Gravitational fields: force, potential, escape and orbits
Key idea: Use one centre-to-centre radial model to connect gravitational force and field strength to negative potential, energy changes, escape and circular orbits.
Before you start: Circular Motion objective chainEnergy & Fields objective chain
By the end, you can
- Use Newton's law of gravitation and derive gravitational field strength for a point or spherical source mass.
- Define gravitational potential, use its negative sign and connect field strength to negative potential gradient.
- Analyse escape speed using conservation of energy rather than a constant-g approximation.
- Relate gravity to centripetal acceleration and identify the conditions and applications of geostationary satellites.
Starting-point self-check
1. Check your starting point
Attempt all four groups without notes and mark the first radial distance, sign, energy boundary or orbit condition you could not justify. Use the recorded topic diagnostic above when you want scoring and a personalised repair plan.
Escape speed through energy stores and transfers 8(i)
Question 1
Use energy conservation to find escape speed from Earth's surface, given GM = 3.99 × 10¹⁴ m³ s⁻² and R = 6.37 × 10⁶ m.
Check the model response
For minimum escape, final kinetic and potential energies at infinity are zero. Thus ½mv² − GMm/R = 0, so v = √(2GM/R) = 1.12 × 10⁴ m s⁻¹.
repair
2. Repair the eight common breaks
Use only the repair matching an error, then redraw the radial reference and energy or force model before retrying.
Escape speed through energy stores and transfers 8(i)
Check this idea
Misconception: Escape speed is the speed needed to keep an engine thrust equal to weight.
Repair: Escape speed is an initial-speed energy condition with no further propulsion: total energy is just zero for arrival at infinity with zero speed.
Check this idea
Misconception: A more massive spacecraft needs a greater escape speed.
Repair: Both kinetic and gravitational potential energy contain spacecraft mass, so it cancels from vesc = √(2GM/r). The required energy still increases with spacecraft mass.
worked example
3. Follow four worked models
Follow how each solution fixes the radial origin, zero-potential reference, system energy and inward resultant before calculating.
Escape speed through energy stores and transfers 8(i)
Model 1
Find the escape speed from r = 7.00 × 10⁶ m around Earth, using GM = 3.99 × 10¹⁴ m³ s⁻².
Check the model response
Set ½mv² − GMm/r = 0 for minimum escape. Then vesc = √(2GM/r) = √[2(3.99 × 10¹⁴)/(7.00 × 10⁶)] = 1.07 × 10⁴ m s⁻¹.
guided practice
4. Guided practice
Use each hint only to choose the radial distance, potential difference, energy boundary or orbit equation.
Escape speed through energy stores and transfers 8(i)
Question 1
Mars has GM = 4.28 × 10¹³ m³ s⁻² and radius 3.39 × 10⁶ m. Find its surface escape speed.
Hint: Use zero total energy at infinity, not constant surface g over an infinite distance.
Check the model response
vesc = √(2GM/R) = √[2(4.28 × 10¹³)/(3.39 × 10⁶)] = 5.03 × 10³ m s⁻¹.
independent practice
5. Independent practice
Solve without repair notes and state the radial origin, sign convention and system boundary used.
Escape speed through energy stores and transfers 8(i)
Question 1
A planet has surface field strength 15.0 N kg⁻¹ and radius 4.00 × 10⁶ m. Derive an expression using g and R, then find its escape speed.
Check the model response
At the surface GM = gR², so vesc = √(2GM/R) = √(2gR). Thus vesc = √[2(15.0)(4.00 × 10⁶)] = 1.10 × 10⁴ m s⁻¹.
Practice exit check
6. Practice assessment
Use this as extra closed-book practice, then complete the separate recorded assessment in your plan.
Escape speed through energy stores and transfers 8(i)
Question 1
A moon has GM = 6.00 × 10¹² m³ s⁻² and radius 1.50 × 10⁶ m. Find the minimum escape speed and explain why it is independent of the launched mass.
Check the model response
vesc = √(2GM/R) = √[2(6.00 × 10¹²)/(1.50 × 10⁶)] = 2.83 × 10³ m s⁻¹. Mass cancels when ½mv² is equated to GMm/R.
Re-test practice
7. Delayed re-test practice
Return after at least three days and solve these fresh contexts without reopening earlier responses. The recorded plan enforces the delay and uses a separate re-test family for selected-response skill-group evidence.
Escape speed through energy stores and transfers 8(i)
Question 1
At a distance where circular orbital speed is 4.00 km s⁻¹, find the escape speed from the same position and justify the relation.
Check the model response
Circular speed satisfies vc² = GM/r, while escape speed satisfies vesc² = 2GM/r. Therefore vesc = √2 vc = 5.66 km s⁻¹.