1) Newton's Universal Law of Gravitation
Every particle in the universe attracts every other particle with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centres.
F = G (m₁ m₂) / r²
G = Universal gravitational constant = 6.67 × 10⁻¹¹ N m²/kg²
Vector Form
The gravitational force on m₁ due to m₂ is: F₁₂ = -G(m₁m₂/r²) r̂₂₁
where r̂₂₁ is the unit vector from m₂ to m₁. The negative sign indicates attraction.
Characteristics of Gravitational Force:
- Always attractive (never repulsive) — unlike electrostatic force which can be repulsive.
- Central force — acts along the line joining the centres of the two bodies.
- Follows inverse square law — F ∝ 1/r².
- Independent of the medium — unlike electrostatic force, it doesn't depend on the intervening medium.
- Weakest force in nature — for small masses, gravitational force is negligible. For example, the gravitational force between two humans is about 10⁻⁷ N — too small to feel!
- Gravitational force is significant only for astronomical bodies (planets, stars, galaxies).
- Long-range force — its range is infinite (theoretically).
(A) F = G m₁ m₂ / r⁴ (B) F = G m₁ m₂ / r² (C) F = G (m₁ + m₂) / r² (D) F = G (m₁ - m₂) / r²
Ans: B
🔥 AMU PYQ: The value of G was first experimentally determined by:
(A) Newton (B) Einstein (C) Cavendish (D) Kepler
Ans: C (Cavendish used a torsion balance in 1798)
🔥 AMU PYQ: Gravitational force between two bodies does NOT depend on:
(A) Product of masses (B) Distance between them (C) Medium between them (D) Universal constant G
Ans: C (gravitational force is independent of the medium)
Importance of the Universal Law of Gravitation
Newton's Universal Law of Gravitation was a revolutionary breakthrough because it unified terrestrial physics (objects falling on Earth) with celestial physics (motion of planets). Before Newton, people thought different laws governed the heavens and the Earth. Newton showed that the same force that makes an apple fall also keeps the Moon in orbit!
Cavendish Experiment — Measuring G
In 1798, Henry Cavendish used a torsion balance to measure the gravitational force between two lead spheres. From this, he calculated G ≈ 6.67 × 10⁻¹¹ N m²/kg². This is why Cavendish is credited with "weighing the Earth" — once G was known, M = gR²/G gave the mass of Earth.
A horizontal rod with small lead spheres is suspended by a thin wire.
Large lead spheres placed near small ones create gravitational attraction.
The rod rotates, twisting the wire. Angle of twist → force → G.
Why G is a Universal Constant
- G has the same value everywhere in the universe (hence "universal").
- G does NOT depend on medium, temperature, or any physical condition.
- G is a scalar quantity with unit N m²/kg².
- G is extremely small (6.67 × 10⁻¹¹), showing gravity is the weakest fundamental force.
2) Gravitational Field and Intensity
The gravitational field at a point is the gravitational force experienced by a unit mass placed at that point. It is a vector quantity.
E = F / m₀ = GM / r²
Direction: Towards the centre of the mass creating the field
SI unit: N/kg (same as m/s² — note that g is the gravitational field intensity!)
So g is actually the gravitational field intensity at Earth's surface. That's why g = 9.8 N/kg = 9.8 m/s².
3) Kepler's Laws of Planetary Motion
Johannes Kepler (1571-1630) gave 3 laws based on Tycho Brahe's precise observations of planetary motion. Newton later used these laws to derive his Universal Law of Gravitation.
Law 1 — The Law of Orbits (Elliptical Orbit Law)
Every planet moves around the Sun in an elliptical orbit, with the Sun at one of the two foci.
An ellipse has two foci. The Sun is at one focus, not the centre.
Eccentricity (e) = distance between foci / length of major axis
Earth's orbit: e ≈ 0.0167 (nearly circular)
Mercury's orbit: e ≈ 0.205 (most elliptical among planets)
Law 2 — The Law of Areas (Equal Area Law)
The line joining the planet to the Sun sweeps out equal areas in equal intervals of time.
Areal velocity = dA/dt = constant
Mathematically: dA/dt = L / 2m = constant
L = angular momentum, m = mass of planet
Consequence: When a planet is closer to the Sun (perihelion), it moves faster. When farther (aphelion), it moves slower. For Earth:
- Perihelion (early January): Earth is closest to Sun → moves fastest
- Aphelion (early July): Earth is farthest from Sun → moves slowest
(A) Conservation of energy (B) Conservation of angular momentum (C) Conservation of linear momentum (D) Conservation of mass
Ans: B (Conservation of angular momentum)
🔥 AMU PYQ: Kepler's second law implies that a planet moves ______ when it is closer to the Sun.
(A) slower (B) faster (C) at same speed (D) stops
Ans: B
Law 3 — The Law of Periods (Harmonic Law)
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 ∝ r³
T² = (4π² / GM) r³
Where: r = semi-major axis (average distance from Sun)
T = time period, M = mass of the central body (Sun)
T² ∝ r³ → T ∝ r^(3/2)
If r → 2r: T → 2^(3/2) = 2√2 ≈ 2.83 times
If r → 3r: T → 3^(3/2) = 3√3 ≈ 5.2 times
(A) r (B) r² (C) r³ (D) r⁴
Ans: C (r³)
🔥 AMU PYQ: If the distance of Jupiter from the Sun is 5.2 times that of Earth, find Jupiter's time period.
Ans: T_J² / T_E² = (r_J / r_E)³ = (5.2)³ = 140.6
T_J = √140.6 × 1 year = 11.86 years ≈ 12 years
4) Acceleration Due to Gravity (g)
The acceleration produced in a body due to the gravitational pull of Earth is called acceleration due to gravity (g).
Derivation of g = GM/R²
Also from Newton's 2nd law: F = mg
Equating: mg = G M m / R²
→ g = G M / R²
G = 6.67 × 10⁻¹¹ N m²/kg²
M = 6 × 10²⁴ kg (mass of Earth)
R = 6.4 × 10⁶ m (radius of Earth)
g = (6.67×10⁻¹¹ × 6×10²⁴) / (6.4×10⁶)²
g = (4.002×10¹⁴) / (4.096×10¹³)
g = 9.8 m/s²
Mass of Earth from g:
M = gR² / G
= (9.8 × (6.4×10⁶)²) / 6.67×10⁻¹¹
≈ 6 × 10²⁴ kg
5) Mass vs Weight
This is a very common AMU trick topic. Know the difference cold.
| Feature | Mass (m) | Weight (W) |
|---|---|---|
| Definition | Amount of matter in a body | Gravitational force on a body |
| Formula | — | W = mg |
| SI unit | kg (kilogram) | N (newton) |
| Nature | Scalar | Vector (towards Earth's centre) |
| Constant? | Yes (same everywhere in universe) | No (varies with location) |
| Can be zero? | No | Yes (free fall, space, centre of Earth) |
| Measuring device | Physical balance / beam balance | Spring balance |
| At Moon | Same as on Earth | ≈ 1/6 of Earth weight |
| At centre of Earth | Same | Zero |
If a person weighs 60 kg (mass), weight on Earth = 60×9.8 = 588 N
Weight on Moon = 60×1.62 = 97.2 N ≈ 1/6 of Earth weight
(A) 10 N (B) 100 N (C) 1 N (D) 1000 N
Ans: B (W = mg = 10 × 10 = 100 N)
🔥 AMU PYQ: Mass of a body on Moon is 60 kg. Its mass on Earth is:
(A) 10 kg (B) 60 kg (C) 360 kg (D) 6 kg
Ans: B (Mass is constant everywhere!)
6) Variation of g with Height, Depth, and Latitude
The value of g is not constant everywhere on or above Earth. It varies with:
6.1 Variation with Height (above Earth's surface)
g_h / g = R² / (R + h)²
Formula: g_h = g [R / (R + h)]²
For small h (h << R), using binomial approximation:
g_h = g (1 - 2h/R)As h increases, g decreases.
6.2 Variation with Depth (below Earth's surface)
Assuming uniform density:
Formula: g_d = g (1 - d/R)
At the centre of Earth (d = R): g = 0
As depth increases, g decreases linearly.
6.3 Variation with Latitude (Rotation of Earth)
geff = g - R ω² cos² φAt equator (φ = 0°): g is minimum (due to maximum centrifugal effect)
At poles (φ = 90°): g is maximum (no centrifugal effect)
Comparison Table — Variation of g:
| Location | Value of g | Reason |
|---|---|---|
| Surface of Earth | g = 9.8 m/s² | Standard value |
| Height h above surface | g_h = g [R/(R+h)]² | Distance from centre increases → g decreases |
| Depth d below surface | g_d = g (1 - d/R) | Mass below reduces → g decreases linearly |
| Centre of Earth | 0 | Equal gravitational pull from all sides cancels |
| Equator | ≈ 9.78 m/s² | Centrifugal effect + larger radius |
| Poles | ≈ 9.83 m/s² | No centrifugal effect + smaller radius |
| Mountains | Less than sea level | Greater distance from Earth's centre |
| Moon | ≈ 1.62 m/s² | Smaller mass and radius |
(A) Increases (B) Decreases (C) Remains same (D) First increases then decreases
Ans: B (Decreases)
🔥 AMU PYQ: At the centre of the Earth, the value of g is:
(A) 9.8 m/s² (B) 4.9 m/s² (C) 0 (D) Infinite
Ans: C (Zero)
🔥 AMU PYQ: Where is the value of g maximum on Earth's surface?
(A) Equator (B) Tropic of Cancer (C) Poles (D) At 45° latitude
Ans: C (Poles)
7) Gravitational Potential Energy (Basic Concept)
Gravitational potential energy (U) of a body at a point is the work done in bringing it from infinity to that point against the gravitational field. It is always negative — which means the force is attractive.
U = -G M m / r
At infinity: U = 0 (by convention)
Near Earth's surface: U = mgh (approximate for small h)
• U is negative because work is done by the gravitational field (attractive force).
• As a body falls towards Earth, its gravitational PE decreases (becomes more negative).
• The kinetic energy increases as PE decreases — conservation of mechanical energy!
8) Free Fall (Detailed)
When a body falls only under the influence of gravity (no air resistance), it is called free fall.
Key Points about Free Fall:
- During free fall, all bodies accelerate downward at the same rate g, regardless of their mass.
- This was famously demonstrated by Galileo's Leaning Tower of Pisa experiment (and later on the Moon by Apollo 15 astronaut David Scott — a feather and a hammer hit the ground at the same time!).
- In free fall, the body experiences weightlessness — its weight appears to be zero.
v = u + g t
h = u t + ½ g t²
v² = u² + 2 g h
Weightlessness
Weightlessness is the feeling of zero weight (g = 0). This happens when:
- A body is in free fall (falling freely under gravity).
- An astronaut in a satellite orbiting Earth (both satellite and astronaut are in free fall towards Earth).
- At the centre of the Earth (gravitational force cancels from all sides).
A feather and a hammer dropped on the Moon hit the ground at the same time because g is same for all bodies in vacuum. On Earth, air resistance makes the feather fall slower — but in vacuum, even on Earth, they'd hit together!
9) Escape Velocity
The minimum velocity required for a body to escape Earth's gravitational pull and never return is called escape velocity.
Derivation:
KE ≥ |PE|
½ m vₑ² ≥ G M m / R
vₑ² ≥ 2 G M / R
vₑ = √(2GM / R) = √(2gR)
vₑ = √(2GM/R) = √(2 × 6.67×10⁻¹¹ × 6×10²⁴ / 6.4×10⁶)
vₑ = √(1.25×10⁸) ≈ 1.12×10⁴ m/s
vₑ = 11.2 km/s
| Celestial Body | Escape Velocity |
|---|---|
| Earth | 11.2 km/s |
| Moon | 2.4 km/s |
| Sun | 618 km/s |
| Jupiter | 59.5 km/s |
| Mars | 5.0 km/s |
| Mercury | 4.3 km/s |
• Escape velocity does NOT depend on the mass of the escaping body!
• Escape velocity depends only on the mass and radius of the planet.
• If Earth's radius shrinks to half, escape velocity becomes √2 times = 15.8 km/s.
• Moon's escape velocity is only 2.4 km/s — that's why it can't hold an atmosphere!
10) Orbital Velocity and Satellites
Orbital Velocity (v₀)
The velocity required for a satellite to revolve around Earth in a circular orbit is called orbital velocity.
Centripetal force = Gravitational force
mv₀²/r = GMm/r²
v₀ = √(GM / r) = √(gR² / r)
For a satellite close to Earth (r ≈ R):
v₀ = √(gR) = √(9.8 × 6.4×10⁶) ≈ 7.9 km/s
Relation Between Escape Velocity and Orbital Velocity:
vₑ = √2 × v₀vₑ / v₀ = √2 ≈ 1.414Escape velocity is 41.4% more than orbital velocity.
Time Period of a Satellite:
T = 2πr / v₀ = 2π √(r³ / GM)
This is Kepler's 3rd law: T² ∝ r³
Types of Satellites:
| Type | Properties |
|---|---|
| Natural satellites | Moons of planets. Example: Moon (natural satellite of Earth). |
| Artificial satellites | Man-made. Used for communication, weather, navigation, research. |
Geostationary (Geosynchronous) Satellites:
- Orbit height ≈ 36,000 km above Earth's surface
- Time period = 24 hours (same as Earth's rotation period)
- Orbits in the equatorial plane, west to east
- Appears stationary from a fixed point on Earth
- Uses: Communication (TV, telephone), weather monitoring, broadcasting
Polar Satellites:
- Orbit at lower altitudes (500-800 km)
- Pass over the poles
- Time period ≈ 90 minutes
- Uses: Remote sensing, mapping, spying, environmental monitoring
(A) √(gR) (B) √(2gR) (C) √(GM/R²) (D) √(2GM/R)
Ans: A (√(gR) ≈ 7.9 km/s)
🔥 AMU PYQ: The time period of a geostationary satellite is:
(A) 12 hours (B) 24 hours (C) 48 hours (D) 1 hour
Ans: B
🔥 AMU PYQ: The ratio of escape velocity to orbital velocity for a satellite near Earth's surface is:
(A) √2 : 1 (B) 1 : √2 (C) 2 : 1 (D) 1 : 2
Ans: A (vₑ = √2 × v₀)
11) Thrust and Pressure (Brief)
Thrust is the force acting perpendicularly to a surface. Pressure is thrust per unit area.
Pressure (P) = Force (F) / Area (A)
SI unit: Pascal (Pa) = N/m²
In the context of gravitation, the weight of a body (thrust) acting on a surface produces pressure. This is why:
- A sharp needle pricks easily (small area → large pressure).
- A camel has broad feet (large area → less pressure → doesn't sink in sand).
- Building foundations are broad (large area → less pressure on ground).
12) Thrust and Pressure (Brief)
Thrust is the force acting perpendicularly to a surface. Pressure is thrust per unit area.
Pressure (P) = Force (F) / Area (A)
SI unit: Pascal (Pa) = N/m²
In the context of gravitation, the weight of a body (thrust) acting on a surface produces pressure. This is why:
- A sharp needle pricks easily (small area → large pressure).
- A camel has broad feet (large area → less pressure → doesn't sink in sand).
- Building foundations are broad (large area → less pressure on ground).
- A tractor has wide tires to reduce pressure on soft ground.
- School bags have broad straps to reduce pressure on shoulders.
(A) Force × Area (B) Force / Area (C) Area / Force (D) Mass × Acceleration
Ans: B (P = F/A)
13) Applications of Gravitation in Daily Life
- Tides in the ocean — caused by the gravitational pull of the Moon (and Sun) on Earth's water.
- Satellites in orbit — communication TV, weather forecasting, GPS navigation — all depend on gravitational forces keeping satellites in orbit.
- Rockets must achieve escape velocity (11.2 km/s) to overcome Earth's gravity.
- Weightlessness experienced by astronauts — they are in continuous free fall around Earth.
- Formation of stars and planets — gravitational collapse of gas clouds leads to star formation.
- Earth's atmosphere is held by gravity — without it, all air would escape into space!
14) Why the Moon has No Atmosphere
The Moon's escape velocity is only 2.4 km/s, which is very small compared to Earth's 11.2 km/s. The average speed of gas molecules at normal temperatures is around 1-2 km/s. Many gas molecules on the Moon can easily exceed 2.4 km/s and escape into space.
Key point: For a planet or moon to retain an atmosphere, its escape velocity must be at least 6 times the root-mean-square speed of gas molecules. Since the Moon's escape velocity is too low, it couldn't hold onto an atmosphere.
Ans: Because its escape velocity (2.4 km/s) is too low to retain gas molecules, which can easily escape into space.
15) Common Mistakes Students Make in Gravitation
Fix: G (universal constant, 6.67×10⁻¹¹) is the same everywhere. g (9.8 m/s² on Earth surface) varies with location.
⚠️ Mistake 2: Thinking mass changes with location.
Fix: Mass is constant everywhere. Weight changes because g changes.
⚠️ Mistake 3: Applying g = GM/R² at height without adjusting R.
Fix: At height h, distance from centre = R + h, so g_h = GM/(R+h)²
⚠️ Mistake 4: Forgetting that Kepler's 3rd law applies to any satellite, not just planets.
Fix: T² ∝ r³ works for any body orbiting a central mass (including Earth satellites).
⚠️ Mistake 5: Thinking escape velocity depends on the mass of the escaping object.
Fix: vₑ = √(2GM/R) — no m term! It's independent of the mass of the escaping body.
⚠️ Mistake 6: Confusing orbital velocity with escape velocity.
Fix: v₀ = √(gR) ≈ 7.9 km/s (orbital), vₑ = √(2gR) ≈ 11.2 km/s (escape). vₑ = √2 × v₀
⚠️ Mistake 7: Thinking g is zero inside Earth only at one point.
Fix: g decreases linearly with depth from surface to centre, where g = 0.
13) Important Formulae — At a Glance
| Quantity | Formula |
|---|---|
| Newton's law of gravitation | F = G m₁ m₂ / r² |
| Gravitational constant | G = 6.67 × 10⁻¹¹ N m²/kg² |
| Acceleration due to gravity | g = GM / R² = 9.8 m/s² |
| Mass of Earth | M = gR² / G ≈ 6 × 10²⁴ kg |
| Weight | W = mg |
| g at height h | g_h = g [R/(R+h)]² ≈ g(1 - 2h/R) for h << R |
| g at depth d | g_d = g (1 - d/R) |
| Kepler's 1st law | Elliptical orbits, Sun at one focus |
| Kepler's 2nd law | Areal velocity = constant (angular momentum conserved) |
| Kepler's 3rd law | T² ∝ r³, T² = (4π²/GM) r³ |
| Gravitational field | E = GM/r² = g |
| Gravitational PE | U = -GMm/r |
| Escape velocity | vₑ = √(2GM/R) = √(2gR) = 11.2 km/s |
| Orbital velocity | v₀ = √(GM/r) = √(gR²/r) |
| Near Earth orbital velocity | v₀ = √(gR) ≈ 7.9 km/s |
| vₑ : v₀ | vₑ = √2 × v₀ |
| Satellite time period | T = 2π √(r³/GM) |
| Geostationary satellite height | ≈ 36,000 km, T = 24 h |
| Pressure | P = F/A, SI unit: Pa (N/m²) |
14) PYQ Battle — 16 AMU Past Questions on Gravitation
(A) Newton's law (B) Faraday's law (C) Kepler's law (D) Kirchhoff's law
Ans: C
(A) Product of masses (B) Distance between them (C) Medium between them (D) G
Ans: C (gravitational force is independent of the medium)
(A) Half (B) Double (C) Four times (D) One-fourth
Ans: C (F ∝ 1/r², so halving r → 4× force)
Ans: g_h = g [R/(R+R/2)]² = g(R/(3R/2))² = g(2/3)² = (4/9)g
W' = (4/9)×63 = 28 N
Ans: g_d = g(1 - d/R) = g(1 - (R/2)/R) = g/2
W' = 250/2 = 125 N
(A) 7.9 km/s (B) 11.2 km/s (C) 3×10⁸ m/s (D) 5 km/s
Ans: B
(A) Same (B) Double (C) Half (D) Four times
Ans: A (if density same, vₑ ∝ R)
(A) √(gR) (B) √(2gR) (C) gR (D) R/√g
Ans: B
(A) Poles (B) Equator (C) Centre of Earth (D) Mountains
Ans: C (g = 0 at centre)
(A) 12 hours (B) 24 hours (C) 48 hours (D) 1 hour
Ans: B
(A) Energy (B) Linear momentum (C) Angular momentum (D) Mass
Ans: C
(A) 2g (B) g/2 (C) 4g (D) Same
Ans: D (g ∝ M/R² → 2M/(2R)² = 2M/4R² = M/2R² — wait, that's g/2. Let me recalculate: g₂ = G(2M)/(2R)² = 2GM/4R² = GM/2R² = g/2. So B)
(A) 7.9 km/s (B) 11.2 km/s (C) 3×10⁸ m/s (D) 9.8 m/s
Ans: A
Ans: v₀ = √(GM/(R+R)) = √(GM/2R) = √(gR/2) = √(9.8×6.4×10⁶/2) = √(31.36×10⁶) ≈ 5.6 km/s
Ans: ΔU = U₂ - U₁ = (-GMm/2R) - (-GMm/R) = GMm(-1/2R + 1/R) = GMm/2R
(A) 100 N (B) 33.3 N (C) 11.1 N (D) 900 N
Ans: C (F ∝ 1/r², triple r → 1/9 × 100 = 11.1 N)
15) Practice Exercise — 15 Problems
Try these. Cover the answers — honest practice is the key.
Answer
g' = G(4M)/(2R)² = 4GM/4R² = GM/R² = g. So g' = 9.8 m/s² (same as Earth)Answer
F = 6.67×10⁻¹¹ × 1 × 1 / 1² = 6.67×10⁻¹¹ N (extremely small!)Answer
vₑ = √(2GM/R) = √(2×6.67×10⁻¹¹×6×10²⁴/6.4×10⁶) = √(1.25×10⁸) = 1.12×10⁴ m/s = 11.2 km/sAnswer
g/2 = g[R/(R+h)]² → 1/2 = R²/(R+h)² → (R+h)/R = √2 → h = (√2 - 1)R = 0.414R ≈ 2650 kmAnswer
T/T₀ = (r/r₀)^(3/2) = ((R+2R)/R)^(3/2) = (3)^(3/2) = 3√3 ≈ 5.2 timesAnswer
vₑ = √(2gR) → gR = vₑ²/2 = (22.4×10³)²/2 = 501.76×10⁶/2 = 2.5088×10⁸ m²/s²Answer
F = 6.67×10⁻¹¹ × 6×10²⁴ × 7.4×10²² / (3.84×10⁸)² ≈ 2×10²⁰ NAnswer
g_d = g(1 - d/R) = g(1 - 1/4) = 3g/4. W' = 60×3/4 = 45 NAnswer
g_h = g(1 - 2h/R) = 9.8(1 - 200/6400) = 9.8(1 - 0.03125) = 9.8×0.96875 ≈ 9.49 m/s²Answer
F = 6.67×10⁻¹¹ × 5 × 10 / 4 = 6.67×10⁻¹¹ × 50/4 = 8.34×10⁻¹⁰ NAnswer
T = 2π√(R³/GM) = 2π√(R/g) = 2π√(6.4×10⁶/9.8) = 2π×808 = 5076 s ≈ 84.6 minAnswer
Slower (because g is less on the mountain, so T = 2π√(L/g) increases, clock ticks slower)Answer
g_J = G(318M)/(11.2R)² = 318/125.44 × g = 2.53 × 9.8 ≈ 24.8 m/s²Answer
Escape velocity = 11.2 km/s. Since 15 > 11.2, YES, it will escape.Answer
Your weight will increase slightly (because the centrifugal effect that reduces apparent weight at equator will be gone).16) Quick Revision — One-liners
- Newton's Law: F = G m₁ m₂ / r² (universal gravitational attraction)
- G = 6.67 × 10⁻¹¹ N m²/kg² (Cavendish determined it experimentally)
- Gravitational force is the weakest, longest-range, always attractive force
- Kepler's 1st: planets move in elliptical orbits with Sun at one focus
- Kepler's 2nd: equal areas in equal time → angular momentum conserved
- Kepler's 3rd: T² ∝ r³ (applies to all satellites, not just planets)
- g = GM/R² = 9.8 m/s² (acceleration due to gravity at Earth's surface)
- g decreases with height (g_h = g[R/(R+h)]²) and depth (g_d = g[1 - d/R])
- At centre of Earth: g = 0; at poles: g max; at equator: g min
- Mass = constant everywhere (kg); Weight = mg (N), varies with g
- Gravitational PE: U = -GMm/r (negative indicates attraction)
- Escape velocity: vₑ = √(2GM/R) = √(2gR) = 11.2 km/s (for Earth)
- Orbital velocity (near Earth): v₀ = √(gR) ≈ 7.9 km/s
- vₑ = √2 × v₀ (escape is 41% more than orbital speed)
- Geostationary: h ≈ 36,000 km, T = 24 h, equatorial orbit, appears stationary
- Free fall: all bodies accelerate at same g (in vacuum)
- Weightlessness: occurs in free fall/space (apparent weight = 0)
- Pressure = F/A, SI unit: Pa (N/m²)