Newton's law of gravitation — one equation that governs falling apples, orbiting moons and the structure of galaxies. Master gravitational field strength and potential to understand how gravity shapes the universe.
🍎State Newton's law of gravitation: F = Gm₁m₂/r²
📐Define gravitational field strength g = F/m and derive g = GM/r²
🌐Draw and interpret gravitational field line diagrams
⚡Define gravitational potential V = −GM/r and explain the negative sign
⚖️Calculate gravitational potential energy using E_p = mV = −GMm/r
🔗Relate field strength to potential gradient: g = −dV/dr
Newton's Law of Gravitation
Every two masses in the universe attract each other with a force given by Newton's Law of Universal Gravitation:
F = Gm₁m₂ / r²
where G = 6.674 × 10⁻¹¹ N m² kg⁻² is the universal gravitational constant, m₁ and m₂ are the two masses (kg), and r is the distance between their centres of mass (m).
Key features of Newton's gravitational force:
Always attractive — there is no repulsive gravitational force.
Acts along the line joining the centres of the two masses.
Inverse-square law: F ∝ 1/r². Double the distance → quarter the force.
Acts on both masses equally and oppositely (Newton's 3rd law): F on m₁ due to m₂ = F on m₂ due to m₁.
Acts at a distance — no physical contact needed.
Universal Gravitational Constant G: G = 6.674 × 10⁻¹¹ N m² kg⁻². Extremely small — gravity is by far the weakest of the four fundamental forces, yet it dominates at large scales because it is always attractive and has infinite range.
Gravitational Field Strength
A gravitational field is a region of space where a mass experiences a gravitational force. We define the gravitational field strength g as the force per unit mass:
g = F / m (N kg⁻¹ or m s⁻²)
For a point mass M (or a uniform sphere) at distance r from its centre, using F = GMm/r²:
g = GM / r²
g is a vector — it points toward the source mass (always attractive). Note that g is numerically equal to the acceleration due to gravity (both have units m s⁻²).
Field lines represent gravitational fields:
Arrows point toward the source mass (inward radially for isolated sphere)
Closer spacing = stronger field
Radial pattern around a point mass / sphere
Uniform parallel field lines near Earth's surface (constant g ≈ 9.81 N kg⁻¹)
At Earth's surface: g = GM_E/R_E² = (6.674×10⁻¹¹ × 5.97×10²⁴)/(6.37×10⁶)² = 9.82 N kg⁻¹ ✓
⚠️ Above Earth's surface g ∝ 1/r² (decreases with distance). Inside a uniform sphere, g ∝ r (decreases linearly toward zero at centre). Real Earth is not uniform, so g varies slightly with location.
Gravitational Potential
The gravitational potential V at a point is defined as the work done per unit mass in bringing a small test mass from infinity to that point:
V = W / m (J kg⁻¹)
For a point mass M at distance r:
V = −GM / r
The negative sign is fundamental:
At infinity, V = 0 (reference point).
Closer to M, r decreases so V becomes more negative — the potential well deepens.
Gravity does positive work on a mass falling toward M, so V decreases (becomes more negative) as potential energy is converted to kinetic energy.
To move a mass from near M to infinity requires doing positive work against gravity — hence V is negative (the mass is "bound").
Gravitational potential energy E_p: The energy of mass m at potential V: E_p = mV = −GMm/r. This is zero at infinity and negative everywhere else — reflecting that bound masses have less energy than free masses.
The relationship between field strength and potential:
g = −dV/dr
Field strength is the negative gradient of potential. Steep potential gradient → strong field. Uniform field → linearly varying potential.
Gravitational Potential and Equipotentials
Equipotential surfaces are surfaces of constant gravitational potential. For an isolated sphere, equipotentials are concentric spheres centred on the mass. Key properties:
No work is done moving along an equipotential.
Field lines are always perpendicular to equipotentials.
Near Earth's surface: equipotentials are approximately horizontal planes (V = constant at constant height). Moving vertically by height h: ΔV = gh, so ΔE_p = mgh — the familiar result from GCSE, now understood as a linear approximation valid when h << R_E.
Location
g (N kg⁻¹)
V (MJ kg⁻¹)
Earth's surface (r = R_E)
9.81
−62.5
Low orbit (r = 1.1R_E)
8.11
−56.8
Geostationary orbit (r = 6.6R_E)
0.225
−9.46
Moon's orbit (r = 60R_E)
0.0027
−1.04
Calculate the gravitational force between Earth (M_E = 5.97 × 10²⁴ kg) and Moon (M_M = 7.35 × 10²² kg) separated by 3.84 × 10⁸ m. Compare this to the weight of a 70 kg person on Earth.
4F = 2.926×10³⁷ / 1.475×10¹⁷ = 1.984×10²⁰ N ≈ 2.0 × 10²⁰ N
5Weight of 70 kg person: W = mg = 70 × 9.81 = 686.7 N ≈ 687 N
F = 2.0 × 10²⁰ N — about 3 × 10¹⁷ times greater than a person's weight! The Moon-Earth force is enormous but spread over astronomical distances.
Calculate the gravitational potential at the surface of Earth and at the International Space Station orbit (altitude 408 km). Hence find the energy needed to move 1000 kg of cargo from Earth's surface to the ISS.
1R_E = 6.37 × 10⁶ m; r_ISS = 6.37×10⁶ + 408×10³ = 6.778×10⁶ m; M_E = 5.97×10²⁴ kg
V_surface = −62.5 MJ kg⁻¹; V_ISS = −58.8 MJ kg⁻¹; Energy needed ≈ 3.76 GJ for 1000 kg
At what altitude above Earth's surface does g = g_surface/4? (R_E = 6.37 × 10⁶ m)
1g = GM/r² — if g halves when r doubles (inverse-square). For g to become 1/4 of surface value:
2g/4 = GM/(r²) and g = GM/R_E² → GM/(r²) = GM/(4R_E²) → r² = 4R_E² → r = 2R_E
3r = 2 × 6.37×10⁶ = 1.274×10⁷ m from Earth's centre
4Altitude h = r − R_E = 1.274×10⁷ − 6.37×10⁶ = 6.37×10⁶ m = 6370 km
Altitude = 6370 km (one Earth-radius above the surface). g is ¼ of surface value at r = 2R_E.
Q1. Two masses are separated by distance r. If the distance is tripled, the gravitational force between them:
Q2. The gravitational potential at a point is −5.0 × 10⁷ J kg⁻¹. What is the potential energy of a 500 kg satellite at this point?
Q3. Why is gravitational potential always negative (or zero)?
Q4. At what distance from Earth's centre does g = 2.45 N kg⁻¹? (g_surface = 9.81 N kg⁻¹, R_E = 6.37 × 10⁶ m)
Q5. Gravitational field lines are always perpendicular to:
Challenge 1. Show that the gravitational field strength at the surface of a planet of density ρ and radius R is g = (4/3)πGρR. Use this to find g on a hypothetical planet with twice Earth's density and twice Earth's radius. (ρ_Earth = 5500 kg m⁻³)
✓ Derivation: g = GM/R². Mass M = ρ × Volume = ρ × (4/3)πR³. So g = G × ρ(4/3)πR³/R² = (4/3)πGρR. QED. For the hypothetical planet: ρ' = 2ρ_E = 11000 kg m⁻³, R' = 2R_E. g' = (4/3)πG(2ρ_E)(2R_E) = 4 × (4/3)πGρ_E R_E = 4g_Earth = 4 × 9.81 = 39.2 N kg⁻¹. Equivalently, g ∝ ρR, so doubling both gives 4× the surface gravity. A person on this planet would weigh 4 times their Earth weight.
Challenge 2. Calculate the change in gravitational potential energy of a 500 kg spacecraft moving from Earth's surface to the Moon's surface. (Data: M_E = 5.97×10²⁴ kg, R_E = 6.37×10⁶ m; M_M = 7.35×10²² kg, R_M = 1.74×10⁶ m; Earth-Moon distance = 3.84×10⁸ m.)
✓ Calculate V at Earth's surface due to BOTH Earth and Moon, then V at Moon's surface due to both. At Earth's surface: V_E = −GM_E/R_E = −(6.674×10⁻¹¹ × 5.97×10²⁴)/6.37×10⁶ = −6.252×10⁷ J kg⁻¹. Contribution from Moon (distance ≈ d = 3.84×10⁸ m from Moon centre ≈ d − R_E ≈ d for Moon): V_M_at_Earth = −GM_M/d = −(6.674×10⁻¹¹ × 7.35×10²²)/3.84×10⁸ = −1.277×10⁴ J kg⁻¹ ≈ negligible. At Moon's surface: V_M = −GM_M/R_M = −(6.674×10⁻¹¹ × 7.35×10²²)/1.74×10⁶ = −2.818×10⁶ J kg⁻¹. V_E_at_Moon = −GM_E/(d−R_M) ≈ −GM_E/d = −(6.674×10⁻¹¹ × 5.97×10²⁴)/3.84×10⁸ = −1.038×10⁶ J kg⁻¹. Total V at Moon surface ≈ −2.818×10⁶ − 1.038×10⁶ = −3.856×10⁶ J kg⁻¹. ΔV = V_Moon − V_Earth ≈ −3.856×10⁶ − (−6.252×10⁷) = +5.866×10⁷ J kg⁻¹. ΔE_p = mΔV = 500 × 5.866×10⁷ = 2.93×10¹⁰ J ≈ 29.3 GJ. (Much of this is recovered as KE during the descent — but getting out of Earth's gravity well requires most of this energy.)
Challenge 3 (Synoptic). Derive the relationship g = −dV/dr for a radial gravitational field. Show that this is consistent with V = −GM/r and g = −GM/r². What is the physical significance of the negative sign?
✓ Derivation: The work done dW against gravity when moving mass m a small distance dr away from M: dW = −F dr = −(−GMm/r²) dr = GMm/r² dr (positive, as we do work against gravity). Change in potential energy dE_p = dW = GMm/r² dr. Change in potential dV = dE_p/m = GM/r² dr. Therefore dV/dr = GM/r² → g = −GM/r² = −dV/dr. QED. Verification: V = −GM/r → dV/dr = GM/r² (differentiating). Then g = −dV/dr = −GM/r². ✓ Physical significance of negative sign: as r increases (moving away from M), V increases (becomes less negative), so dV/dr > 0. The field g is directed toward M (decreasing r), which is the direction of decreasing r, hence g = −dV/dr gives a negative (inward) result for the radial direction, confirming gravity acts inward. In general: field always points "downhill" on the potential energy surface, toward lower potential.