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.
Gravitational force and field strength 8(a)–(d)
Question 1
Earth and the Moon have masses 5.97 × 10²⁴ kg and 7.35 × 10²² kg, with centre separation 3.84 × 10⁸ m. Calculate their gravitational force using G = 6.67 × 10⁻¹¹ N m² kg⁻².
Check the model response
F = GMm/r² = 6.67 × 10⁻¹¹(5.97 × 10²⁴)(7.35 × 10²²)/(3.84 × 10⁸)² = 1.98 × 10²⁰ N. Each body experiences this attractive force towards the other.
Question 2
Derive the field strength g due to a point mass M from Newton's law and g = F/m, then explain why near Earth's surface g also equals free-fall acceleration.
Check the model response
F = GMm/r², so g = F/m = GM/r²; the test mass cancels. Near Earth's surface, changes in r are small compared with Earth's radius, so g is approximately constant. A freely falling mass has F = mg = ma, hence a = g.
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.
Gravitational force and field strength 8(a)–(d)
Check this idea
Misconception: Gravitational force is inversely proportional to separation.
Repair: It is inversely proportional to the square of centre-to-centre separation: doubling r reduces force to one quarter.
Check this idea
Misconception: The r in GM/r² is altitude above a planet's surface.
Repair: For a spherical planet, r is measured from its centre. At altitude h, use r = R + h.
Check this idea
Misconception: A heavier test mass experiences a stronger gravitational field strength.
Repair: It experiences a larger force, but g = F/m = GM/r² is a property of the source and position, not the test 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.
Gravitational force and field strength 8(a)–(d)
Model 1
A planet's field strength is 4.00 N kg⁻¹ at 3.00 × 10⁶ m from its centre. Estimate its mass using G = 6.67 × 10⁻¹¹ N m² kg⁻².
Check the model response
From g = GM/r², M = gr²/G = 4.00(3.00 × 10⁶)²/(6.67 × 10⁻¹¹) = 5.40 × 10²³ kg.
guided practice
4. Guided practice
Use each hint only to choose the radial distance, potential difference, energy boundary or orbit equation.
Gravitational force and field strength 8(a)–(d)
Question 1
A spherical planet has mass 6.00 × 10²⁴ kg and radius 6.50 × 10⁶ m. Find g at altitude 5.00 × 10⁵ m.
Hint: The inverse-square distance is measured from the planet's centre.
Check the model response
r = 6.50 × 10⁶ + 5.00 × 10⁵ = 7.00 × 10⁶ m. Thus g = GM/r² = 6.67 × 10⁻¹¹(6.00 × 10²⁴)/(7.00 × 10⁶)² = 8.17 N kg⁻¹ towards the centre.
independent practice
5. Independent practice
Solve without repair notes and state the radial origin, sign convention and system boundary used.
Gravitational force and field strength 8(a)–(d)
Question 1
At a planet's surface, g = 12.0 N kg⁻¹ and radius R = 8.00 × 10⁶ m. Find g at altitude R above the surface.
Check the model response
The new centre distance is r = R + R = 2R. Since g ∝ 1/r², g = 12.0(R/2R)² = 3.00 N kg⁻¹.
Practice exit check
6. Practice assessment
Use this as extra closed-book practice, then complete the separate recorded assessment in your plan.
Gravitational force and field strength 8(a)–(d)
Question 1
A 4.00 kg mass and a 7.00 kg mass are 0.250 m apart. Find their gravitational force and the field strength at the 4.00 kg mass due to the 7.00 kg mass.
Check the model response
F = Gm₁m₂/r² = 6.67 × 10⁻¹¹(4.00)(7.00)/(0.250)² = 2.99 × 10⁻⁸ N. The field due to 7.00 kg is g = F/4.00 = 7.47 × 10⁻⁹ N kg⁻¹ towards it, equivalently G(7.00)/r².
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.
Gravitational force and field strength 8(a)–(d)
Question 1
A planet has mass 4.80 × 10²⁴ kg and radius 5.60 × 10⁶ m. Find the field strength at altitude 1.40 × 10⁶ m and the force on a 1200 kg probe there.
Check the model response
r = 7.00 × 10⁶ m. Thus g = GM/r² = 6.67 × 10⁻¹¹(4.80 × 10²⁴)/(7.00 × 10⁶)² = 6.53 N kg⁻¹ and F = mg = 7.84 × 10³ N inward.