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.
Circular and geostationary orbits 8(j)–(k)
Question 1
A satellite is in a circular orbit of radius 7.00 × 10⁶ m around Earth. Find its speed using GM = 3.99 × 10¹⁴ m³ s⁻² and state why it accelerates.
Check the model response
GMm/r² = mv²/r gives v = √(GM/r) = √(3.99 × 10¹⁴/7.00 × 10⁶) = 7.55 × 10³ m s⁻¹. Gravity continually changes the velocity direction, providing inward centripetal acceleration.
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.
Circular and geostationary orbits 8(j)–(k)
Check this idea
Misconception: Any satellite with a 24-hour period is geostationary.
Repair: It must also have a circular equatorial orbit and move west-to-east with Earth's rotation. Then it stays above one longitude and can support continuous communication or weather observation.
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.
Circular and geostationary orbits 8(j)–(k)
Model 1
A satellite orbits Earth at r = 4.22 × 10⁷ m. Find its speed and period, then state the additional conditions and one use needed for a geostationary interpretation.
Check the model response
v = √(GM/r) = 3.07 × 10³ m s⁻¹. T = 2πr/v = 8.64 × 10⁴ s = 24.0 h. It must orbit circularly in the equatorial plane, west-to-east. Remaining over one longitude supports continuous communications or weather monitoring.
guided practice
4. Guided practice
Use each hint only to choose the radial distance, potential difference, energy boundary or orbit equation.
Circular and geostationary orbits 8(j)–(k)
Question 1
Find the speed and period of an Earth satellite in a circular orbit of radius 8.00 × 10⁶ m. Use GM = 3.99 × 10¹⁴ m³ s⁻².
Hint: Set gravity equal to mv²/r, then use T = 2πr/v.
Check the model response
v = √(GM/r) = 7.06 × 10³ m s⁻¹. T = 2π(8.00 × 10⁶)/(7.06 × 10³) = 7.12 × 10³ s = 119 min, so it is not geostationary.
independent practice
5. Independent practice
Solve without repair notes and state the radial origin, sign convention and system boundary used.
Circular and geostationary orbits 8(j)–(k)
Question 1
Show that the period of a circular gravitational orbit satisfies T² = 4π²r³/(GM), then state why a geostationary satellite is useful.
Check the model response
GMm/r² = m(4π²r/T²), so T² = 4π²r³/(GM). A geostationary satellite stays above one longitude, allowing a fixed ground antenna and continuous coverage of the same region.
Practice exit check
6. Practice assessment
Use this as extra closed-book practice, then complete the separate recorded assessment in your plan.
Circular and geostationary orbits 8(j)–(k)
Question 1
A satellite has a circular-orbit radius 9.00 × 10⁶ m around a planet with GM = 5.00 × 10¹⁴ m³ s⁻². Find its speed, then state the period requirement and two further conditions required for it to be geostationary.
Check the model response
v = √(GM/r) = √(5.00 × 10¹⁴/9.00 × 10⁶) = 7.45 × 10³ m s⁻¹. Its period must equal the planet's rotation period; it must also be circular and equatorial, orbiting west-to-east.
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.
Circular and geostationary orbits 8(j)–(k)
Question 1
A satellite's circular-orbit radius is doubled around the same planet. State the factors by which its speed and period change, and explain whether this alone makes it geostationary.
Check the model response
Since v ∝ r⁻¹ᐟ², speed changes by 1/√2. Since T ∝ r³ᐟ², period changes by 2√2. Radius alone is insufficient: the period must match rotation and the orbit must also be circular, equatorial and in the rotation direction.