Knowledge That Ignites
Complete notes with derivations, worked examples, and formula sheet
Gravitation: Universal attractive force between every pair of masses. This fundamental force governs a wide range of phenomena from everyday experiences to cosmic events.
Statement: 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 centers.
F = G(m₁m₂)/r²
Where:
Direction: The force acts along the line joining the centers of the two masses. It is always attractive and follows Newton's Third Law (equal and opposite).
The gravitational field strength at a point is defined as the force per unit mass experienced by a small test mass placed at that point.
g = F/m = GM/r²
(Units: N/kg or m/s²)
Gravitational potential at a point is the work done in bringing a unit mass from infinity to that point.
V = -GM/r
(Units: J/kg)
The negative sign indicates that gravitational potential is always negative (work is done by the field).
Standard near-surface value: g ≈ 9.8 m/s² (often approximated as 10 m/s² for calculations).
g = GMₑ/Rₑ²
Where Mₑ = mass of Earth (5.97 × 10²⁴ kg), Rₑ = radius of Earth (6.37 × 10⁶ m)
At height h above Earth's surface:
gₕ = g(R/(R+h))²
Given: R = 6370 km, h = 500 km, g = 9.8 m/s²
gₕ = 9.8 × (6370/(6370+500))²
gₕ ≈ 9.8 × (6370/6870)² ≈ 9.8 × 0.860 ≈ 8.43 m/s²
At depth d below Earth's surface (assuming uniform density):
g_d = g(1 - d/R)
Note: g decreases linearly with depth and becomes zero at Earth's center.
When an object falls under gravity alone (neglecting air resistance), it is in free fall. The acceleration is constant (g = 9.8 m/s²).
1. v = u + gt
2. s = ut + ½gt²
3. v² = u² + 2gs
Where: u = initial velocity, v = final velocity, s = displacement, t = time
Given: Stone dropped (u = 0), s = 80 m, g = 9.8 m/s²
Find: Time to reach ground (t) and final velocity (v)
Solution:
1. Time: t = √(2s/g) = √(2×80/9.8) ≈ √(16.33) ≈ 4.04 s
2. Final velocity: v = gt ≈ 9.8 × 4.04 ≈ 39.6 m/s
| Property | Mass (m) | Weight (W) |
|---|---|---|
| Definition | Amount of matter in a body | Gravitational force on the body |
| SI Unit | Kilogram (kg) | Newton (N) |
| Nature | Scalar (magnitude only) | Vector (magnitude and direction) |
| Variation | Constant everywhere | Varies with location (g changes) |
| Measuring Device | Beam balance, physical balance | Spring balance |
| Relationship | W = mg (Weight = mass × acceleration due to gravity) | |
U = -GMm/r
The negative sign indicates bound systems; zero potential at infinity.
For a small mass m orbiting a large mass M at radius r:
Centripetal force = Gravitational force
mvₒ²/r = GMm/r²
vₒ = √(GM/r)
Minimum speed to escape gravitational field (reach infinity with zero velocity):
vₑ = √(2GM/r) = √2 × vₒ
Escape velocity is √2 times orbital velocity at the same point.
Artificial Satellite: Human-made object placed into orbit by rockets.
Altitude: ~35,786 km above Earth
Period: 24 hours (matches Earth's rotation)
Characteristics: Fixed above a point on equator; used for communication, broadcasting
Altitude: 160-2,000 km above Earth
Period: 90-120 minutes
Characteristics: Used for imaging, ISS, Hubble telescope, some communication satellites
Escape Condition: If a satellite's speed ≥ escape velocity at that radius, it will leave Earth's gravitational field.
P = Force (thrust) perpendicular / Area
SI Unit: Pascal (Pa) = N/m²
Statement: A body immersed in a fluid experiences an upward buoyant force equal to the weight of the fluid displaced.
Buoyant Force (F_b) = Weight of displaced fluid = ρVg
Where: ρ = density of fluid, V = volume of displaced fluid, g = acceleration due to gravity
Given: Block volume = 0.02 m³, density = 800 kg/m³, Water density = 1000 kg/m³
Find: Will it float? If yes, what fraction is submerged?
Solution:
1. Mass of block = density × volume = 800 × 0.02 = 16 kg
2. Weight of block = mg = 16 × 9.8 ≈ 156.8 N
3. Buoyant force if fully immersed = ρ_water × V × g = 1000 × 0.02 × 9.8 = 196 N
4. Since 196 N > 156.8 N, the block floats
5. Fraction submerged = ρ_block/ρ_water = 800/1000 = 0.8 or 80%
| Quantity | Formula | Value (for Earth) |
|---|---|---|
| Escape Velocity | vₑ = √(2GM/R) | ≈ 11.18 km/s |
| Orbital Velocity at surface | vₒ = √(GM/R) | ≈ 7.91 km/s |
| g at 500 km altitude | gₕ = g(R/(R+h))² | ≈ 8.43 m/s² |
| Stone from 80 m height | t = √(2s/g) | t ≈ 4.04 s, v ≈ 39.6 m/s |
| Block buoyancy example | Fraction submerged = ρ_block/ρ_water | 80% submerged |
Q1. State Newton's law of gravitation
Answer: Write the statement, formula (F = Gm₁m₂/r²), and explain all symbols with units.
Q2. Derive escape velocity
Answer: Equate kinetic energy (½mv²) to the gain in potential energy to reach infinity (GMm/R). Show steps clearly: ½mv² = GMm/R → v = √(2GM/R).
Q3. Explain Archimedes' principle
Answer: State the principle, give mathematical form (F_b = ρVg), and mention one application (ship design, hydrometers, submarines).
Q4. Why does g change with altitude?
Answer: Give formula gₕ = g(R/(R+h))² and explain: As altitude increases, distance from Earth's center increases, reducing gravitational force and thus g.
Q5. Compare mass & weight
Answer: Use a table format comparing definition, SI unit, nature, variation, and measuring device.
F = G(m₁m₂)/r²
G = 6.674×10⁻¹¹ N·m²/kg²
g = GM/r²
Units: N/kg or m/s²
V = -GM/r
Units: J/kg
gₕ = g(R/(R+h))²
g_d = g(1 - d/R)
vₒ = √(GM/r)
vₑ = √(2GM/r)
vₑ = √2 × vₒ
U = -GMm/r
W = mg
F_b = ρVg
ρ = fluid density
v = u + gt
s = ut + ½gt²
v² = u² + 2gs
ρ_object/ρ_fluid
= Fraction submerged