Gravitational fields
Use Newton's law of gravitation, field strength g = GM/r², potential φ = −GM/r and energy to solve problems on escape speed, circular orbits and geostationary satellites.
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Before you begin
One law of gravitation describes a falling apple, the Moon’s orbit and the satellites that carry our phone signals. The lessons start with Newton’s law and the field strength of a point mass, then define gravitational potential and potential energy and link them to the field through the potential gradient. They use energy to find the escape speed, and circular motion to find orbital speeds, Kepler’s third law and the geostationary orbit.
Be comfortable with: centripetal force, because every orbit calculation sets gravity equal to mv²/r.
Learning goals
- Apply Newton's law of gravitation to point and spherical masses.
- Derive and apply gravitational field strength, including the near-surface model.
- Derive the gravitational field strength due to a point mass from Newton's law of gravitation and the definition of field strength.
- Relate gravitational potential, potential energy and field gradient.
- Analyse escape speed using conservation of energy.
- Analyse circular gravitational orbits and geostationary satellite conditions.
Syllabus statements covered
- recall and use Newton’s law of gravitation in the form F = Gm₁m₂/r²
- derive, from Newton’s law of gravitation and the definition of gravitational field strength, the field strength due to a point mass, g = GM/r²
- recall and use g = GM/r² for the gravitational field strength due to a point mass to solve problems
- show an understanding that near the surface of the Earth, gravitational field strength is approximately constant and equal to the acceleration of free fall
- define gravitational potential at a point as the work done per unit mass by an external force in bringing a small test mass from infinity to that point
- solve problems using the equation φ = −GM/r for the gravitational potential in the field due to a point mass
- show an understanding that that the gravitational potential energy of a system of two point masses is Uᴳ = −GMm/r
- recall that gravitational field strength at a point is equal to the negative potential gradient at that point and use this to solve problems
- analyse problems related to escape velocity by considering energy stores and transfers
- analyse circular orbits in inverse square law fields by relating the gravitational force to the centripetal acceleration it causes
- show an understanding of satellites in geostationary orbit and their applications
Lessons
Work through them in order.
Core lessons
- Gravitational force and field strengthUse Newton's law of gravitation, derive g = GM/r² for a point mass and explain why g is nearly constant near the Earth's surface.
- Gravitational potential and energyDefine gravitational potential, use φ = −GM/r and U = −GMm/r, and find the work needed to move a mass in a field.
- Reading gravitational field and potential graphsUse g = −dφ/dr to find field strength from a potential graph, including between two masses where the fields cancel.
- Escape speed through energy stores and transfersDerive the escape speed from energy conservation, state its assumptions and use total energy to decide whether a body escapes.
- Circular-orbit speed and periodTreat gravity as the centripetal force to derive the orbital speed and Kepler's third law, T² ∝ r³, and compare satellites.
- Geostationary orbits and their usesState the conditions for a geostationary orbit, calculate its radius and height, and describe the uses of geostationary satellites.
Practise and check
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Topic reference
Quick reference
| Idea | Relation | Notes |
|---|---|---|
| Newton’s law of gravitation | F = Gm₁m₂/r² | G = 6.67 × 10⁻¹¹ N m² kg⁻²; r centre to centre |
| field strength of a point mass | g = GM/r² | N kg⁻¹; towards the mass |
| gravitational potential | φ = -GM/r | J kg⁻¹; zero at infinity |
| gravitational potential energy | UG = -GMm/r | of the two masses |
| field and potential | g = -dφ/dr | field points to lower potential |
| escape speed | v_esc = square root of (2GM/r) | total energy zero |
| circular orbit | GMm/r² = mv²/r, v = square root of (GM/r) | gravity is the centripetal force |
| Kepler’s third law | T² = (4π²/GM)r³ | same central mass |
Core knowledge to remember
- Newton’s law of gravitation: two point masses attract each other with a force proportional to the product of their masses and inversely proportional to the square of their separation.
- Gravitational field strength at a point is the gravitational force per unit mass on a mass placed at that point. Near the Earth’s surface it is approximately constant and equal to the acceleration of free fall.
- Gravitational potential at a point is the work done per unit mass by an external force in bringing a small test mass from infinity to that point.
- The gravitational field strength at a point is equal to the negative potential gradient at that point.
- A geostationary satellite has a period equal to the Earth’s rotation period, moves west to east in a circular orbit above the equator, and so stays above one point on the Earth.
Common mistakes
- Using the height above the surface instead of the distance from the centre.
- Treating -20 MJ kg⁻¹ as a higher potential than -5 MJ kg⁻¹.
- Using v² = 2gh with constant g to find the escape speed.
- Thinking a higher orbit needs a higher speed.
- Thinking astronauts in orbit are weightless because there is no gravity there.