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.
Potential, potential energy and field gradient 8(e)–(h)
Question 1
At r = 7.00 × 10⁶ m from Earth's centre, use GM = 3.99 × 10¹⁴ m³ s⁻² to find gravitational potential and the potential energy of a 500 kg spacecraft.
Check the model response
φ = −GM/r = −3.99 × 10¹⁴/(7.00 × 10⁶) = −5.70 × 10⁷ J kg⁻¹. U = mφ = 500(−5.70 × 10⁷) = −2.85 × 10¹⁰ J.
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.
Potential, potential energy and field gradient 8(e)–(h)
Check this idea
Misconception: Gravitational potential is positive because gravity is attractive.
Repair: With zero at infinity, φ = −GM/r. A bound position has negative potential because external work is required to move a unit mass to infinity.
Check this idea
Misconception: Field strength equals the positive slope of a potential–distance graph.
Repair: Radially, g = −dφ/dr. Potential rises towards zero as r increases, so its positive outward slope corresponds to an inward field.
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.
Potential, potential energy and field gradient 8(e)–(h)
Model 1
A 600 kg spacecraft moves slowly from Earth's surface at 6.37 × 10⁶ m to r = 7.00 × 10⁶ m. Find the minimum external work, using GM = 3.99 × 10¹⁴ m³ s⁻².
Check the model response
φ₁ = −GM/r₁ = −6.26 × 10⁷ J kg⁻¹ and φ₂ = −5.70 × 10⁷ J kg⁻¹. For negligible kinetic-energy change, external work = ΔU = m(φ₂ − φ₁) = 600(5.64 × 10⁶) = 3.38 × 10⁹ J. Potential becomes less negative.
guided practice
4. Guided practice
Use each hint only to choose the radial distance, potential difference, energy boundary or orbit equation.
Potential, potential energy and field gradient 8(e)–(h)
Question 1
Potential rises from −6.00 × 10⁷ to −5.80 × 10⁷ J kg⁻¹ over an outward radial interval of 2.00 × 10⁵ m. Estimate the radial field strength.
Hint: Calculate dφ/dr, then apply the negative-gradient sign.
Check the model response
dφ/dr ≈ [2.00 × 10⁶]/[2.00 × 10⁵] = +10.0 J kg⁻¹ m⁻¹. Hence g = −dφ/dr = −10.0 N kg⁻¹, where the negative sign means inward.
independent practice
5. Independent practice
Solve without repair notes and state the radial origin, sign convention and system boundary used.
Potential, potential energy and field gradient 8(e)–(h)
Question 1
For GM = 3.20 × 10¹⁴ m³ s⁻², find φ at r = 8.00 × 10⁶ m and the energy needed to move a 250 kg mass slowly from there to infinity.
Check the model response
φ = −GM/r = −4.00 × 10⁷ J kg⁻¹ and U = mφ = −1.00 × 10¹⁰ J. The minimum external work to reach zero potential energy at infinity is +1.00 × 10¹⁰ J.
Practice exit check
6. Practice assessment
Use this as extra closed-book practice, then complete the separate recorded assessment in your plan.
Potential, potential energy and field gradient 8(e)–(h)
Question 1
Define gravitational potential and explain, using φ = −GM/r and g = −dφ/dr, the signs of potential and radial field outside an isolated spherical mass.
Check the model response
Potential is external work done per unit mass in bringing a small test mass slowly from infinity to the point. With zero at infinity, φ = −GM/r is negative. It increases outward, so dφ/dr is positive and g = −dφ/dr points inward.
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.
Potential, potential energy and field gradient 8(e)–(h)
Question 1
A 400 kg probe moves slowly from r = 8.00 × 10⁶ m to r = 1.20 × 10⁷ m around a body with GM = 4.00 × 10¹⁴ m³ s⁻². Find its potential-energy change.
Check the model response
φ₁ = −5.00 × 10⁷ and φ₂ = −3.33 × 10⁷ J kg⁻¹. ΔU = m(φ₂ − φ₁) = 400(1.67 × 10⁷) = +6.67 × 10⁹ J.