleration, average velocity = (u + v)/2.

Rate of change of velocity. Unit: m/s². Deceleration (retardation) is negative acceleration.

Velocity after time t, starting from initial velocity u with constant acceleration a.

Displacement in time t under constant acceleration.

Useful when time t is not given. Solve for v or s.

Distance travelled in the nth second (not first n seconds).

Speed constant, direction changes ⇒ velocity changes ⇒ acceleration exists (centripetal).

$W$: Work done (J) · $F$: Applied force (N) · $s$: Displacement (m) · $\theta$: Angle between force and displacement

θ is angle between force and displacement. When θ = 90°, W = 0.

Energy due to motion. Scalar. Always positive.

Energy due to position above a reference. Valid near Earth's surface where g is constant.

Net work done on a body equals its change in kinetic energy.

Rate of doing work. Unit: watt (W) = 1 J/s.

Useful when force and velocity are constant along the same line.

In absence of friction, total mechanical energy is conserved.

Electricity bills use kWh (1 unit). Convert to joules for SI calculations.

Rate of flow of charge. Unit: ampere (A) = 1 C/s.

$V$: Potential difference (V) · $I$: Current (A) · $R$: Resistance (Ω)

For a metallic conductor at constant temperature, V ∝ I.

$\rho$: Resistivity (Ω·m) · $L$: Length of conductor (m) · $A$: Cross-section area (m²)

R depends on length, area and material. ρ is a material constant.

Same current through each. Voltages add. Equivalent resistance is larger than largest.

Same voltage across each. Currents add. Equivalent resistance is smaller than smallest.

Three equivalent forms. Choose the one using known variables.

Heat produced when current I flows for time t through R. Unit: joule.

1 unit of electricity = 1 kWh. 1 kWh = 3.6×10⁶ J.

1 calorie is the heat needed to raise the temperature of 1 g of water by 1 °C (from 14.5 °C to 15.5 °C). The SI unit of heat is the joule (J).

$Q$: Heat absorbed or given out (cal or J) · $m$: Mass (g or kg) · $c$: Specific heat capacity (cal/g °C or J/kg °C) · $\Delta T$: Change in temperature (°C)

Heat needed to change the temperature of a body without a change of state. ΔT is the change in temperature (°C or K — same number).

$c$: Specific heat capacity (cal/g °C; SI: J/kg °C)

Heat needed to raise the temperature of unit mass of a substance by 1 °C. Water has c = 1 cal/g °C, one of the highest of common substances — which is why it is used in hot-water bags and as a coolant.

Holds when the two bodies are in contact, no heat is lost to the surroundings, and heat flows until both reach the same temperature.

$m_1,c_1,T_1$: Mass, specific heat and initial temperature of the hot body · $m_2,c_2,T_2$: Mass, specific heat and initial temperature of the cold body · $T$: Final common temperature (°C)

Hot body at T₁, cold body at T₂, common final temperature T. Write each bracket as (higher − lower) so that both sides are positive.

Rearranged form of the heat exchange equation. Valid only if neither body changes state. For two samples of water, c cancels: T = (m₁T₁ + m₂T₂)/(m₁ + m₂).

$m,c,T_1$: Mass, specific heat and initial temperature of the hot solid · $m_w,c_w$: Mass and specific heat of water · $m_c,c_c$: Mass and specific heat of the calorimeter (with stirrer) · $T_2,T$: Initial temperature of water and calorimeter; final temperature

To find the specific heat c of a solid: the hot solid gives heat to both the water and the calorimeter. Ignore the calorimeter term only if the question says so.

$Q$: Heat absorbed during melting/boiling or released during freezing/condensation (cal or J) · $m$: Mass that changes state (g or kg) · $L$: Specific latent heat (cal/g or J/kg)

Heat absorbed or given out during a change of state at constant temperature. Specific latent heat L is the heat needed to change the state of unit mass.

In SI units the textbook table gives 333 kJ/kg (3.33 × 10⁵ J/kg) and 2256 kJ/kg (2.256 × 10⁶ J/kg). The same heat is released when water freezes at 0 °C or steam condenses at 100 °C.

Ice below 0 °C to steam at 100 °C: warm the ice, melt it, warm the water, boil it. Add the heat for each stage separately. c of ice is not in the textbook table; use the value given in the question (typically 0.5 cal/g °C).

From 0 °C to 4 °C water CONTRACTS on heating; above 4 °C it expands like other liquids. So water at 4 °C sinks to the bottom of lakes and aquatic life survives under the ice.

Mass of water vapour present in unit volume of air. Unit: kg/m³ (often quoted in g/m³).

Compare at the SAME temperature. RH has no unit; it is a percentage. When the air is cooled to its dew point, it becomes saturated and RH = 100%.

The temperature at which the water vapour in air becomes saturated. Cooling below it makes vapour condense as dew, fog or mist.

Angle of incidence = angle of reflection. Both measured from the normal.

u = object distance, v = image distance, f = focal length. Use sign convention.

Negative m ⇒ inverted (real); positive ⇒ upright (virtual).

For spherical mirrors, focal length is half the radius of curvature.

Refractive index of medium 2 with respect to medium 1.

$c$: Speed of light in vacuum ≈ 3×10⁸ m/s · $v$: Speed of light in the medium (m/s)

Ratio of speed of light in vacuum to speed in the medium. Always ≥ 1.

Note the MINUS sign (unlike mirror formula). Applies to thin lenses.

Unit: dioptre (D). Convex lens ⇒ +P, concave lens ⇒ −P.

Positive for erect (virtual), negative for inverted (real).

Optical centre is the origin. Distances in the direction of incident light (left to right) are positive, against it negative. Heights above the principal axis are positive, below negative.

$u$: Object distance from the optical centre (cm or m) · $v$: Image distance from the optical centre (cm or m) · $f$: Focal length (cm or m)

Valid for both convex and concave lenses with Cartesian signs. Note the MINUS sign — this is not the mirror formula.

$M$: Magnification (no unit) · $h_1$: Height of the object (cm) · $h_2$: Height of the image (cm)

M negative ⇒ real and inverted image; M positive ⇒ virtual and erect. |M| > 1 magnified, |M| < 1 diminished. M has no unit.

$P$: Power (dioptre, D) · $f$: Focal length in metres

SI unit: dioptre (D); 1 D = 1 m⁻¹. Convex lens ⇒ positive power, concave lens ⇒ negative power. A shorter focal length means a more powerful lens.

Same formula, with the conversion built in. f = −40 cm ⇒ P = 100/(−40) = −2.5 D.

$f_1,f_2$: Focal lengths of the individual lenses · $f$: Equivalent focal length of the combination

For thin lenses kept touching each other. Use the sign of each focal length.

Powers add (with signs). Focal lengths do NOT add. +5 D and −2 D in contact give +3 D.

25 cm is the least distance of distinct vision. The eye adjusts the focal length of its lens (accommodation) using the ciliary muscles to see objects anywhere in this range.

$x$: Distance of the defective eye's far point

A myopic eye sees near objects but not distant ones. A concave lens forms a virtual image of a distant object at the far point x. From the lens formula with u = −∞, v = −x (spectacle lens taken to be at the eye; the textbook treats this qualitatively).

$N$: Near point of the defective eye (cm), N > 25 cm

A hypermetropic eye cannot see near objects clearly; its near point N is beyond 25 cm. A convex lens forms a virtual image, at N, of an object kept at 25 cm. Lens formula with u = −25 cm, v = −N (spectacle lens taken to be at the eye).

$m$: Mass of the satellite (kg) · $M$: Mass of the Earth = 6 × 10²⁴ kg · $R$: Radius of the Earth = 6.4 × 10⁶ m · $h$: Height of the satellite above the Earth's surface (m)

The Earth's gravitational force on the satellite provides exactly the centripetal force needed for circular motion at height h.

$v_c$: Critical (orbital) velocity (m/s) · $G$: Universal gravitational constant = 6.67 × 10⁻¹¹ N m²/kg² · $R+h$: Radius of the orbit, from the Earth's centre (m)

The tangential speed a satellite must have to move in a circular orbit at height h. It does not depend on the satellite's own mass, and it decreases as h increases.

Put h ≈ 0 and use g = GM/R². With g = 9.8 m/s² and R = 6.4 × 10⁶ m this is about 7.9 km/s.

$T$: Period of revolution (s)

Time for one revolution = circumference of the orbit ÷ speed. Gives T in seconds when lengths are in m and v_c in m/s.

Substitute v_c = √(GM/(R + h)) into T = 2π(R + h)/v_c. Higher orbit ⇒ longer period.

For satellites of the same planet, T²/r³ is the same. If r becomes 4 times, T becomes 4^(3/2) = 8 times.

A satellite in a higher orbit moves more slowly. If r becomes 9 times, v_c becomes one-third.

$v_{esc}$: Escape velocity (m/s) · $R$: Radius of the planet (m)

Minimum velocity with which an object must be projected from the surface to escape the planet's gravity. For the Earth, about 11.2 km/s.

11.2 km/s ≈ 1.414 × 7.9 km/s. A launch speed between 7.9 and 11.2 km/s cannot make the object escape the Earth.

Heights above the Earth's surface. Low Earth orbit: satellites for scientific experiments and atmospheric studies, the International Space Station and the Hubble telescope (one revolution in about 90 min). Medium: elliptical polar orbits for studying polar regions (period 2 to 24 h), and GPS satellites at about 20,200 km. High: geosynchronous satellites (period about 24 h), used for meteorology and for telephone, TV and radio signals.