Newton's Law of Gravitation, Gravitational Potential, Kepler's 3rd Law & Geostationary Orbits
Newton's Law of Universal Gravitation states that every point mass attracts every other point mass with a force directly proportional to the product of their masses and inversely proportional to the square of their separation: F = GMm / r². Surrounding any mass M is a radial gravitational field of strength g = F / m = GM / r² (a vector pointing inward toward M) and a scalar gravitational potential V = -GM / r, which is always negative because gravitational potential is defined as zero at infinity and gravity is strictly attractive. Equating gravitational attraction to centripetal force (GMm / r² = mv² / r = mω²r) yields orbital speed v = √(GM / r), Kepler's Third Law T² = (4π² / GM)r³, total satellite energy E_total = -GMm / (2r), and escape velocity v_esc = √(2GM / R). Examiners regularly test potential gradient g = -ΔV / Δr, geostationary vs low-polar orbits, and why gravitational potential is negative.
- Newton's Law of Gravitation (F = GMm / r²) & Field Strength (g = GM / r²)
- Force of attraction between two point masses (or uniform spheres with distance r measured between their centres): F = GMm / r², where G = 6.67 × 10⁻¹¹ N m² kg⁻². Gravitational field strength g at a point is the gravitational force per unit mass placed at that point: g = F / m (units: N kg⁻¹ or m s⁻²). In a radial field outside a sphere of mass M and radius R (for r ≥ R): g = GM / r² (inverse-square law). Inside a uniform planet of constant density ρ (for r < R), enclosed mass M_enc ∝ r³, so g increases linearly from 0 at the centre to g_surface at r = R.
- Gravitational Potential (V = -GM / r) & Why V is Negative
- Gravitational potential V at a point is the work done per unit mass in bringing a small test mass from infinity to that point: V = -GM / r (scalar, units: J kg⁻¹). Why is V always negative? 1. Gravitational potential is defined to be zero at infinity (V_∞ = 0). 2. Because the gravitational force is always attractive, work is done BY the field (energy is released) as a mass moves from infinity toward M, and an external agent must do positive work to pull a mass from r out to infinity (where it reaches 0 J kg⁻¹). Gravitational potential energy of mass m is E_p = mV = -GMm / r, and work done moving mass m between two points is ΔW = mΔV.
- Potential Gradient (g = -ΔV / Δr), Equipotentials & Area Under g-r Graph
- Gravitational field strength g equals the negative of the potential gradient: g = -ΔV / Δr. Graphical rules: 1. The gradient of a Gravitational Potential V vs r graph at any radius r equals -g. 2. The area under a Gravitational Field Strength g vs r graph between r₁ and r₂ equals the change in gravitational potential ΔV (in J kg⁻¹; multiply by m to get work done ΔW). Equipotential surfaces are concentric spherical shells around M perpendicular (at 90°) to the radial field lines; zero work is done moving a mass along an equipotential surface because ΔV = 0.
Newton's Law of Gravitation, Gravitational Potential, Kepler's 3rd Law & Geostationary Orbits: Wave Equation (v = fλ) & Force Vector (F = ma) Simulator
Adjust physical variables on the sliders below to see real-time changes in wave geometry, resultant forces, and acceleration.
F_net = 80 N − 25 N = 55 NAcceleration: a = F_net ÷ m = 55 ÷ 10 = 5.50 m/s²Key Concepts & Exam Definitions (6)
- Newton's Law of Gravitation (F = GMm / r²) & Field Strength (g = GM / r²)
- Force of attraction between two point masses (or uniform spheres with distance r measured between their centres): F = GMm / r², where G = 6.67 × 10⁻¹¹ N m² kg⁻². Gravitational field strength g at a point is the gravitational force per unit mass placed at that point: g = F / m (units: N kg⁻¹ or m s⁻²). In a radial field outside a sphere of mass M and radius R (for r ≥ R): g = GM / r² (inverse-square law). Inside a uniform planet of constant density ρ (for r < R), enclosed mass M_enc ∝ r³, so g increases linearly from 0 at the centre to g_surface at r = R.
- Gravitational Potential (V = -GM / r) & Why V is Negative
- Gravitational potential V at a point is the work done per unit mass in bringing a small test mass from infinity to that point: V = -GM / r (scalar, units: J kg⁻¹). Why is V always negative? 1. Gravitational potential is defined to be zero at infinity (V_∞ = 0). 2. Because the gravitational force is always attractive, work is done BY the field (energy is released) as a mass moves from infinity toward M, and an external agent must do positive work to pull a mass from r out to infinity (where it reaches 0 J kg⁻¹). Gravitational potential energy of mass m is E_p = mV = -GMm / r, and work done moving mass m between two points is ΔW = mΔV.
- Potential Gradient (g = -ΔV / Δr), Equipotentials & Area Under g-r Graph
- Gravitational field strength g equals the negative of the potential gradient: g = -ΔV / Δr. Graphical rules: 1. The gradient of a Gravitational Potential V vs r graph at any radius r equals -g. 2. The area under a Gravitational Field Strength g vs r graph between r₁ and r₂ equals the change in gravitational potential ΔV (in J kg⁻¹; multiply by m to get work done ΔW). Equipotential surfaces are concentric spherical shells around M perpendicular (at 90°) to the radial field lines; zero work is done moving a mass along an equipotential surface because ΔV = 0.
- Escape Velocity Derivation: v_esc = √(2GM / R) = √(2gR)
- Escape velocity is the minimum launch speed required for an unpowered object on the surface of a planet (mass M, radius R) to reach infinity (r = ∞, where V = 0 and final v = 0) and never return. By conservation of energy, Gain in E_p = Loss in E_k (or E_total = E_k + E_p = 0 at infinity): ½m(v_esc)² + (-GMm / R) = 0 → ½m(v_esc)² = GMm / R (notice mass m of the projectile cancels) → v_esc = √(2GM / R). Since g = GM / R² (so GM = gR²), this also equals v_esc = √(2gR).
- Circular Satellite Orbits, Kepler's 3rd Law (T² ∝ r³) & Satellite Energy
- For a satellite of mass m in a circular orbit of radius r around mass M, the gravitational force provides the centripetal force: 1. Orbital speed: GMm / r² = mv² / r → v = √(GM / r) (lower orbits move faster). 2. Kepler's Third Law: since v = 2πr / T (or ω = 2π / T in GMm / r² = mω²r), substituting gives v² = 4π²r² / T² = GM / r → T² = [(4π²) / (GM)] r³, so T² ∝ r³. 3. Satellite Energy: Kinetic energy E_k = ½mv² = +GMm / (2r); Potential energy E_p = -GMm / r; Therefore, Total Energy E_total = E_k + E_p = -GMm / (2r) = -E_k. (If atmospheric drag lowers E_total, r decreases, E_p becomes more negative by 2ΔE, and E_k and speed v increase).
- Geostationary vs Low Polar Orbits
- 1. Geostationary Satellites (used for telecommunications and satellite TV so ground dish aerials do not need to track them): Must satisfy three conditions: (a) Orbital time period T = exactly 24 hours (86,400 s, matching Earth's sidereal rotation period), (b) Orbit directly above the equator in the equatorial plane, and (c) Orbit in the same direction as Earth's rotation (West to East), at a fixed orbital radius r ≈ 4.22 × 10⁷ m (~36,000 km altitude above Earth's surface). 2. Low Polar Orbit Satellites (used for weather monitoring, military surveillance, and Earth mapping): orbit over the North and South poles at low altitude (~200-1000 km) with short periods (~90-100 min), scanning the entire Earth's surface as the planet rotates beneath them with high image resolution.
Quick-Fire Flashcards (6)
Click any card to check the answer, then mark it as Known to track your progress.
Define gravitational potential V at a point in a gravitational field, and explain why it always has a negative value.
What do the gradient of a V-r graph and the area under a g-r graph represent?
Derive the formula for escape velocity v_esc from the surface of a planet of mass M and radius R.
Derive Kepler's Third Law (T² = (4π² / GM)r³) for a planet or satellite in a circular orbit of radius r.
State the three essential features of a geostationary satellite orbit and state one practical use.
Why is zero work done by the gravitational field on a satellite moving in a circular orbit around the Earth?
Exam-Style Questions & Mark Schemes (4)
Write or think through your answer first, then click Check Mark Scheme to see the exact points examiners award marks for.
- Q1
The Earth has a mass of 5.97 × 10²⁴ kg and a mean radius of 6.37 × 10⁶ m (G = 6.67 × 10⁻¹¹ N m² kg⁻²). Calculate: (a) the gravitational potential at the Earth's surface, (b) the minimum escape velocity from the Earth's surface, and (c) the work done in lifting a 450 kg communications satellite from the Earth's surface to an orbit at an altitude of 12,000 km above the Earth's surface (ignoring kinetic energy). (5 marks)
Click “Check Mark Scheme” to compare your answer, 1. (a) Gravitational potential at Earth's surface (r₁ = R = 6.37 × 10⁶ m): V_surface = -GM ÷ R = -(6.67 × 10⁻¹¹ × 5.97 × 10²⁴) ÷ (6.37 × 10⁶) = -6.25 × 10⁷ J kg⁻¹ (must include minus sign) [1 mark]. 2. (b) Escape velocity v_esc = √(2GM ÷ R) = √(-2 × V_surface) = √(2 × 6.251 × 10⁷) = 1.12 × 10⁴ m s⁻¹ (11.2 km s⁻¹) [1 mark]. 3. (c) Calculate total orbital radius r₂ from the centre of the Earth: r₂ = R + h = (6.37 × 10⁶) + (12.0 × 10⁶) = 1.837 × 10⁷ m [1 mark]. 4. Calculate gravitational potential at r₂: V₂ = -GM ÷ r₂ = -(6.67 × 10⁻¹¹ × 5.97 × 10²⁴) ÷ (1.837 × 10⁷) = -2.168 × 10⁷ J kg⁻¹ [1 mark]. 5. Work done lifting the 450 kg satellite: ΔW = mΔV = m(V₂ - V_surface) = 450 × [-2.168 × 10⁷ - (-6.251 × 10⁷)] = 450 × (4.083 × 10⁷) = 1.84 × 10¹⁰ J [1 mark].
- Q2
Show from first principles that the orbital period T and orbital radius r of a satellite in a circular orbit around a planet of mass M satisfy T² = (4π² / GM)r³, and hence calculate the altitude above the Earth's surface of a geostationary satellite (M_Earth = 5.97 × 10²⁴ kg, R_Earth = 6.37 × 10⁶ m, G = 6.67 × 10⁻¹¹ N m² kg⁻²). (5 marks)
Click “Check Mark Scheme” to compare your answer, 1. The gravitational attraction provides the centripetal force required for circular motion: GMm ÷ r² = mv² ÷ r, which simplifies to v² = GM ÷ r [1 mark]. 2. Substitute the circular orbital speed v = 2πr ÷ T (so v² = 4π²r² ÷ T²): 4π²r² ÷ T² = GM ÷ r, which rearranges to T² = (4π² ÷ GM) r³ [1 mark]. 3. For a geostationary satellite, T = 24 hours = 24 × 3600 = 86,400 s [1 mark]. 4. Rearrange for r³ and substitute values: r³ = (G M T²) ÷ (4π²) = (6.67 × 10⁻¹¹ × 5.97 × 10²⁴ × 86,400²) ÷ (4π²) = 7.528 × 10²² m³ → r = (7.528 × 10²²)¹/³ = 4.22 × 10⁷ m [1 mark]. 5. Subtract Earth's radius to find altitude h above the surface: h = r - R_Earth = (4.222 × 10⁷) - (6.37 × 10⁶) = 3.59 × 10⁷ m (35,900 km) [1 mark].
- Q3
A low-Earth-orbit spy satellite of mass m is in a circular orbit of radius r around the Earth. Show that its total mechanical energy is E_total = -GMm / (2r), and explain what happens to its orbital radius, gravitational potential energy, kinetic energy, and orbital speed when it encounters slight atmospheric drag in the upper atmosphere. (5 marks)
Click “Check Mark Scheme” to compare your answer, 1. From GMm / r² = mv² / r, we have mv² = GMm / r, so the satellite's kinetic energy is E_k = ½mv² = +GMm / (2r) [1 mark]. 2. Adding gravitational potential energy E_p = -GMm / r gives total mechanical energy: E_total = E_k + E_p = +GMm / (2r) + (-GMm / r) = -GMm / (2r) [1 mark]. 3. Work done against atmospheric drag removes mechanical energy from the satellite, making E_total = -GMm / (2r) more negative, which means the orbital radius r decreases (the satellite spirals into a lower orbit) [1 mark]. 4. As r decreases, gravitational potential energy E_p = -GMm / r decreases (becomes more negative) by twice the energy lost to drag [1 mark]. 5. Half of the lost E_p is dissipated as heat against drag and the other half is converted into kinetic energy E_k = +GMm / (2r) (where v = √(GM / r)), so kinetic energy and orbital speed actually increase as the satellite drops lower [1 mark].
- Q4
Compare and contrast gravitational fields and electrostatic fields in terms of field strength formulas, potential formulas, and nature of the force. (4 marks)
Click “Check Mark Scheme” to compare your answer, 1. Similarity in inverse-square field strength: Both radial fields obey an inverse-square law with distance r from a point source: g = GM / r² for gravity and E = Q / (4πε₀r²) for electric fields [1 mark]. 2. Similarity in inverse-distance potential and gradient: Both potentials vary as 1/r (V_g = -GM / r and V_e = Q / (4πε₀r)) and both satisfy field strength = -potential gradient (g = -ΔV_g / Δr and E = -ΔV_e / Δr) [1 mark]. 3. Difference in nature of force: Gravitational forces act on mass and are always attractive (so V_g is always negative), whereas electrostatic forces act on electric charge and can be either attractive OR repulsive (so V_e can be positive or negative) [1 mark]. 4. Difference in shielding/magnitude: Electric fields are vastly stronger at the subatomic scale than gravitational fields and can be screened/shielded by conductors (Faraday cage), whereas gravitational fields cannot be shielded [1 mark].
Common Exam Questions (4)
What is the #1 mistake students make in satellite orbit and gravitational field calculations?
Using the altitude h above the planet's surface instead of the distance r from the centre of the planet. Every formula in gravitation (F = GMm/r², g = GM/r², V = -GM/r, T² ∝ r³) requires r = R_planet + h (measured from the centre of mass of the planet). Also remember you cannot use ΔE_p = mgΔh over large distances because g is not constant.
Why do astronauts inside the International Space Station (ISS) experience 'weightlessness' even though Earth's gravitational field strength at 400 km altitude is still ~8.7 N kg⁻¹ (nearly 90% of surface gravity)?
Astronauts in orbit are not in zero gravity. Both the astronauts and the space station are in continuous free-fall toward the Earth with the exact same centripetal acceleration (a = g = 8.7 m s⁻²). Because they accelerate together at the same rate, the floor of the station exerts zero normal contact reaction force (R = 0) on the astronauts, creating the sensation of apparent weightlessness.
How do you calculate the position of the neutral point (zero resultant gravitational field strength) between the Earth and the Moon?
Let the neutral point be at distance x from the Earth's centre and distance (D - x) from the Moon's centre (where D is the Earth-Moon separation). Equate the magnitudes of the two opposing field strengths: G M_Earth / x² = G M_Moon / (D - x)², cancel G, and take the square root of both sides before solving for x: √(M_Earth) / x = √(M_Moon) / (D - x).
Does the escape velocity of a rocket depend on the mass of the rocket?
No. In ½m(v_esc)² = GMm / R, the mass m of the escaping object cancels out completely, so a 10⁻¹⁰ kg dust grain and a 10⁶ kg spacecraft have the exact same escape velocity (11.2 km s⁻¹ on Earth), although the heavier spacecraft requires much more kinetic energy (½mv²) to reach that speed.
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