Pen colours: blue = bodies and diagrams, red = forces, laws and key steps, green = answers and motion, pencil = construction lines. Boxes: red-edged box with “learn it word for word” = a definition the markscheme wants exactly; Why? = the reasoning; Mistake = what students get wrong; yellow note = exam tip or important note. Use G = 6.67 × 10⁻¹¹ N m² kg⁻² unless told otherwise; for the Earth, M = 6.0 × 10²⁴ kg and R = 6.4 × 10⁶ m.
The big picture: a field of force
In A.2 the weight of a body was mg, with g = 9.8 N kg⁻¹. D.1 explains where that force comes from: every mass attracts every other mass. A mass creates a gravitational field in the space around it, and any other mass placed in that field feels a force. The same idea, with charge instead of mass, gives the electric field of D.2.
A mass makes a field; a small test mass placed in it feels a force towards the mass.
Newton’s universal law of gravitationF = G m₁m₂/r² for bodies treated as point masses; r is the distance between their centres; G = 6.67 × 10⁻¹¹ N m² kg⁻². The force is always attractive and acts along the line joining the two masses.
Each body pulls the other with a force of the same size (Newton’s third law), even when the masses are very different.
When can a real body be treated as a point mass? A sphere of uniform density acts as if all its mass were at its centre, so r is measured centre to centre. Any body can also be treated as a point mass if it is very small compared with the distance between the bodies (a satellite seen from a planet).
Exam tip: “State the conditions…” (2 marks)The planet is spherical (and of uniform density), so it acts as a point mass at its centre; the satellite is much smaller than the planet / the distance between them.
Worked example 1A star of mass 3.0 × 10³⁰ kg and a planet of mass 4.0 × 10²⁴ kg are 2.0 × 10¹¹ m apart. Find the gravitational force between them and the acceleration of the planet.
The planet’s acceleration a = F/m = 2.0 × 10²²/4.0 × 10²⁴ = 5.0 × 10⁻³ m s⁻² (towards the star). The star feels the same force but, with its huge mass, hardly accelerates.
Worked example 2The distance between two bodies is tripled. What happens to the force between them?
F ∝ 1/r², so tripling r divides F by 3² = 9.
The force becomes one ninth of its value. (Doubling r would give one quarter.)
Gravitational field strength g at a point · learn it word for word the force per unit mass experienced by a small point mass at that point. g = F/m = GM/r² unit N kg⁻¹ (= m s⁻²); g is a vector pointing towards the mass M.
Why is g the same as the acceleration of free fall? The only force on a falling body is its weight F = mg. Newton’s second law gives a = F/m = g. So the field strength in N kg⁻¹ and the free-fall acceleration in m s⁻² are the same number.
Mistake Writing “gravity” in a definition. Write force per unit mass on a small point mass: “the strength of gravity” earns nothing.
Worked example 3A planet has 8 times the mass of the Earth and 4 times its radius. Find the gravitational field strength at its surface (g on Earth = 9.8 N kg⁻¹).
Write the formula in Earth terms: g = G(8M)/(4R)² = (8/16) × GM/R².
g = 0.5 × 9.8 = 4.9 N kg⁻¹. Do not forget to square the 4.
Worked example 4Find the gravitational field strength 400 km above the Earth’s surface.
Add the height to the radius: r = 6.4 × 10⁶ + 0.4 × 10⁶ = 6.8 × 10⁶ m.
g = GM/r² = 6.67 × 10⁻¹¹ × 6.0 × 10²⁴/(6.8 × 10⁶)² = 8.7 N kg⁻¹. Gravity is far from zero up there: about 90% of its surface value.
A field line shows the direction of the force on a small mass. For gravity the lines always point towards the mass. Where the lines are closer together, the field is stronger.
Around a sphere the field is radial and weakens with distance; close to a planet’s surface it is (nearly) uniform: parallel, equally spaced lines.Two masses: the lines from both bend towards the masses. At the neutral point N between them the two pulls cancel.
Exam tip: sketching field linesDraw arrows towards the mass, perpendicular to the surface, never crossing, with spacing that grows with distance for a radial field and stays equal for a uniform field.
Between two masses, each pulls a test mass towards itself. Somewhere on the line joining them the two pulls are equal in size and opposite in direction: the net field is zero. The pull from the left equals the pull from the right. This point is always closer to the smaller mass.
At P the two pulls balance.
Net field zero at distance x from M: GM/x² = Gm/(d − x)² so x/(d − x) = √(M/m)
Worked example 5A planet of mass 6.0 × 10²⁴ kg has a moon of mass 1/64 of its mass. Their centres are 4.5 × 10⁸ m apart. (a) Find where the net gravitational field is zero. (b) Find the gravitational potential there (HL). (c) Find the potential energy of a 3000 kg probe placed there (HL).
x = 8(d − x) → 9x = 8d → x = 8 × 4.5 × 10⁸/9 = 4.0 × 10⁸ m from the planet (0.5 × 10⁸ m from the moon).
(b) Potential is a scalar: add the two values. V = −GM/x − G(M/64)/(d − x) = −1.0 × 10⁶ − 0.13 × 10⁶ = −1.1 × 10⁶ J kg⁻¹.
(c) Ep = mV = 3000 × (−1.13 × 10⁶) = −3.4 × 10⁹ J.
Important note: zero field on a V–r graphThe field is minus the gradient of the potential–distance graph (see Field and potential). So the net field is zero exactly where the V–r graph is flat: its tangent is horizontal. Between a planet and its moon this is the top of the hump of the V–r curve. Questions ask it both ways: as a word problem (equal pulls) and from a graph (zero gradient).
Mistake Adding the two field strengths at the zero point. They point in opposite directions, so you set them equal; you add them only when finding the potential (a scalar).
Kepler found three rules from careful observations of the planets, decades before Newton explained them with his law of gravitation.
Kepler’s first law: each planet moves on an ellipse with the Sun at one focus. Kepler’s second law: the line joining a planet to the Sun sweeps out equal areas in equal times. Kepler’s third law: the square of the orbital period is proportional to the cube of the semi-major axis: T² ∝ a³ (for a circular orbit a = r).
Equal areas in equal times: near the Sun (perihelion) the planet must move faster to sweep the same area.
Mistake Saying “equal arc lengths” (it is equal areas), putting the Sun at the centre of the ellipse (it is at a focus), or writing T ∝ r³ (it is T² ∝ r³).
Worked example 6Two moons orbit the same planet in circular orbits. Moon A has radius 4.0 × 10⁸ m and period 2.0 days. Moon B has radius 1.0 × 10⁹ m. Find the period of moon B.
Same central mass, so T²/r³ is the same for both: T_B = T_A × (r_B/r_A)^(3/2).
r_B/r_A = 2.5, and 2.5^1.5 = 3.95.
T_B = 2.0 × 3.95 = 7.9 days.
Worked example 7A comet is 1.0 × 10¹¹ m from the Sun at its closest point and 4.0 × 10¹¹ m at its furthest. Find the ratio (speed at the closest point)/(speed at the furthest point).
In a short time t the areas swept are about ½ × (vt) × distance. Equal areas: ½ v_P t × r_P = ½ v_A t × r_A.
v_P/v_A = r_A/r_P = 4.0 × 10¹¹/1.0 × 10¹¹ = 4. The comet is 4 times faster near the Sun.
For a circular orbit the only force on the satellite is gravity, and it points to the centre. So gravity provides the centripetal force. This one idea gives every orbit formula, and deriving Kepler’s third law from it is one of the most common IB questions.
Gravity is the centripetal force.
Show that (learn this derivation)Show that the period T of a satellite in a circular orbit of radius r around a body of mass M is given by T² = 4π²r³/(GM).
Gravity provides the centripetal force: GMm/r² = mv²/r.
Cancel m and one r: v² = GM/r, so v = √(GM/r).
One orbit is a distance 2πr in a time T: v = 2πr/T, so v² = 4π²r²/T².
Equate: 4π²r²/T² = GM/r → T² = 4π²r³/(GM). This is Kepler’s third law: T² ∝ r³ with the constant 4π²/GM.
How the marks go (2–3 marks)✔ Clear evidence that the centripetal force is equated to the gravitational force (GMm/r² = mv²/r or mω²r). ✔ v = 2πr/T (or ω = 2π/T) substituted. ✔ Rearranged to the result. Do not start from an equation of simple harmonic motion or from the answer.
Worked example 8A geostationary satellite orbits above the equator with a period of 24 hours. Find its orbit radius, its height above the surface and its speed.
T = 24 × 3600 = 86 400 s. From T² = 4π²r³/GM: r³ = GMT²/4π².
r = (6.67 × 10⁻¹¹ × 6.0 × 10²⁴ × 86 400²/4π²)^(1/3) = 4.2 × 10⁷ m.
Height = r − R = 4.2 × 10⁷ − 0.64 × 10⁷ = 3.6 × 10⁷ m (36 000 km).
v = 2πr/T = 3.1 × 10³ m s⁻¹. It stays above the same point on the equator, so a fixed dish can point at it.
Worked example 9A moon orbits a planet with radius 6.0 × 10⁸ m and period 3.0 days. Find the mass of the planet.
Rearrange Kepler’s third law: M = 4π²r³/(GT²), with T = 3.0 × 86 400 s.
M = 4π² × (6.0 × 10⁸)³ / (6.67 × 10⁻¹¹ × (2.59 × 10⁵)²) = 1.9 × 10²⁷ kg. Watching a moon is how we “weigh” a planet.
Worked example 10Find the orbital speed of a satellite 500 km above the Earth, and of one that just grazes the surface. Find the period of the grazing orbit.
r = 6.9 × 10⁶ m: v = √(GM/r) = 7.6 × 10³ m s⁻¹.
Grazing, r = 6.4 × 10⁶ m: v = 7.9 × 10³ m s⁻¹: lower orbits are faster.
T = 2πr/v = 2π × 6.4 × 10⁶/7.9 × 10³ = 5.1 × 10³ s ≈ 85 minutes.
Mistake Using the height above the surface as r. In every formula r is measured from the centre of the planet: r = R + h. Markschemes give [1] max when the height is ignored.
Gravitational potential energy Ep of a system · learn it word for word the work done to assemble the system from infinite separation of the components of the system. Ep = −G m₁m₂/r where r is the separation between the centres of mass of the two bodies. Unit J. Ep belongs to the pair of bodies, not to one of them.
Worked example 11Find the change in gravitational potential energy when a 1200 kg satellite is taken from the Earth’s surface to a height of 600 km.
At the surface: Ep = −GMm/R = −6.67 × 10⁻¹¹ × 6.0 × 10²⁴ × 1200/6.4 × 10⁶ = −7.50 × 10¹⁰ J.
At r = 7.0 × 10⁶ m: Ep = −6.86 × 10¹⁰ J.
ΔEp = (−6.86 × 10¹⁰) − (−7.50 × 10¹⁰) = +6.4 × 10⁹ J. Positive: the energy goes up as the satellite moves away, even though both values are negative.
Mistake Using mgh for a height of hundreds of kilometres. g changes with height, so mgh is wrong for large heights (markschemes award [0]). Use −GMm/r at both points and subtract.
Why gravitational potential is always negative (HL)
Gravity attracts, so it takes positive work to pull the mass out to infinity, where the energy is zero.
Important noteGravitational potential and gravitational potential energy are always negative, because gravity is always attractive. 1. Zero is defined at infinity. 2. To take a mass from any point out to infinity we must do positive work against the attraction. 3. So the mass starts with less energy than at infinity: less than zero, which means negative. 4. As r increases, V rises towards zero (it gets less negative), but it never becomes positive. Compare D.2: like charges repel, so the electric potential of a positive charge is positive.
The “potential well”: V = −GM/r is below zero everywhere and rises towards zero far away.
How the marks go: “Explain why the gravitational potential is negative” (2 marks)✔ The potential is defined as zero at infinity. ✔ The force is attractive, so positive work must be done (energy supplied) to move a mass to infinity: the potential everywhere else is less than zero.
Gravitational potential Vg at a point · learn it word for word the work done per unit mass in bringing a small mass from infinity to that point. Vg = −GM/r unit J kg⁻¹; Vg is a scalar. A mass m at that point has Ep = mVg.
Mistake Leaving out “per unit mass”, or saying “from the point to infinity”. Both lose the mark. And never drop the minus sign: “must see negative sign”.
Worked example 12The gravitational potential at a point near a planet is −2.5 × 10⁷ J kg⁻¹. Find the potential energy of a 500 kg body placed there.
Ep = mV = 500 × (−2.5 × 10⁷) = −1.25 × 10¹⁰ J.
Worked example 13Point P is 4.0 × 10⁵ m from the centre of an asteroid of mass 5.0 × 10¹² kg and 1.5 × 10⁵ m from the centre of a smaller asteroid of mass 8.0 × 10¹⁰ kg. Find the gravitational potential at P.
Potential is a scalar: find each one, then add (no directions).
W = mΔVg the work done against the field (by an external force, at a small constant speed) to move a mass m between two points. It does not depend on the path.
Moving from A to B, away from the planet: the external force and the gravitational force point opposite ways.
Work done AGAINST the field (by an external force)
Work done BY the field (by gravity)
Equation
W = mΔV = m(V_B − V_A)
W = −mΔV
Moving away from the planet (V rises)
positive: we supply energy
negative
Moving towards the planet (V falls)
negative
positive: gravity speeds the body up
Worked example 14The gravitational potential at A is −6.0 × 10⁷ J kg⁻¹ and at B it is −2.0 × 10⁷ J kg⁻¹. Find the work done in moving a 150 kg probe from A to B (a) by an external agent, (b) by the gravitational force.
Move a mass m a small distance Δr against the field: the work is mgΔr, and it is also mΔV. So g = ΔV/Δr in size. With the direction (the field points the way V decreases):
g = −ΔVg/Δr the field strength is minus the gradient (minus the slope of the tangent) of the graph of V against r. Steep V–r graph → strong field. Flat V–r graph (zero gradient) → zero field.
Potential between a planet and its moon. At the top of the hump the tangent is horizontal: the net field is zero there (r = 0.75 d in this sketch).
Important noteThe net field is zero where the V–r graph has zero gradient. To read where g = 0 from a graph, find where the tangent is horizontal (the maximum of V between two bodies). To read g at any point, draw the tangent and find its gradient.
Worked example 15The graph shows how the gravitational potential V varies with distance r from the centre of a planet of radius 2.0 × 10⁶ m. (a) Find the work needed to move a 2000 kg probe from the surface to r = 8.0 × 10⁶ m. (b) Find g at the surface from the graph.V–r graph for worked example 15.
W = mΔV = 2000 × (−1.25 × 10⁷ + 5.0 × 10⁷) = 7.5 × 10¹⁰ J.
(b) Draw the tangent at r = 2.0 × 10⁶ m. Its gradient is 25 J kg⁻¹ per m (it rises about 5.0 × 10⁷ J kg⁻¹ over 2.0 × 10⁶ m).
g = −gradient in direction, size 25 N kg⁻¹, pointing towards the planet.
Worked example 16On the V–r graph between a planet and its moon, the potential at the planet’s surface is −6.0 × 10⁷ J kg⁻¹ and at the top of the hump it is −0.5 × 10⁷ J kg⁻¹. Find the minimum energy a 900 kg probe needs to reach the moon from the planet’s surface.
The probe only needs to reach the top of the hump (the zero-field point). After that the moon pulls it in.
E = mΔV = 900 × (−0.5 × 10⁷ − (−6.0 × 10⁷)) = 4.95 × 10¹⁰ J.
An equipotential surface joins points with the same potential. Around a sphere they are spheres (circles on paper).
Equipotentials and field lines around a sphere.
1. Field lines cross equipotentials at 90°. 2. Moving along an equipotential, ΔV = 0, so no work is done. 3. Drawn with equal steps of V, the surfaces are closer together where the field is stronger.
For a satellite of mass m in a circular orbit of radius r: from v² = GM/r, the kinetic energy is Ek = ½mv² = GMm/2r. Add the potential energy:
Ek = GMm/2r Ep = −GMm/r ET = Ek + Ep = −GMm/2r Ep = −2Ek and ET = −Ek. A bound orbit always has negative total energy.
The three energies against orbit radius.
Worked example 17An 800 kg satellite moves from a circular orbit of radius 7.0 × 10⁶ m to one of radius 8.0 × 10⁶ m. Find its total energy in the first orbit and the energy the engines must supply.
New orbit: ET = −2.0 × 10¹⁰ J. Energy supplied = ΔET = +2.9 × 10⁹ J. The satellite ends up slower (Ek falls) but with more potential energy.
Worked example 18A probe of mass m is launched from the surface of a planet (mass M, radius R) with kinetic energy 5GMm/6R. It later settles into a circular orbit with no further work done by its engines. Find the orbit radius.
Total energy at launch: ET = 5GMm/6R − GMm/R = −GMm/6R. It is negative, so the probe cannot escape.
In orbit ET = −GMm/2r. Energy is conserved: −GMm/2r = −GMm/6R → r = 3R.
The total energy decides the path (no air, no engines after launch): ET < 0: the body is bound to the planet. It cannot escape: it falls back to the ground, or it moves in a circle or an ellipse. We see it again. ET = 0: it just escapes on a parabola, reaching infinity with zero speed. It never returns. ET > 0: it escapes on a hyperbola, still moving at infinity. It never returns. Launched straight up, the path is a straight line: it falls back if ET < 0 and escapes if ET ≥ 0. ▶ See it move: Newton’s cannon in the visual guide
Fired sideways from the same mountain with more and more speed: the sign of the total energy decides the path.
Escape speed · learn it word for word the minimum speed a body must have to reach infinity (to escape the gravitational field of the planet), with no further energy supplied. vesc = √(2GM/r) It does not depend on the mass of the body.
Show that (learn this derivation)Show that the escape speed from a point a distance r from the centre of a planet of mass M is vesc = √(2GM/r).
At infinity the body has Ep = 0 and, at the minimum launch speed, Ek = 0: its total energy there is zero.
Energy is conserved, so the total energy at launch is also zero: ½mv² − GMm/r = 0.
vesc = √(2GM/r). In terms of the potential: v = √(−2V).
Mistake Equating forces (GMm/r² = mv²/r) and then putting in a factor of 2. That is the orbit equation, not escape: escape is an energy argument, and the markscheme gives no marks for the force method.
Worked example 19Find the escape speed from the Earth’s surface, and the extra speed a satellite in a 300 km orbit needs to escape.
Surface: v = √(2GM/R) = √(2 × 6.67 × 10⁻¹¹ × 6.0 × 10²⁴/6.4 × 10⁶) = 1.1 × 10⁴ m s⁻¹ (11 km s⁻¹).
At r = 6.7 × 10⁶ m: vesc = √(2GM/r) = 1.09 × 10⁴ m s⁻¹, and it already moves at vorb = √(GM/r) = 7.73 × 10³ m s⁻¹.
Extra speed = 3.2 × 10³ m s⁻¹.
Worked example 20A ball is launched upwards from the surface of a planet (mass M, radius R, no air) with a speed of vesc/√2. How far from the centre does it get?
At the highest point Ek = 0: −GMm/r = −GMm/2R → r = 2R (a height of R above the surface).
A step further: black holesIf a star is so small that its escape speed reaches the speed of light c, nothing can escape: R = 2GM/c². For a star of 10 solar masses (2.0 × 10³¹ kg) this is about 3.0 × 10⁴ m: the whole star squeezed into a 30 km ball.
How the marks go: “Outline the effect of a small drag force on the orbit” (3 marks)✔ The drag force does negative work: the satellite’s total energy decreases (becomes more negative). ✔ ET = −GMm/2r, so the orbit radius decreases. ✔ v = √(GM/r), so the speed increases (kinetic energy increases; potential energy decreases by twice as much). Do not write “the kinetic energy decreases”: it is the total energy that falls.
At 400 km, g is still about 8.7 N kg⁻¹ (worked example 4), so the astronaut is not “out of gravity”. The astronaut and the spacecraft are both in free fall with the same acceleration towards the Earth. The floor does not need to push on the astronaut, so the contact force N = 0 and the astronaut feels weightless. The weight (the gravitational force) is still there: it provides the centripetal force.
Writing the explanations
Many D.1 marks are for definitions and short derivations. Markschemes look for the same key words every time.
“Define gravitational field strength” (1–2 marks)✔ Force per unit mass ✔ acting on a small (test / point) mass placed at that point. Not “the acceleration due to gravity” and not “gravity”.
“Define gravitational potential at a point” (2 marks)✔ The work done per unit mass ✔ in bringing a small mass from infinity to that point. “Energy” is accepted for “work done”. “From the point to infinity” loses the mark.
“Show that T² = 4π²r³/GM” (2–3 marks)✔ Centripetal force equated to the gravitational force: GMm/r² = mv²/r. ✔ v = 2πr/T substituted and rearranged. Show every line: the answer is given, so the marks are for the working.
“Show that the escape speed is √(2GM/R)” / “Define escape speed” (2 marks)✔ Definition: the minimum speed to reach infinity / escape the planet’s gravitational field. Not “to leave the orbit” or “to escape the atmosphere”. ✔ Derivation: total energy = 0 (kinetic energy gained = potential energy lost): ½mv² − GMm/R = 0.
“Explain why the gravitational potential is negative” (2 marks)✔ Potential is zero at infinity. ✔ The force is attractive, so work must be done (energy supplied) to move a mass to infinity.
“Outline the effect of air resistance on a satellite’s orbit” (3 marks)✔ Total energy decreases ✔ radius decreases ✔ speed increases.
“Explain why the planet moves in a circular orbit” (2 marks)✔ A centripetal force is needed for circular motion ✔ and it is provided by the gravitational force between the planet and the star.
“Determine where the gravitational field is zero” (2 marks)✔ Read where the gradient of the V–r graph is zero (the maximum) — or — ✔ equate the two field strengths GM/x² = Gm/(d − x)² and solve.
Common mistakes
Ten traps in D.1 1. Using the height above the surface as r. Always r = R + h. 2. Forgetting to square the factor for the radius in ratio problems (4 times the radius → 16 times smaller g). 3. Dropping the minus sign in V = −GM/r or Ep = −GMm/r. 4. Using mgh for large heights. Use −GMm/r at both points. 5. Deriving escape speed with forces. Escape is an energy argument: total energy = 0. 6. Defining escape speed as “the speed to leave the orbit” or “to escape the atmosphere”. 7. Saying the kinetic energy of a satellite falls because of drag. The total energy falls; the speed rises. 8. Adding field strengths at the point of zero field. Set them equal; add only potentials. 9. Kepler: “equal arc lengths”, “Sun at the centre”, “T ∝ r³”. Correct: equal areas, Sun at a focus, T² ∝ r³. 10. Writing V without a unit. Its unit is J kg⁻¹; field strength is N kg⁻¹.
D.1 and D.2 side by side
Coming next in D.2: the same structure with different letters. Swap G for k = 1/4πε₀, mass for charge, and g for E:
Gravitational (D.1)
Electric (D.2)
F = GMm/r²
F = kQq/r²
g = F/m = GM/r²
E = F/q = kQ/r²
Vg = −GM/r
Ve = kQ/r
Ep = −GMm/r
Ep = kQq/r
g = −ΔVg/Δr
E = −ΔVe/Δr
W = mΔVg
W = qΔVe
always attractive → V always negative
like charges repel → V has the sign of the charge
Practise next
You have read the notes. Now practise, in this order:
Levelled practice: first a ★ at Level 2 in every topic (the goal), then Levels 3 and 4 for a 6 or 7. Hints and worked answers help you as you go.
IB-format practice (Paper 1A · Paper 2): timed, one attempt, and your report goes to your teacher. Do these when the practice goal is reached.