A2 Circular MotionGravitational Fields 5.2.1(a)5.4.1(a)3.6.1.1(a)3.7.1(a)

~17 min Difficulty: 7/10 12 marks

Prior knowledge Circular motionNewton’s law of gravitationorbital mechanics

Problem structure

Part (a): gravitational force expression (1 mark). Part (b): orbital speed derivation (3 marks). Part (c): stellar mass calculation (5 marks). Part (d): direction of force explanation (2 marks). Part (e): assumption (1 mark).

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Two identical stars, each of mass m, orbit their common centre of mass in circular orbits of radius r. The stars are separated by a distance 2r. Their orbital period is T. (a) Write down an expression for the gravitational force between the two stars. (1) (b) Show that the orbital speed v of each star is given by v = √(Gm/4r). (3) (c) The stars are separated by 4.0 × 10¹¹ m and have an orbital period of 2.0 years. Calculate the mass of each star. The gravitational constant G = 6.67 × 10⁻¹¹ N m² kg⁻². (5) (d) Explain why the gravitational force on each star is directed towards the common centre of mass, even though the other star is at a distance 2r away. (2) (e) State one assumption made in this model. (1)

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(a) F = Gm² / (2r)² = Gm² / 4r². (b) Gravitational force provides centripetal force for circular motion of radius r: Gm² / 4r² = mv² / r v² = Gm / 4r ⇒ v = √(Gm/4r). (c) Period T = 2.0 years = 2.0 × 365.25 × 24 × 3600 = 6.31 × 10⁷ s. Separation d = 2r = 4.0 × 10¹¹ m, so r = 2.0 × 10¹¹ m. Using v = 2πr / T and v² = Gm/4r: (2πr/T)² = Gm/4r m = 16π²r³ / (GT²) m = 16π² × (2.0 × 10¹¹)³ / [6.67 × 10⁻¹¹ × (6.31 × 10⁷)²] m = 16π² × 8.0 × 10³³ / [6.67 × 10⁻¹¹ × 3.98 × 10¹⁵] m = 1.26 × 10³⁶ / 2.66 × 10⁵ = 4.74 × 10³⁰ kg. (d) Each star orbits the common centre of mass, which lies midway between them because the stars have equal mass. The gravitational force acts along the line joining the stars and is always directed towards the other star, which is exactly towards the centre of mass for circular motion. (e) Assumption: stars are point masses / spherical / orbit in same plane / no external gravitational influences.

Marking guidance

(a) [1] - Correct expression Gm²/4r² (1) (b) [3] - Equates gravitational and centripetal force (1) - Correct radius of orbit is r (1) - Correct derivation of v = √(Gm/4r) (1) (c) [5] - Converts period to seconds (1) - Correct orbital radius (1) - Correct combination of v = 2πr/T and v² = Gm/4r (1) - Correct substitution (1) - Correct mass ≈ 4.7 × 10³⁰ kg (1) (d) [2] - Centre of mass is midway (1) - Gravitational force directed along line of centres, so towards centre of mass (1) (e) [1] - Valid assumption (1)