This physics grades 11-12 lesson on gravitational field notes gives the key ideas first, then worked examples and diagrams. After that, practise with the worksheet on gravitational field and check yourself with the quiz.
Quick answer: Newton’s law of gravitation says two masses attract with force F = G m₁m₂/r². The field strength of a planet is g = GM/r², its potential is V = −GM/r, and a satellite in circular orbit has speed v = √(GM/r) and period T² = 4π²r³/(GM), which is Kepler’s third law.
Level: Physics grades 11-12 (A Level · IB · Class 11-12) · Curriculum: IB D.1
Gravitational field notes: the key ideas
What does Newton’s law of gravitation say?
Also, every mass attracts every other mass. The force is tiny for everyday objects but huge for planets. Newton showed that it depends on the product of the masses and falls with the square of the distance between their centres.
Newton’s law of gravitation: F = G m₁ m₂ ÷ r², where G = 6.67 × 10⁻¹¹ N m² kg⁻². In addition, the force acts along the line joining the centres of the masses, and the two forces are equal and opposite.
Earth (5.97 × 10²⁴ kg) and the Moon (7.35 × 10²² kg) have centres 3.84 × 10⁸ m apart.
F = 6.67 × 10⁻¹¹ × 5.97 × 10²⁴ × 7.35 × 10²² ÷ (3.84 × 10⁸)² = 1.98 × 10²⁰ N.
What is gravitational field strength?
Gravitational field strength g is the gravitational force per unit mass on a small test mass: g = F ÷ m, in N/kg. Moreover, it is a vector that points towards the source mass. Outside a spherical body of mass M, g = G M ÷ r².
At the Earth’s surface g = G M ÷ R² = 9.81 N/kg, which is the same as 9.81 m/s². In addition, the field is not uniform far from the planet. Moreover, it follows an inverse-square law, as the graph shows.
The International Space Station orbits 400 km above the surface, so r = 6.37 × 10⁶ + 4.0 × 10⁵ = 6.77 × 10⁶ m.
g = G M ÷ r² = 3.98 × 10¹⁴ ÷ (6.77 × 10⁶)² = 8.69 N/kg, about 89 % of its surface value. Astronauts float because they and the station are in free fall, not because gravity is missing.
What are gravitational potential and escape speed?
Also, the gravitational potential V at a point is the work done per unit mass to bring a small mass from infinity to that point. It is zero at infinity and negative everywhere nearer, because gravity does positive work as the mass comes in. In addition, the potential energy of a mass m is Ep = m V.
V = −G M ÷ r, Ep = −G M m ÷ r, vesc = √(2 G M ÷ R)
V = potential (J/kg), Ep = potential energy (J), M = mass of the planet (kg), m = mass of the object (kg), R = radius of the planet (m), vesc = escape speed (m/s).
The escape speed is the launch speed for which the kinetic energy exactly cancels the potential energy, ½ m v² = G M m ÷ R, so the object just reaches infinity with no speed left.
Escape speed from the Earth: v = √(2 G M ÷ R) = √(2 × 3.98 × 10¹⁴ ÷ 6.37 × 10⁶) = √(1.25 × 10⁸) = 1.12 × 10⁴ m/s, or 11.2 km/s. Moreover, the potential at the surface is V = −3.98 × 10¹⁴ ÷ 6.37 × 10⁶ = −6.25 × 10⁷ J/kg.
How do satellites and Kepler’s third law work?
For a circular orbit, gravity provides the centripetal force: G M m ÷ r² = m v² ÷ r. This gives v = √(G M ÷ r) and, since v = 2π r ÷ T, T² = 4π² r³ ÷ (G M). This is Kepler’s third law: T² is proportional to r³ for all bodies orbiting the same mass.
| Quantity | Formula | Unit |
|---|---|---|
| Force between masses | G m₁ m₂ ÷ r² | N |
| Field strength | G M ÷ r² | N/kg |
| Potential | −G M ÷ r | J/kg |
| Orbital speed | √(G M ÷ r) | m/s |
| Period | 2π √(r³ ÷ G M) | s |
Also, a geostationary satellite has a period of 24 h = 86 400 s.
r³ = G M T² ÷ 4π² = 3.98 × 10¹⁴ × 86 400² ÷ 39.5 = 7.53 × 10²² m³, so r = 4.22 × 10⁷ m, which is 3.59 × 10⁷ m above the surface.
v = 2π r ÷ T = 3.07 km/s.
Students use the height above the surface as r. In every formula, r is the distance from the centre of the planet, so add the planet’s radius to the altitude. In addition, a faster orbit is a lower orbit, not a higher one.
Memorise one equation, G M ÷ r² = v² ÷ r, and derive the rest. Also remember G M = 3.98 × 10¹⁴ m³/s² for the Earth, so you can skip the multiplication each time.
- Newton’s law: F = G m₁ m₂ ÷ r², with G = 6.67 × 10⁻¹¹ N m² kg⁻².
- Field strength g = F/m = G M ÷ r², directed towards the source; 9.81 N/kg at Earth’s surface.
- Potential V = −G M ÷ r (zero at infinity); escape speed v = √(2 G M ÷ R).
- Circular orbit: v = √(G M ÷ r) and T² = 4π² r³ ÷ (G M).
- r is always the distance from the centre of the planet, not the height above the surface.
Gravitational force, potential and satellite orbits: frequently asked questions
Why is gravitational potential negative?
Potential is defined as zero at infinity. Gravity is attractive, so it does positive work when a mass moves in from infinity, and the work you must do to take the mass back is positive. Therefore the potential at any finite distance is below zero.
Why do astronauts feel weightless in orbit?
Moreover, the astronauts and their spacecraft fall freely around the Earth with the same acceleration. There is no contact force between them and the floor, so they feel weightless although gravity is still about 89 % of its surface value.
What is the difference between g and G?
G is the universal gravitational constant, 6.67 × 10⁻¹¹ N m² kg⁻², the same everywhere. g is the field strength at a particular place, which depends on the mass of the planet and the distance, and is 9.81 N/kg at the Earth’s surface.
Does the escape speed depend on the mass of the rocket?
No. The kinetic energy and the potential energy are both proportional to the mass of the rocket, so it cancels, and the escape speed depends only on the planet’s mass and radius.
What does Kepler’s third law say?
Also, the square of the orbital period is proportional to the cube of the orbital radius for bodies orbiting the same central mass. For circular orbits, T² = 4π² r³ ÷ (G M).
