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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

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.

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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]

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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]

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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).

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