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A-Level Newton's gravitation, gravitational field strength, potential and energy, escape velocity and orbits
What the A-Level syllabus expects for Newton's gravitation, gravitational field strength, potential and energy, escape velocity and orbits, and how to practise it.
What the syllabus expects
- Recall Newton's law of gravitation and use it in calculations
Scope: in the form F = Gm1m2/r2 - Take Newton's law of gravitation, pair it with how field strength is defined, and so obtain g = GM/r2 for a point mass
- Put g = GM/r2, the field strength around a point mass, to work in problems
- Understand that just above Earth's surface the gravitational field strength stays nearly fixed and matches the free-fall acceleration
- Define gravitational potential as the work an outside agent does per unit mass to carry a tiny test mass in from infinity
- Handle calculations with φ = -GM/r, the potential around a point mass
- Recognise that a pair of point masses together holds gravitational potential energy U = -GMm/r
- Remember that field strength equals minus the gradient of potential at a location, and use this in problems
- Reason about escape velocity through the lens of energy stores and their transfers
- Examine circular orbits within inverse-square fields by equating gravity to the centripetal acceleration it supplies
- Understand geostationary satellites and where they get put to use
How it's examined
Questions on this topic most often ask you to show, find, define, evaluate. About 5% of the past-paper style questions in Rae's bank for this subject sit in this topic.
Worked examples
Example 1 (3 marks)
A body P of mass m travels around a body Q of mass M on an elliptical path. Show that its total energy E equals E = −GMm/(2a), where 2a is the major-axis length of the ellipse.
Show the worked answer
Total energy E = (1/2)mv² - GMm/r is conserved, and angular momentum L = m v r is conserved. Evaluate at perihelion (r_p = a(1-e), speed v_p) and aphelion (r_a = a(1+e), speed v_a), where r_p + r_a = 2a. Angular momentum: v_p r_p = v_a r_a, so v_a = v_p r_p/r_a. Energy equality: (1/2)v_p² - GM/r_p = (1/2)v_a² - GM/r_a. Substitute v_a: (1/2)v_p²(1 - r_p²/r_a²) = GM(1/r_p - 1/r_a). (1/2)v_p² (r_a² - r_p²)/r_a² = GM(r_a - r_p)/(r_p r_a). Cancel (r_a - r_p): (1/2)v_p² (r_a + r_p)/r_a² = GM/(r_p r_a), giving v_p² = 2GM r_a/[r_p(r_a + r_p)]. Hence E = (1/2)m v_p² - GMm/r_p = (GMm/r_p)[r_a/(r_a + r_p) - 1] = (GMm/r_p)[-r_p/(r_a + r_p)] = -GMm/(r_a + r_p) = -GMm/(2a).
Example 2 (3 marks)
For the binary system, P lies a distance d from star A's centre, the two centres are 2.8 × 10⁸ km apart, and M_A/M_B = 3.0. Find d. [3]
Show the worked answer
Principle: in a binary system the two stars orbit their common centre of mass P, staying on opposite sides of it, so they go round with the same angular velocity ω and the same period. The single gravitational force between them acts on both stars, so by Newton's third law each star feels a centripetal force of the same size. Step 1 - define the two orbit radii. Let the separation of the centres be D = 2.8 × 10⁸ km. Star A orbits at radius d from P (this is what we want). Since P lies on the line joining the centres, star B must orbit at radius D − d = (2.8 × 10⁸ − d) km. Step 2 - write the centripetal force on each star, using F = M ω² r. For star A: F_A = M_A ω² d For star B: F_B = M_B ω² (D − d) Step 3 - set them equal, since the same gravitational force provides both. M_A ω² d = M_B ω² (D − d) The ω² is common to both sides, so cancel it: M_A d = M_B (D − d) Step 4 - bring in the mass ratio. Divide both sides by M_B: (M_A / M_B) d = D − d and M_A / M_B = 3.0, so 3.0 d = 2.8 × 10⁸ − d (Notice this says the heavier star sits closer to P, which is what you would expect.) Step 5 - solve for d. Add d to both sides: 3.0 d + d = 2.8 × 10⁸ 4.0 d = 2.8 × 10⁸ km Divide both sides by 4.0: d = 2.8 × 10⁸ / 4.0 d = 7.0 × 10⁷ km So P lies 7.0 × 10⁷ km from the centre of star A.
Example 3 (3 marks)
For the binary system (ω ≈ 4.98 × 10⁻⁸ rad s⁻¹, centre separation 2.8 × 10⁸ km, M_A/M_B = 3.0, d = 7.0 × 10⁷ km), find the mass M_B of star B. [3]
Show the worked answer
Principle: the gravitational attraction between the two stars is what keeps each of them in its circular orbit about the common centre of mass P. So for star A we can write Newton's law of gravitation = centripetal force needed by A. Newton's law of gravitation: F = G M_A M_B / r², where r is the separation of the two CENTRES and G = 6.67 × 10⁻¹¹ N m² kg⁻². Centripetal force in terms of angular velocity: F = M_A ω² d, where d is the radius of star A's own orbit (its distance from P), not the separation. Step 1 - convert every length to metres before substituting (the data are in km, and 1 km = 10³ m). separation r = 2.8 × 10⁸ km = 2.8 × 10¹¹ m star A's orbit radius d = 7.0 × 10⁷ km = 7.0 × 10¹⁰ m Step 2 - equate the two expressions for the force on star A. G M_A M_B / r² = M_A ω² d The mass M_A of star A appears on both sides, so cancel it (this is why we do not need M_A itself, only the ratio was needed earlier): G M_B / r² = ω² d Step 3 - rearrange for M_B. Multiply both sides by r², then divide by G: M_B = ω² d r² / G Step 4 - substitute, working out one factor at a time. ω² = (4.98 × 10⁻⁸)² = 2.48 × 10⁻¹⁵ rad² s⁻² ω² d = 2.48 × 10⁻¹⁵ × 7.0 × 10¹⁰ = 1.74 × 10⁻⁴ m s⁻² r² = (2.8 × 10¹¹)² = 7.84 × 10²² m² ω² d r² = 1.74 × 10⁻⁴ × 7.84 × 10²² = 1.36 × 10¹⁹ Step 5 - divide by G. M_B = 1.36 × 10¹⁹ / 6.67 × 10⁻¹¹ M_B = 2.04 × 10²⁹ kg The mass of star B is about 2.0 × 10²⁹ kg (roughly a tenth of the Sun's mass).
More worked questions on this topic
- A lone uniform spherical planet has surface gravitational potential φ. A particle of mass m is (3 marks)
More A-Level H2 Physics topics
Physical quantities, units, measurement uncertainty and vector basics · Types of force, turning effects and conditions for equilibrium · Kinematics, uniformly accelerated motion, momentum and Newton's laws · Energy stores and transfers, work, kinetic and potential energy, fields, power and efficiency · Falling freely, the gravitational potential energy of a uniform field, and how air resistance changes the motion · Impulse and the conservation of momentum and energy · all of A-Level H2 Physics