Educerie
Start free
Level

4 higher-level sections hidden.

Educerie · IB Diploma · Physics

Theme D Fields · D.1 Gravitational fields

Level
SL and HL. Sections 9, 10, 11 and 12 are HL only. If you are SL, skip them; nothing in your papers tests them.
Themes (key concepts)
forces, energy. Gravity is the force every mass exerts on every other, described by a field that fills space; at HL the same field is described by energy, through potential, which is what decides the fuel a rocket needs and whether a satellite stays in orbit.
The question this unit answers
how is a gravitational field measured and described, and how does that description let us put satellites in orbit and send spacecraft across the solar system?
Where it is examined
Paper 1A multiple choice on inverse-square scaling and field lines; Paper 1B, where orbital data are linearised as T² against r³; Paper 2 parts of 2 to 7 marks: calculate a force or a field strength, find where two fields cancel, derive Kepler's third law. HL Paper 2 adds potential, equipotentials, escape and orbital speeds, and the energy changes of satellites, often in one extended question.

What you must be able to do

You must be able toLevelWhat it looks like in the exam
State and apply Kepler's three lawsSL, HL"State Kepler's third law" (1 mark); explain why a planet is fastest nearest the Sun (2 marks)
Use F = Gm₁m₂/r² for point massesSL, HLCalculate the force between two bodies (2 marks); scaling questions in Paper 1A
State when extended bodies can be treated as point massesSL, HL"Outline why the Earth may be treated as a point mass" (1–2 marks)
Define g and use g = F/m = GM/r²SL, HLg at a height above a planet (2 marks)
Sketch and interpret gravitational field linesSL, HLRadial and uniform fields (2 marks)
Find the resultant g on the line joining two bodiesSL, HLLocate the point where g = 0 (3 marks)
Derive and use T² ∝ r³ for circular orbitsSL, HL"Show that T² = 4π²r³/GM" (3 marks); mass of a planet from a T²–r³ graph (Paper 1B)
Define gravitational potential energy and potential; use Ep = −Gm₁m₂/r and Vg = −GM/rHL only"Explain why Ep is negative" (2 marks); calculate Vg
Use g = −ΔVg/Δr and W = mΔVgHL onlyg from the gradient of a Vg–r graph; work to raise a mass (2–3 marks)
Draw equipotentials and relate them to field linesHL onlySketch equipotentials round a planet (2 marks)
Use vorbital = √(GM/r) and vesc = √(2GM/r)HL onlyOrbital speed; escape speed (2–3 marks)
Calculate energy changes: launching to orbit, changing orbit, escapingHL onlyExtended Paper 2 question (4–7 marks)
Explain the effect of atmospheric drag on an orbitHL only"Explain why the satellite's speed increases" (3 marks)

Before you start

You need circular motion from A.2: a body moving in a circle at speed v needs a centripetal force mv²/r, and v = 2πr/T. At HL you need work and energy from A.3, including the idea that work done on a system increases its energy. Have the data booklet open at D.1 for the equations and for G = 6.67 × 10⁻¹¹ N m² kg⁻².

1The idea in one paragraph

Every mass attracts every other mass. Kepler described the result for the planets: elliptical orbits, faster when closer to the Sun, and a fixed link between orbit size and period. Newton explained it with one law: the force is proportional to both masses and falls with the square of the distance. To describe the pull at a point without naming the body being pulled, we use the gravitational field strength g, the force per kilogram, and draw it with field lines. For circular orbits, gravity supplies the centripetal force, and Kepler's third law follows. At HL the field is also described by energy: gravitational potential is the work per kilogram needed to bring a mass in from infinity, it is negative everywhere, its gradient is the field strength, and it decides orbital energies and the speed needed to escape.

2Kepler's three laws

Kepler found these laws from observations of the planets, before anyone could explain them. Figure 1 shows the first two.

Figure 1 · Kepler's first and second laws Figure 1 · Kepler's first and second laws Sun, at one focus other focus planet short, wide sector: fast long, thin sector: slow the two shaded areas are equal; each is swept out in the same time Equal times sweep out equal areas, so the planet moves fastest nearest the Sun. Ellipse exaggerated.
Figure 1 · Kepler's first and second laws
  1. The law of orbits. Each planet moves in an ellipse with the Sun at one focus. (A circle is an ellipse with both foci at the centre. Most planetary orbits are nearly circular; Figure 1 exaggerates the shape.)
  2. The law of areas. The line from the Sun to a planet sweeps out equal areas in equal times. Near the Sun the line is short, so to sweep the same area it must turn through a larger angle: the planet moves fastest when closest to the Sun and slowest when furthest away.
  3. The law of periods. The square of the orbital period is proportional to the cube of the mean distance from the Sun: T² ∝ r³, with the same constant for every planet of the Sun.

The third law holds for anything orbiting one central body, such as Jupiter's moons; the constant depends only on the central mass (section 8).

3Newton's universal law of gravitation

Newton proposed that every pair of point masses attracts each other with a force

F = G m₁ m₂ / r²

where m₁ and m₂ are the masses, r is the distance between them, and G = 6.67 × 10⁻¹¹ N m² kg⁻² is the gravitational constant.

  • The force is always attractive, along the line joining the two masses.
  • It is proportional to each mass: double one mass, double the force.
  • It is an inverse-square law: double the distance, a quarter of the force; triple it, a ninth.
  • The two bodies pull on each other with forces equal in size and opposite in direction, a Newton's third law pair (A.2). The Earth pulls the Moon exactly as hard as the Moon pulls the Earth; the Moon simply has much less mass to accelerate.

G is tiny, so gravity between everyday objects is almost undetectable; it dominates astronomy because the masses are enormous.

Worked example: the Earth and the Moon. The Earth's mass is 5.97 × 10²⁴ kg, the Moon's is 7.35 × 10²² kg, and their centres are 3.84 × 10⁸ m apart.

F = G m1 m2 / r2
F = 6.67 × 10-11 × 5.97 × 1024 × 7.35 × 1022 / (3.84 × 108)2
F = 1.98 × 1020 Nthe same size on each body

4When can a body be treated as a point mass?

Newton's law is written for point masses. You can still use it for planets in two situations:

  • A sphere of uniform density (or any spherically symmetric body) attracts a mass outside it as if all its mass were concentrated at its centre. So r is always measured centre to centre: for a satellite at height h above a planet of radius R, r = R + h.
  • Any body whose size is very small compared with the separation can be treated as a point. The Sun and the Earth are both small compared with the 1.5 × 10¹¹ m between them.

5Gravitational field strength

A gravitational field is the region round a mass where another mass feels a gravitational force. We measure it with a small test mass m placed at the point.

g = F / m = G M / r²

The gravitational field strength g at a point is the gravitational force per unit mass on a small point mass placed there (small, so it does not disturb the field). Its unit is N kg⁻¹, the same as m s⁻², since a freely falling mass there accelerates at g. It is a vector, pointing towards the mass that creates the field.

Dividing Newton's law by m gives the second form: for a point or spherical mass M, g = GM/r². Note that g depends on M and r, not on the mass being pulled.

Worked example: g at the surface and at the height of a space station.

at the surface: g = GM / R2 = 6.67 × 10-11 × 5.97 × 1024 / (6.37 × 106)2 = 9.81 N kg-1
at h = 400 km: r = 6.37 × 106 + 4.00 × 105 = 6.77 × 106 mcentre to centre
g = 6.67 × 10-11 × 5.97 × 1024 / (6.77 × 106)2 = 8.69 N kg-1

At the station's height g is still almost 90% of its surface value. Astronauts float because they and the station are falling freely together round the Earth, not because gravity has vanished.

Figure 2 draws g against distance from the Earth's centre. Outside the Earth it falls as 1/r².

Figure 2 · Gravitational field strength above the Earth's surface Figure 2 · Gravitational field strength above the Earth's surface distance from the Earth's centre, r (in Earth radii) g / N kg⁻¹ surface 1R 2R 3R 4R 5R 2 4 6 8 10 9.81 2.45 1.09 Outside the Earth, g falls as 1/r²: twice as far from the centre, a quarter of the field.
Figure 2 · Gravitational field strength above the Earth's surface

Scaling without numbers. A planet with twice the Earth's mass and twice its radius has a surface field 2 ÷ 2² = ½ of the Earth's. Paper 1A tests this ratio method constantly: write g ∝ M/r², change each quantity, and multiply.

6Field lines

A field line shows the direction of the force on a small mass placed on it. Figure 3 shows the two fields the guide names.

Figure 3 · Gravitational field lines Figure 3 · Gravitational field lines (a) Around a planet: radial M (b) Close to the surface: uniform planet's surface parallel and equally spaced: g ≈ constant Lines point the way a small mass would be pulled. Closer lines mean a stronger field.
Figure 3 · Gravitational field lines
  • Round a point or spherical mass the field is radial: straight lines pointing into the centre, spreading apart with distance. The lines get further apart as the field weakens.
  • Close to the surface of a planet, over a region small compared with the planet, the lines are parallel and equally spaced, pointing down: a uniform field, where g is the same everywhere. This is the field of A.1 and A.3, where g = 9.81 N kg⁻¹ is treated as constant.

Rules for drawing: arrows always point towards the mass; lines never cross; the closer the lines, the stronger the field; lines meet the surface of a spherical body at right angles.

7The resultant field of two bodies

Field strength is a vector, so where two bodies both pull, the resultant is the vector sum. The guide restricts this to points on the line joining the two bodies, where the fields are either in the same direction or opposite, and you add or subtract magnitudes.

Between the Earth and the Moon the two fields point in opposite directions. Figure 4 plots the resultant along the line.

Figure 4 · The resultant field on the line from the Earth to the Moon Figure 4 · The resultant field on the line from the Earth to the Moon distance from the Earth's centre, as a fraction of the Earth–Moon distance d resultant g / N kg⁻¹ -0.004 0.004 0.008 0.012 0 0.2d 0.4d 0.6d 0.8d d N Earth's field dominates Moon's field dominates close to the Moon → resultant Earth alone (dashed) Moon alone (dashed) E Moon Positive means towards the Earth. The fields cancel at N, 90% of the way to the Moon.
Figure 4 · The resultant field on the line from the Earth to the Moon

At one point, N, the two fields are equal and opposite, and the resultant g is zero. Because the Earth is 81 times more massive, N is much closer to the Moon.

Worked example: where does g = 0? Let x be the distance from the Earth's centre and d = 3.84 × 10⁸ m.

G ME / x2 = G MM / (d − x)2equal magnitudes, G cancels
x / (d − x) = √(ME / MM) = √(5.97 × 1024 / 7.35 × 1022) = 9.01
x = 9.01 (d − x) so x = 9.01 d / 10.01 = 0.900 d
x = 0.900 × 3.84 × 108 = 3.46 × 108 m from the Earth's centre

Taking the square root first avoids a quadratic. Only a point between the bodies works; outside them the fields point the same way.

8Circular orbits and Kepler's third law

For a satellite, moon or planet in a circular orbit, gravity provides the centripetal force. That single idea, from A.2, gives Kepler's third law.

G M m / r2 = m v2 / rgravity is the centripetal force
v = 2πr / Tone orbit in one period
G M m / r2 = m (2πr / T)2 / r = 4π2 m r / T2
T2 = (4π2 / GM) r3m cancels: T² ∝ r³

The orbiting mass m cancels, so the period does not depend on it, and the constant 4π²/GM depends only on the central mass. So T and r for anything orbiting a body give that body's mass: this is how the masses of the Sun and planets are known.

Worked example (Paper 1B style): the mass of Jupiter. The four largest moons of Jupiter have these orbital radii and periods (values rounded).

Moonr / 10⁶ mT / daysr³ / 10²⁷ m³T² / 10¹² s²
Io421.71.7690.07500.0234
Europa671.03.5510.3020.0941
Ganymede10707.1551.230.382
Callisto188316.696.682.08

Plotting T² against r³, Figure 5, gives a straight line through the origin, which confirms T² ∝ r³.

Figure 5 · T² against r³ for the four largest moons of Jupiter Figure 5 · T² against r³ for the four largest moons of Jupiter r³ / 10²⁷ m³ T² / 10¹² s² Io Europa Ganymede Callisto 1 2 3 4 5 6 7 0.5 1.0 1.5 2.0 Δ(r³) = 4.0 × 10²⁷ m³ Δ(T²) = 1.25 × 10¹² s² A straight line through the origin: Kepler's third law. Gradient = 4π²/GM, which gives Jupiter's mass.
Figure 5 · T² against r³ for the four largest moons of Jupiter
gradient = Δ(T2) / Δ(r3) = 1.25 × 1012 / 4.0 × 1027 = 3.11 × 10-16 s2 m-3
gradient = 4π2 / GM so M = 4π2 / (G × gradient)
M = 39.5 / (6.67 × 10-11 × 3.11 × 10-16) = 1.9 × 1027 kg

Convert days to seconds before squaring, and use a large triangle on the line, not two data points.

9HLGravitational potential energy and potential

Gravitational potential energy of a system is the work done to assemble the system from infinite separation of its components. For two masses a distance r apart (centre to centre):

Ep = −G m₁ m₂ / r

Why negative? The zero of potential energy is chosen at infinite separation. Gravity is attractive, so as two masses come together from infinity, the field does positive work on them, and you would have to hold them back: the external work done to assemble them is negative. So Ep is less than zero at every finite separation, and it rises towards zero as the masses separate. More negative means more tightly bound.

The gravitational potential Vg at a point is the work done per unit mass in bringing a small mass from infinity to that point:

Vg = −G M / runit: J kg-1; zero at infinity
Ep = m Vgenergy of mass m at that point
W = m ΔVgwork done moving m between two points

Potential is a scalar, so the potentials of several bodies simply add, with no directions to worry about.

Worked example: lifting 1 kg to 600 km.

Vg at surface = −6.67 × 10-11 × 5.97 × 1024 / 6.37 × 106 = −6.25 × 107 J kg-1
Vg at 600 km = −6.67 × 10-11 × 5.97 × 1024 / 6.97 × 106 = −5.71 × 107 J kg-1
W = m ΔVg = 1 × (−5.71 × 107 − (−6.25 × 107)) = 5.38 × 106 J
compare mgh = 1 × 9.81 × 6.00 × 105 = 5.89 × 106 Jtoo big: g falls with height

mgh assumes g stays 9.81 N kg⁻¹ all the way up; over large heights, use potential.

10HLPotential gradient and equipotentials

Figure 6 plots Vg against r for the Earth. The curve rises towards zero with distance.

Figure 6 · HL · Gravitational potential around the Earth Figure 6 · HL · Gravitational potential around the Earth distance from the Earth's centre, r (in Earth radii) Vg / MJ kg⁻¹ 1R 2R 3R 4R 5R 6R -60 -40 -20 tangent at r = 2R: gradient = ΔVg/Δr = 2.45 N kg⁻¹, so g = 2.45 N kg⁻¹ towards the Earth −62.5 at the surface Vg is zero at infinity and negative everywhere else. The gradient of the curve gives the size of g.
Figure 6 · HL · Gravitational potential around the Earth

The field strength is the potential gradient, with a minus sign:

g = −ΔVg / Δr

The minus sign says the field points the way potential decreases, towards the Earth. In practice: draw the tangent to the Vg–r graph at a point and measure its gradient; its size is the size of g there. At r = 2R the tangent in Figure 6 has gradient 2.45 J kg⁻¹ m⁻¹, and g there is 2.45 N kg⁻¹ towards the Earth, a quarter of 9.81 as the inverse-square law requires.

An equipotential surface joins points with the same potential. Figure 7 draws them with the field lines. This is what the guide means by mapping a field using potential.

Figure 7 · HL · Equipotentials and field lines Figure 7 · HL · Equipotentials and field lines (a) Around a planet (Vg in MJ kg⁻¹) −50 −40 −30 −20 Earth (b) Close to the surface surface higher Vg lower Vg equal steps in Vg, equally spaced Equipotentials (amber, dashed) cross field lines (teal) at right angles. Around a planet, equal steps in Vg get further apart with distance.
Figure 7 · HL · Equipotentials and field lines
  • No work is done moving along an equipotential, since ΔVg = 0 and W = mΔVg.
  • Field lines cross equipotentials at right angles. If the field had a component along an equipotential, moving along it would take work.
  • Round a planet the equipotentials are spheres (circles in the drawing). For equal steps in Vg they get further apart with distance, because the field weakens, and a weaker field needs a longer distance for the same change in potential.
  • In a uniform field the equipotentials are parallel planes, equally spaced for equal steps, perpendicular to the field.

11HLOrbits, orbital energy and escape

Orbital speed. From section 8, gravity as the centripetal force gives

vorbital = √(GM / r)

A lower orbit is faster. The International Space Station, about 400 km up, moves at √(GM/6.77 × 10⁶ m) = 7.67 km s⁻¹.

Energy in a circular orbit. Multiply v² = GM/r by ½m:

Ek = ½ m v2 = G M m / 2r
Ep = −G M m / r
Etotal = Ek + Ep = −G M m / 2r
so Ek = −Etotal and Ep = 2 Etotal

Figure 8 plots all three. Etotal is negative: the satellite is bound. A higher orbit has a less negative Etotal, so climbing needs energy, even though the satellite ends up slower.

Figure 8 · HL · The energies of a satellite in a circular orbit Figure 8 · HL · The energies of a satellite in a circular orbit orbital radius r (in planet radii) energy / (GMm/R) 1 2 3 4 5 -1.0 -0.5 0.5 Ek = GMm/2r E total = −GMm/2r Ep = −GMm/r Ek = −E and Ep = 2E at every radius. A higher orbit has more total energy but less kinetic energy.
Figure 8 · HL · The energies of a satellite in a circular orbit

Worked example: an 800 kg satellite. It is launched from rest on the surface of a non-rotating Earth into a circular orbit 600 km up (r = 6.97 × 10⁶ m), and later moved to 1000 km (r = 7.37 × 10⁶ m).

GM = 6.67 × 10-11 × 5.97 × 1024 = 3.98 × 1014 N m2 kg-1
vorbital = √(3.98 × 1014 / 6.97 × 106) = 7.56 × 103 m s-1
E in orbit = −GMm / 2r = −3.98 × 1014 × 800 / (2 × 6.97 × 106) = −2.29 × 1010 J
E at rest on surface = Ep = −GMm / R = −3.98 × 1014 × 800 / 6.37 × 106 = −5.00 × 1010 J
energy to reach orbit = −2.29 × 1010 − (−5.00 × 1010) = 2.72 × 1010 J
energy to move up = GMm/2 × (1/r1 − 1/r2) = 1.24 × 109 J

Escape speed. A body escapes if it can reach infinity, where Ep = 0. The least launch speed that does it leaves the body with zero kinetic energy at infinity, so its total energy is exactly zero:

½ m vesc2 − G M m / r = 0
vesc = √(2GM / r)
from the Earth's surface: vesc = √(2 × 3.98 × 1014 / 6.37 × 106) = 1.12 × 104 m s-1

So a body escapes if its total energy is zero or positive. A satellite in orbit, with E = −GMm/2r, needs exactly +GMm/2r more, and at every radius vesc = √2 × vorbital. Neither speed depends on the launched mass; both ignore air resistance.

12HLDrag on a low orbit

A low satellite passes through thin upper atmosphere, which exerts a small viscous drag force, with a surprising result.

  1. Drag does negative work, so the satellite's total energy decreases: E = −GMm/2r becomes more negative.
  2. More negative E means a smaller r: the satellite gradually spirals lower.
  3. At smaller r, Ek = GMm/2r is larger: the satellite speeds up.

Ep falls by twice as much as Ek rises; drag takes the other half as internal energy. As the orbit sinks into denser air the decay speeds up until the satellite re-enters, which is why low satellites fire thrusters now and then to climb back.

13Where marks are lost

Measuring r from the surface. In F = Gm₁m₂/r², g = GM/r², Vg and every orbit formula, r is from the centre. At height h, r = R + h.

Treating g as constant at large heights. g = 9.81 N kg⁻¹ only near the surface. Above that use GM/r², and at HL use potential, not mgh.

Squaring r in the wrong place. g ∝ 1/r², but Vg ∝ 1/r. Double the distance: g quarters, Vg halves.

Saying astronauts are weightless because there is no gravity. At 400 km, g is still 8.7 N kg⁻¹. They are in free fall with the station.

Looking for the null point outside the two bodies. Two attracting fields cancel only between the masses, nearer the smaller one.

Forgetting to convert days to seconds in Kepler's third law. Convert, then square.

HL: dropping the minus sign, or saying a higher orbit has more kinetic energy. Ep and Vg are negative, zero at infinity. A higher orbit has more total energy but less kinetic energy.

HL: saying drag slows a satellite down. Drag lowers its orbit, and a lower orbit is faster.

14Draw it right

  1. Radial field lines: straight, evenly spread round the sphere, arrows pointing in, meeting the surface at right angles, never crossing.
  2. Uniform field: parallel, equally spaced lines with arrows towards the surface; only over a region small compared with the planet.
  3. g against r: starts at the surface value at r = R and falls as 1/r², approaching zero but never reaching it; at 2R it is a quarter.
  4. Kepler's second law: a short, wide sector near the Sun and a long, thin one far away, labelled equal areas in equal times.
  5. T² against r³: straight line through the origin; say what the gradient equals.
  6. HL Vg against r: negative everywhere, rising towards zero, steepest near the surface; a tangent for g.
  7. HL equipotentials: concentric circles round a planet, spacing increasing outwards for equal steps; perpendicular to every field line; equally spaced parallel lines in a uniform field.
  8. HL orbital energies: Ek positive, Ep and E negative, with E halfway between Ep and zero at every r.

15Try it

Marks in brackets. Answers and marker's notes are at the end. G = 6.67 × 10⁻¹¹ N m² kg⁻²; mass of the Earth 5.97 × 10²⁴ kg; radius of the Earth 6.37 × 10⁶ m.

Q1. Planet X has twice the mass of the Earth and twice its radius. What is the gravitational field strength at its surface? 1 mark

A. 2.5 N kg⁻¹

B. 4.9 N kg⁻¹

C. 9.8 N kg⁻¹

D. 20 N kg⁻¹

Q2. A planet of mass 8.0 × 10²⁴ kg has a moon of mass 2.0 × 10²³ kg. Their centres are 4.0 × 10⁸ m apart.

(a) Sketch the gravitational field lines round the planet alone. 2 marks

(b) Determine the distance from the planet's centre, on the line between them, at which the resultant gravitational field strength is zero. 3 marks

Q3. A geostationary satellite orbits above the equator with a period of 24.0 hours.

(a) Show that the radius of its orbit is about 4.2 × 10⁷ m. 3 marks

(b) Calculate the gravitational field strength at the satellite. 2 marks

Q4. An astronomer measures the orbits of four moons of a distant planet (invented data).

r / 10⁸ m2.03.55.08.0
T / days1.132.614.459.01

(a) Explain why a graph of T² against r³ should be a straight line through the origin. 2 marks

(b) Use the data to determine the mass of the planet. 3 marks

(c) State one assumption about the moons' orbits. 1 mark

Q5 (HL). A satellite of mass 800 kg is in a circular orbit 600 km above the Earth's surface.

(a) Calculate its orbital speed. 2 marks

(b) Show that its total energy is about −2.3 × 10¹⁰ J. 2 marks

(c) Determine the minimum energy needed to launch it into this orbit from rest on the surface of a non-rotating Earth. 2 marks

(d) Determine the additional energy it would need to escape from this orbit. 1 mark

Q6 (HL).

(a) Explain why gravitational potential is negative at every point near a planet. 2 marks

(b) Explain why equipotential surfaces are always perpendicular to field lines. 2 marks

(c) A satellite in a low orbit experiences a small drag force from the atmosphere. Explain what happens to its height and to its speed. 3 marks

16In one breath

Kepler: elliptical orbits with the Sun at a focus, equal areas in equal times, T² ∝ r³. Newton: F = Gm₁m₂/r², attractive, inverse-square, a third-law pair, with r centre to centre for spheres and for bodies small compared with their separation. g = F/m = GM/r², in N kg⁻¹, towards the mass; field lines radial round a sphere, parallel near the surface. Between two bodies the fields cancel at a point nearer the smaller one. Gravity as centripetal force gives T² = 4π²r³/GM, so orbits reveal the central mass. HL: Ep = −Gm₁m₂/r and Vg = −GM/r, zero at infinity; g = −ΔVg/Δr; W = mΔVg; equipotentials perpendicular to field lines. vorbital = √(GM/r), vesc = √(2GM/r); in orbit E = −GMm/2r = −Ek. Escape needs E ≥ 0. Drag lowers E, so the orbit shrinks and the satellite speeds up.

Answers

Q1. B. g ∝ M/r², so gX = 9.81 × 2 ÷ 2² = 4.9 N kg⁻¹. A ignores the doubled mass; C uses 1/r instead of 1/r²; D ignores the radius.

Q2. (a) Radial lines, evenly spaced round the planet, arrows pointing in towards the centre, not crossing. (b) G(8.0 × 10²⁴)/x² = G(2.0 × 10²³)/(d − x)², so x/(d − x) = √40 = 6.32. x = 6.32d ÷ 7.32 = 0.864 × 4.0 × 10⁸ = 3.5 × 10⁸ m from the planet's centre. (a) 1 for radial lines, 1 for inward arrows. (b) M1 for equating the two field strengths, M1 for the square-root step or a correct quadratic, A1 for 3.5 × 10⁸ m. A second root beyond the moon must be rejected; giving it as the answer scores A0.

Q3. (a) GMm/r² = m(2π/T)²r, so r³ = GMT²/4π². T = 24.0 × 3600 = 8.64 × 10⁴ s. r³ = 6.67 × 10⁻¹¹ × 5.97 × 10²⁴ × (8.64 × 10⁴)² ÷ 4π² = 7.53 × 10²² m³, so r = 4.22 × 10⁷ m. (b) g = GM/r² = 3.98 × 10¹⁴ ÷ (4.22 × 10⁷)² = 0.22 N kg⁻¹. (a) M1 for gravity = centripetal force, M1 for T in seconds substituted, A1 for r to at least 3 s.f. ("show that" needs more figures than the question gives). (b) M1, A1.

Q4. (a) For a circular orbit, gravity provides the centripetal force: GMm/r² = 4π²mr/T², so T² = (4π²/GM) r³. 4π²/GM is constant for one planet, so T² is proportional to r³: a straight line through the origin. (b) With T in seconds, T² runs from 9.53 × 10⁹ to 6.06 × 10¹¹ s² as r³ runs from 8.0 × 10²⁴ to 5.12 × 10²⁶ m³; the gradient of the line is 1.18 × 10⁻¹⁵ s² m⁻³. M = 4π²/(G × gradient) = 39.5 ÷ (6.67 × 10⁻¹¹ × 1.18 × 10⁻¹⁵) = 5.0 × 10²⁶ kg. (c) The orbits are circular (or: the moons' masses are negligible compared with the planet's). (a) 1 for gravity as centripetal force, 1 for the constant 4π²/GM. (b) M1 for T in seconds and T², r³, M1 for gradient = 4π²/GM, A1 for 5.0 × 10²⁶ kg (accept 4.9–5.1). (c) 1. Leaving T in days gives a mass about 10¹⁰ times too big: M1 only.

Q5 (HL). (a) r = 6.37 × 10⁶ + 6.00 × 10⁵ = 6.97 × 10⁶ m. v = √(GM/r) = √(3.98 × 10¹⁴ ÷ 6.97 × 10⁶) = 7.56 × 10³ m s⁻¹. (b) E = −GMm/2r = −3.98 × 10¹⁴ × 800 ÷ (2 × 6.97 × 10⁶) = −2.29 × 10¹⁰ J. (c) At rest on the surface E = −GMm/R = −5.00 × 10¹⁰ J. Energy needed = −2.29 × 10¹⁰ − (−5.00 × 10¹⁰) = 2.7 × 10¹⁰ J. (d) To escape, E must reach zero: 2.3 × 10¹⁰ J more. (a) M1 for r = R + h, A1. (b) M1 for E = −GMm/2r (or Ek + Ep), A1 for −2.29 × 10¹⁰ J. (c) M1 for surface energy −GMm/R, A1 for 2.7 × 10¹⁰ J. (d) A1. Using r = 6.00 × 10⁵ m anywhere scores M0 for that part.

Q6 (HL). (a) Potential is defined as zero at infinity. Gravity is attractive, so bringing a mass in from infinity the field does positive work and the external work done is negative; potential, the work per unit mass, is therefore negative at every finite distance. (b) No work is done moving a mass along an equipotential, since ΔVg = 0. If the field had a component along the surface, moving along it would take work, so the field, and the field lines, must be at right angles to it. (c) Drag does negative work, reducing the total energy E = −GMm/2r, so E becomes more negative and r decreases: the satellite loses height. Ek = GMm/2r increases as r decreases, so its speed increases; the potential energy falls by twice the gain in kinetic energy. (a) 1 for zero at infinity, 1 for attractive force / negative work. (b) 1 for no work along an equipotential, 1 for no field component along it. (c) 1 for total energy decreasing, 1 for lower orbit, 1 for faster, with Ek = GMm/2r or v = √(GM/r). "It slows down and falls" scores 1 for falling only.

Educerie · written from the published IB Diploma Programme Physics guide, first assessment 2025, section D.1 Gravitational fields. Original text, examples and questions. Diagrams drawn by Educerie. Last reviewed 25 September 2026.

Mocks: in the future, hold tight!