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D.1 Gravitational fields

Kepler's three laws, Newton's law of gravitation, gravitational field strength and field lines, orbits, and (HL) gravitational potential and potential energy, equipotentials, escape speed, orbital speed and energy.

Kepler's lawsNewton's law of gravitationField strength gField linesOrbitsPotential (HL)Escape & orbital speed (HL)
RSby Riyaz Sir
mentorriyaz.in
No tag = SL and HLHL ONLY HL students onlyDATA BOOKLET given in the examM1A1 how marks are awardedhighlight = comes up again and againTest yourself at the end

In 30 seconds

  • Every pair of masses attracts: F = Gm1m2/r2. Spheres act as if all their mass were at the centre (for points outside them).
  • Field strength g = F/m = GM/r2, in N kg−1. Field lines show the direction of the force on a test mass.
  • Kepler: orbits are ellipses with the Sun at a focus; equal areas in equal times; T2 ∝ r3, which follows from gravity providing the centripetal force.
  • HL: gravitational potential Vg = −GM/r (zero at infinity), potential energy Ep = −GMm/r, g = −ΔV/Δr, escape speed √(2GM/r) and orbital speed √(GM/r).

By the end you can

  • State Kepler's three laws and Newton's law of gravitation.
  • Calculate gravitational forces and field strengths, and draw field lines.
  • Derive and use T2 ∝ r3 for circular orbits.
  • Use gravitational potential, potential energy and work done in a field. HL
  • Relate field strength to the potential gradient and sketch equipotentials. HL
  • Calculate escape speed, orbital speed and orbital energy, and describe the effect of atmospheric drag. HL

1Kepler and Newton

Sun (at a focus)far: slownear: fastEqual areas swept in equal times (Kepler 2). T² ∝ r³ (Kepler 3).
The planet sweeps out equal areas in equal times, so it moves fastest when it is closest to the Sun.
  • Kepler 1: planets move in ellipses with the Sun at one focus. Kepler 2: a line from the Sun to the planet sweeps equal areas in equal times. Kepler 3: T2 ∝ r3 (r = mean orbital radius).
F = Gm1m2r2 DATA BOOKLET
G = 6.67 × 10−11 N m2 kg−2; r between centres
Worked example

Find the gravitational force between the Earth (5.97 × 1024 kg) and the Moon (7.35 × 1022 kg), 3.84 × 108 m apart.

  1. F = 6.67 × 10−11 × 5.97 × 1024 × 7.35 × 1022/(3.84 × 108)2 M1
  2. = 2.0 × 1020 N A1
  • Point masses: a uniform sphere behaves as a point mass at its centre for anything outside it. Two bodies whose separation is large compared with their sizes can also be treated as points.

2Gravitational field strength

g = Fm = GMr2 DATA BOOKLET
field strength, N kg−1 (the same as m s−2)
Earthradial field: lines point to the centrenear the surface: almost uniform(parallel, equally spaced)
Radial far away, nearly uniform close to the surface.
rgR2R9.8 at the surface¼ of it at 2Routside a sphere:g = GM/r²
Outside the planet g falls as 1/r2. Double the distance from the centre and g drops to a quarter.
Worked example

Find g at the Earth's surface (R = 6.37 × 106 m) and at the height of the International Space Station (400 km).

  1. surface: g = GM/R2 = 6.67 × 10−11 × 5.97 × 1024/(6.37 × 106)2 = 9.8 N kg−1 A1
  2. ISS: r = 6.77 × 106 m, g = 8.7 N kg−1 M1A1
  3. astronauts float because they are in free fall with the station, not because g = 0
Common mistakeUsing the height above the surface as r. In GM/r2, r is measured from the centre of the planet.

3Circular orbits

  • Gravity provides the centripetal force: GMm/r2 = mv2/r = m(4π2r/T2). So T2 = (4π2/GM) r3, Kepler's third law, and the orbiting mass m cancels.
Worked example

Find the orbital radius of a geostationary satellite (T = 24 h).

  1. r3 = GMT2/4π2 = 6.67 × 10−11 × 5.97 × 1024 × 86 4002/4π2 M1
  2. r = 4.2 × 107 m (about 36 000 km above the surface) A1

4Gravitational potential and energy HL

Gravitational potential energy Ep — the work done in bringing the masses together from infinite separation. It is negative because gravity is attractive.

Gravitational potential Vg — the work done per unit mass in bringing a small test mass from infinity to that point. A scalar, zero at infinity.

Ep = −Gm1m2r DATA BOOKLET
J
Vg = −GMr DATA BOOKLET
J kg−1
g = −ΔVgΔr DATA BOOKLET
field strength is minus the potential gradient
W = mΔVg DATA BOOKLET
work done moving mass m between two points
rVRV = −GM/r→ 0 at infinityalways negativeg = −ΔV/Δr(gradient of this graph,with a minus sign)
The potential rises towards zero as you move away; the slope of this curve gives g.
Worked example

How much energy is needed to lift a 1000 kg satellite from the Earth's surface to 400 km (ignore the kinetic energy it needs)?

  1. ΔVg = GM(1/R − 1/r) = 3.98 × 1014 × (1/6.37 × 106 − 1/6.77 × 106) M1
  2. = 3.69 × 106 J kg−1 A1
  3. W = mΔV = 3.7 × 109 J (mgh with g = 9.8 would give 3.9 × 109 J, an overestimate) A1
Mequipotentials (purple)are spheres, alwaysperpendicular tofield linesno work movingalong one
Equipotentials are spaced further apart further out, because the field is weaker there.
  • Equipotential surfaces join points of equal potential. They are always perpendicular to field lines, and no work is done moving along one.

5Escape speed and orbital speed HL

vesc = √2GMr DATA BOOKLET
minimum launch speed to reach infinity (no air resistance)
vorbital = √GMr DATA BOOKLET
speed of a circular orbit of radius r
Worked example

Find the escape speed from the Earth's surface and the orbital speed of the ISS.

  1. vesc = √(2 × 3.98 × 1014/6.37 × 106) = 11.2 km s−1 A1
  2. vorb = √(3.98 × 1014/6.77 × 106) = 7.7 km s−1 A1
  • Orbital energy: Ek = GMm/2r, Ep = −GMm/r, so the total E = −GMm/2r. A lower orbit has more negative total energy but higher speed.
  • Atmospheric drag on a low satellite removes energy. Its total energy becomes more negative, so it falls to a lower orbit, where it actually speeds up, until it re-enters the atmosphere.
Common mistakeDeriving escape speed from the orbital speed. They are different: vesc = √2 × vorbital at the same radius.
Exam tipFor "show that" orbital questions, start from GMm/r2 = mv2/r and cancel m. It is worth one method mark on its own.
Linking questionThe inverse-square law here has exactly the same mathematics as Coulomb's law in D.2: compare g = GM/r2 with E = kQ/r2, and Vg = −GM/r with Ve = kQ/r.
Idea for your IA / explorationUse orbital data for Jupiter's moons (from a planetarium app or published tables) to test Kepler's third law with a log–log graph, and find Jupiter's mass from the gradient.

Test yourself — cover the answers first

Quick check

  1. State Newton's law of gravitation.
  2. What is the unit of g?
  3. If r doubles, what happens to g?
  4. Why is a geostationary satellite always above the same point?
  5. HL Why is gravitational potential negative?
  6. HL What is the ratio vesc/vorbital at the same radius?

Exam-style

  1. A planet has twice the Earth's mass and twice its radius. Find g at its surface.
  2. HL A 500 kg satellite orbits at r = 8.0 × 106 m around the Earth (GM = 3.98 × 1014 m3 s−2). Find its speed, period and total energy.
Show answers
  1. F = Gm1m2/r2, attractive, along the line between centres
  2. N kg−1 (= m s−2)
  3. it falls to a quarter
  4. its period equals the Earth's rotation period (24 h) and it orbits above the equator
  5. it is zero at infinity and work must be done to move a mass out to infinity
  6. √2
  7. g ∝ M/R2
    = 9.8 × 2/4
    = 4.9 N kg−1
  8. v = √(GM/r) = 7.05 km s−1
    T = 2πr/v = 7.1 × 103 s (about 2 h)
    E = −GMm/2r = −3.98 × 1014 × 500/(1.6 × 107) = −1.2 × 1010 J
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Independent notes written by Riyaz Sir. Not produced, endorsed or approved by the International Baccalaureate Organization. Syllabus codes refer to the IB DP subject guides.