Lesson 8: Gravitation and OrbitsAP

The same inverse-square law that makes an apple fall also holds the Moon, steers probes to Mars, and sets the timetable of every satellite. Gravitation (CED Unit 3) is AP Physics 1's first genuinely universal law, and its exam questions reward proportional reasoning more than calculator work.

Newton's law of universal gravitation

Every mass attracts every other mass with F_g = G m_1 m_2 / r², where G = 6.67 × 10⁻¹¹ N·m²/kg² and r is the distance between centres. The force is mutual (a third-law pair), always attractive, and falls with the square of separation: double the distance, quarter the force. Near a planet's surface, collecting all the constants gives the familiar g = GM/r², and an object's weight mg is just the gravitational force evaluated at that r. Because r is measured from the centre, altitude problems must add the planet's radius — the single most common numerical slip in this unit.

Orbits: gravity as the centripetal force

A circular orbit is free fall with enough sideways speed: gravity alone supplies the centripetal requirement, GMm/r² = mv²/r. The orbiting mass cancels, giving v = √(GM/r) — orbit speed depends only on the central mass and radius. Combining with v = 2πr/T yields Kepler's third law: T² = (4π²/GM) r³, so T² ∝ r³ for any one central body. Kepler's first two laws (elliptical orbits; equal areas in equal times — really conservation of angular momentum in disguise) complete the set the CED expects you to use qualitatively.

Higher orbits are slower — counter-intuitive but inescapable from v = √(GM/r) — and take longer still because the circle is bigger. Geostationary satellites exploit T = 24 h: Kepler's law then fixes r ≈ 4.2 × 10⁷ m from Earth's centre (about 6.6 Earth radii), which is why they hover over one longitude and why there is exactly one geostationary ring.

Fields and apparent weightlessness

It is often cleaner to think of a planet as producing a gravitational field g = GM/r² at each point; any mass placed there feels F = mg_local. Astronauts on the International Space Station are not "beyond gravity": at 400 km altitude g is still about 8.7 m/s², nearly 90% of its surface value. They float because station, crew and contents are all in continuous free fall around Earth together — the Lesson 5 elevator with N = 0, extended indefinitely.

Worked example 1 — g at the space station's altitude

Earth: M = 5.97 × 10²⁴ kg, radius 6.37 × 10⁶ m. The ISS orbits 4.0 × 10⁵ m above the surface. Find g there. Given: r = 6.37 × 10⁶ + 0.40 × 10⁶ = 6.77 × 10⁶ m (centres, not surface!).

g = GM/r² = (6.67 × 10⁻¹¹ × 5.97 × 10²⁴) / (6.77 × 10⁶)²
  = 3.98 × 10¹⁴ / 4.58 × 10¹³ ≈ 8.7 m/s²

Check: the answer is a little below the surface value 9.8 m/s², reduced by the factor (6.37/6.77)² ≈ 0.885 — consistent, and nowhere near zero.

Worked example 2 — orbital speed and period of the ISS

Using the same orbit radius, find the ISS's speed and period. Given: r = 6.77 × 10⁶ m, GM = 3.98 × 10¹⁴ m³/s².

v = √(GM/r) = √(3.98 × 10¹⁴ / 6.77 × 10⁶) = √(5.88 × 10⁷) ≈ 7,670 m/s
T = 2πr/v = 2π(6.77 × 10⁶) / 7,670 ≈ 5,550 s ≈ 92 min

Check: about 7.7 km/s and roughly an hour and a half per orbit — matching the well-known ~16 sunrises per day aboard the station (24 h / 1.53 h ≈ 16 ✓).

Real-world application: GPS and the geostationary ring

GPS satellites sit in ~12-hour orbits (T = ½ day), so Kepler's law puts them at r ≈ 2.66 × 10⁷ m — about 20,000 km up — moving at 3.9 km/s. Their clocks must be corrected for both special and general relativity to keep metre-level accuracy, a striking case of orbital mechanics meeting modern physics. Weather and communications satellites instead use the geostationary radius computed above; a satellite placed even slightly inside or outside it drifts east or west within weeks, which is why station-keeping fuel limits satellite lifetimes.

Experimental design note

You cannot drop a planet in the lab, but Kepler's third law is testable with data: given a table of moons or planets (period and orbital radius), plot T² against r³. A straight line through the origin confirms the law, and its slope 4π²/(GM) yields the central mass — this is genuinely how astronomers "weigh" planets that have moons, and a favourite data-analysis FRQ format.

Common mistakes (from College Board scoring commentary)

• Using altitude instead of centre-to-centre distance in F = Gm₁m₂/r² and g = GM/r².
• Claiming orbiting astronauts experience no gravity — they experience ~90% of surface gravity; they are in free fall.
• Believing higher orbits are faster; v = √(GM/r) says the opposite.
• Applying T² ∝ r³ across different central masses (the Moon's orbit and a Jupiter-moon orbit have different constants).

AP Exam Experience

MCQ 1. A satellite of mass m orbits Earth (mass M) in a circle of radius 2R, where R is Earth's radius. The gravitational force on the satellite is:

• (A) GMm/R²
• (B) GMm/(2R²)
• (C) GMm/(4R²)
• (D) 2GMm/R²

Answer: (C). r in the law is the centre distance 2R, and the square gives (2R)² = 4R². (B) is the unsquared trap.

MCQ 2. Planet X has twice Earth's mass and twice Earth's radius. The surface gravity on X is:

• (A) 2g
• (B) g
• (C) g/2
• (D) g/4

Answer: (C). g = GM/r²: the mass factor doubles g, the radius factor quarters it, so g_X = (2/4)g = g/2.

MCQ 3. Two satellites orbit the same planet, one at radius r and one at radius 4r. The ratio of their periods (T at 4r : T at r) is:

• (A) 2 : 1
• (B) 4 : 1
• (C) 8 : 1
• (D) 16 : 1

Answer: (C). T² ∝ r³, so T ∝ r^1.5; a factor of 4 in r gives 4^1.5 = 8 in T.

FRQ. A planet of mass M has a moon in a circular orbit of radius r with period T. A second moon orbits the same planet with period 8T. (a) Determine the orbital radius of the second moon in terms of r. (b) Determine the ratio of the moons' orbital speeds (v_second : v_first). (c) A probe lands on the planet's surface (radius R_p) and measures surface gravity g_p. Write an expression for the planet's mass M in terms of g_p, R_p and G. (d) Explain why the mass of a moon does not appear in the expression for its orbital period. Scoring guideline. (a) 1 point: T² ∝ r³, so r_second = r × (8)^(2/3) = 4r. (b) 1 point: v = 2πr/T, so v_second/v_first = (4r/8T)/(r/T) = 1/2. (c) 1 point: g_p = GM/R_p² → M = g_p R_p²/G. (d) 1 point: in GMm/r² = mv²/r (or the period derivation) the moon's mass m cancels — gravity's pull and the moon's inertia both scale with m, so the orbit is fixed by the planet's mass and the radius alone.

Worked example 3 — building the geostationary orbit

Find the radius and speed of a satellite whose period equals Earth's rotation (T = 86,400 s), using GM = 3.98 × 10¹⁴ m³/s².

Kepler: T² = 4π²r³/(GM) → r³ = GM T²/(4π²) = (3.98 × 10¹⁴)(7.46 × 10⁹)/39.5 ≈ 7.5 × 10²² m³
r ≈ 4.22 × 10⁷ m from Earth's centre → altitude ≈ 4.22 × 10⁷ − 6.37 × 10⁶ ≈ 3.59 × 10⁷ m (≈ 35,900 km)
v = 2πr/T = 2π × 4.22 × 10⁷/86,400 ≈ 3,070 m/s

Check: the speed (3.1 km/s) is much less than low-orbit speed (7.7 km/s) — higher orbit, slower satellite, exactly as v = √(GM/r) demands; yet it keeps pace with the rotating ground because its circle is bigger. The altitude ~36,000 km is the famous figure quoted for communications satellites ✓.

Fields, Kepler's second law, and a glimpse ahead

The field picture — each planet wrapped in a field g = GM/r² pointing inward — repays the small effort of adopting it: force questions become two-step (field at the point, then F = m g_field for whatever sits there), and comparing planets becomes comparing fields. Kepler's second law (a planet sweeps equal areas in equal times, so it moves faster when nearer the Sun) looks like geometry but is secretly Lesson 14: gravity exerts no torque about the Sun, so angular momentum L = mvr is conserved, and a smaller r forces a larger v. The examiners know students who see that connection; it is the kind of cross-unit thread the translation questions are built from.

Worked example 4 — weighing in on Mars

Mars has mass 6.42 × 10²³ kg and radius 3.39 × 10⁶ m. Find (a) the surface gravity on Mars, (b) the weight of a 70 kg astronaut there, (c) the mass of the astronaut there.

(a) g = GM/r² = (6.67 × 10⁻¹¹ × 6.42 × 10²³)/(3.39 × 10⁶)² = 4.28 × 10¹³/1.15 × 10¹³ ≈ 3.7 m/s²
(b) Weight = mg = 70 × 3.7 ≈ 260 N (versus ≈ 690 N on Earth)
(c) Mass is unchanged: 70 kg — mass is not a force and does not care which planet hosts it

Check: 3.7/9.8 ≈ 0.38 — matching the widely quoted "Mars gravity is about 38% of Earth's" ✓. An astronaut who can jump 0.50 m vertically on Earth (takeoff speed fixed by her legs) would rise v²/(2g) with the same takeoff speed: 0.50 × 9.8/3.7 ≈ 1.3 m on Mars — the arithmetic behind every "low-gravity bounce" video from Apollo (the Moon's 1.6 m/s² does even better).

Escape velocity: when the orbit never closes

Give a projectile enough speed and gravity, though it bends its path forever, never quite closes it: at v_esc = √(2GM/r) the object escapes to infinity with nothing left over. For Earth that is √(2 × 3.98 × 10¹⁴/6.37 × 10⁶) ≈ 11.2 km/s — note it is exactly √2 times the surface-skimming orbital speed (7.9 km/s), a ratio worth remembering because it recurs in exam options. Escape velocity is an energy statement in disguise (½mv² = GMm/r), so it previews Unit 5's methods; a black hole is the object whose escape velocity, computed this Newtonian way, exceeds the speed of light.