Gravitation Notes – Complete Chapter

Newton's Law of Gravitation

Properties of Gravitational Force

Every particle of matter attracts every other particle with a force
Always attractive — never repulsive
Weakest force in nature (F_gravity << F_{electrostatic})
Long range — infinite range
Follows inverse square law
Central force (acts along line joining the two bodies)
Does NOT depend on medium
Conservative force; follows action-reaction pair (Newton's 3rd law)
Mediated by gravitational field; reason for stability of Universe
Valid for point masses and spherical masses with symmetry

F = GM₁M₂/r² (Newton's Law of Gravitation)

G = 6.67 × 10⁻¹¹ N·m²/kg² (Universal Gravitational Constant); dim [M⁻¹L³T⁻²]

Neutral Point (Between Two Masses)

Point where net gravitational field = 0
For masses M₁ and M₂ separated by distance d, neutral point from M₁: x = d/(1 + √(M₂/M₁))
Neutral point is always closer to the smaller mass
For equal masses: neutral point is at the midpoint

Q: Four masses m at corners of a square of side a. Find net force on one mass.

Forces from two adjacent masses (at distance a): F₁ = Gm²/a² each; Force from diagonal mass (at √2 a): F₂ = Gm²/2a²; F_net = √(F₁²+F₁²) + F₂ = F₁√2 + F₂ = (2√2+1)Gm²/2a²

Ans: F = (2√2+1)Gm²/(2a²) ✓

Gravitational Field Intensity (I)

Definition

Gravitational field intensity at a point = force per unit mass = I = F/m
Vector quantity; direction: towards the mass (attractive)
Unit = N/kg; Dimensional formula [LT⁻²]
For point mass: I = GM/r²

Body	Region	Field Intensity
Point mass M	At distance r	I = GM/r²
Hollow Sphere (R)	Outside (r > R)	I = GM/r²
	On surface (r = R)	I = GM/R²
	Inside (r < R)	I = 0
Solid Sphere (R)	Outside (r > R)	I = GM/r²
	On surface (r = R)	I = GM/R²
	Inside (r < R)	I = GMr/R³
Ring (mass M, radius R)	On axis at distance x	I = GMx/(R²+x²)^3/2
Ring — maximum field	At x = R/√2	I_max = GM/(R²·3√3/2)

Acceleration Due to Gravity — Variation

g on Surface of Earth

g₀ = GM_E/R_E² ≈ 9.8 m/s²
M_E = 6 × 10²⁴ kg; R_E = 6400 km; ρ_E = 5.5 × 10³ kg/m³
g ∝ 1/R² (mass constant); g ∝ M (radius constant)

Variation	Formula	Note
Height H above surface	g_H = g₀R²/(R+H)² = g₀(1 − 2H/R) for H << R	Always decreases
Depth d below surface	g_d = g₀(1 − d/R)	Always decreases
Rotation of Earth	g_eff = g₀ − ω²R cos²λ	λ = latitude
At Equator	g_eq = g₀ − ω²R (minimum)	λ = 0°
At Poles	g_pole = g₀ (maximum)	λ = 90°
Shape of Earth	g_pole > g_equator	R_pole < R_equator

⚡ Key Tricks

g at centre of Earth = 0 (d = R in g_d = g₀(1−d/R))
g at height H = g at depth d → 2H/R = d/R → d = 2H
% decrease at height H = 2H/R × 100
% decrease at depth d = d/R × 100
If Earth rotates 17 times faster → body at equator feels weightless
T = 2π/ω → at weightlessness: T ≈ 84.6 min (current period would need to be reduced to this)

Q: At what height above surface is g reduced by 1%? (R = 6400 km)

ΔgH/g × 100 = 2H/R; 1 = 2H/6400 → H = 32 km

Ans: H = 32 km ✓

Q: At what depth is g same as at height H = R/2?

g_H = g(1−2H/R) = g(1−1) = 0? No: g_H = gR²/(R+H)² = gR²/(3R/2)² = 4g/9; g_d = g(1−d/R) = 4g/9 → d/R = 5/9 → d = 5R/9

Ans: d = 5R/9 ✓

Gravitational Potential (V) and Potential Energy (U)

Gravitational Potential

V = work done in bringing unit mass from ∞ to that point = U/m
Scalar quantity; always negative (since gravity is attractive)
Unit = J/kg; dimensional formula [L²T⁻²]
I = −dV/dr (field = negative potential gradient)
Change in potential ΔV does NOT depend on reference

Body	Region	Potential V
Point mass M	At distance r	V = −GM/r
Hollow Sphere (R)	Outside (r > R)	V = −GM/r
	On surface	V = −GM/R
	Inside	V = −GM/R (constant)
Solid Sphere (R)	Outside (r > R)	V = −GM/r
	On surface	V = −GM/R
	Inside (r < R)	V = −GM(3R²−r²)/(2R³)
Solid Sphere — Centre	r = 0	V_centre = −3GM/(2R) = (3/2)V_surface
Ring (R) — on axis at x	—	V = −GM/√(R²+x²)
Ring — Centre	x = 0	V = −GM/R

U = mV (Potential energy = mass × potential)

U = −GMm/r (for two point masses)

Total PE for N bodies = N(N−1)/2 terms (one for each pair)

ΔU = mΔV = mgH/(1+H/R) (moving mass m from surface to height H)

Escape Velocity

Definition and Key Points

Minimum speed required to escape gravitational field of a planet (reach infinity)
Depends on mass and radius of planet — NOT on mass of object or angle of projection
v_e = √(2GM/R) = √(2gR) = R√(2G × density × 4π/3)
For Earth: v_e = 11.2 km/s
v_e = √2 × v_orbital (surface) → v_e = √2 × v₀
If ρ = constant → v_e ∝ R
If M = constant → v_e ∝ 1/√R

⚡ When projected with v > v_e

Speed at infinity: v_∞ = √(v² − v_e²)

If projected with v_e: takes parabolic path (v_∞ = 0)

If v < v_e: follows elliptical path; returns to surface

Q: A planet has radius R and mean density ρ. Find escape velocity.

M = ρ × 4πR³/3; v_e = √(2GM/R) = √(2G × 4πρR²/3) = R√(8πGρ/3)

Ans: v_e = R√(8πGρ/3) ✓

Orbital Velocity and Satellites

Orbital velocity: v₀ = √(GM/(R+H)) = √(g₀R²/(R+H))

At surface (H << R): v₀ = √(gR) ≈ 7.9 km/s

Time period: T = 2π(R+H)/v₀ = 2π√((R+H)³/GM)

Relation: v_e = √2 × v₀ (at surface)

Energy of Satellite at Orbital Height (R+H)

Energy	Formula	Sign
Kinetic Energy	KE = GMm/[2(R+H)]	Always +ve
Potential Energy	PE = −GMm/(R+H)	Always −ve
Total Energy	TE = −GMm/[2(R+H)]	Always −ve
Binding Energy	BE = +GMm/[2(R+H)]	Always +ve
Ratio KE : PE : TE	1 : −2 : −1	—

Important Satellite Results

TE is negative → satellite is bound to planet
|PE| = 2|KE| = 2|TE|
BE = |TE| = GMm/[2(R+H)] = minimum energy needed to free satellite
If orbital speed increased by √2 times (41.4%) → satellite escapes to infinity
BE of body on surface = mgh = GMm/R (for H = 0)
BE_{body on surface} = BE_{satellite at surface orbit}

Shifting Orbital Height — Energy Required

To shift satellite from radius r₁ to r₂ (r₂ > r₁):

ΔE = TE_f − TE_i = −GMm/(2r₂) − (−GMm/(2r₁)) = GMm/2 × (1/r₁ − 1/r₂)

Energy must be given (positive) to move to higher orbit

Even though satellite slows down in higher orbit, energy is put in!

Types of Satellites

Property	Geostationary Satellite	Polar Satellite
Orbit height from surface	≈ 36000 km = 6R	~500–800 km
Orbit height from centre	7R	Close to R
Time period	24 hours (= Earth's rotation)	~100 minutes
Direction of orbit	West to East (equatorial)	Pole to pole (passes over poles)
Angular speed vs Earth	Same as Earth	Different
Appears stationary?	Yes (relative to Earth)	No
Earth coverage	1/3rd of Earth's surface	Full Earth over time
Uses	Communication, TV relay	Weather, spy, GPS

Circular vs Elliptical Orbit

Circular orbit: Speed = const, KE = const, PE = const, TE = const
Elliptical orbit: Speed = variable, KE = variable, PE = variable, TE = constant (COME)
r_max = a(1+e); r_min = a(1−e) (a = semi-major axis, e = eccentricity)
At perihelion (closest): v_max, KE_max, PE_min
At aphelion (farthest): v_min, KE_min, PE_max

Kepler's Laws of Planetary Motion

Kepler's Three Laws

① Law of Orbits (1st Law):
Every planet moves in an elliptical orbit with the Sun at one focus. Path of planet: ellipse (not circle).

② Law of Areas (2nd Law):
Line joining planet to Sun sweeps equal areas in equal time intervals.
→ dA/dt = constant → L (angular momentum) = constant
→ Planet moves faster near Sun (perihelion) and slower at aphelion

③ Law of Periods (3rd Law):
T² ∝ r³ (square of time period ∝ cube of semi-major axis)
T² = (4π²/GM) × r³ → T²/r³ = 4π²/GM = constant (K)

T² = Kr³ where K = 4π²/(GM)

v_max/v_min = r_max/r_min = (1+e)/(1−e) (from conservation of L: rv = const at aphelion/perihelion)

Kepler's laws are valid for all bodies in gravitational orbit (satellites, comets, planets)

Q: A satellite has time period T at height H from Earth's surface. What will be time period at height 4H?

T ∝ r^(3/2); r₁ = R+H, r₂ = R+4H; T₂/T₁ = [(R+4H)/(R+H)]^(3/2); For H = R: T₂/T = (5R/2R)^(3/2) = (5/2)^(3/2)

Ans: T₂ = T[(R+4H)/(R+H)]^(3/2) ✓ (use given R value)

Q: A planet takes 8 years to complete one orbit. Another takes 1 year. Find ratio of their distances from Sun.

T² ∝ r³ → r₁/r₂ = (T₁/T₂)^(2/3) = (8/1)^(2/3) = 4

Ans: r₁/r₂ = 4 ✓

Q: A body projected with velocity v from Earth's surface. At infinity its speed is 5 km/s. Find v. (v_e = 11.2 km/s)

v_∞² = v² − v_e² → 25 = v² − 125.44 → v² = 150.44 → v ≈ 12.27 km/s

Ans: v ≈ 12.27 km/s ✓