IB DP Physics: Gravitational fields: Kepler, Newton's law and field strength (DP Physics) – Revision Notes

These notes condense the standard level and higher level part of sub-topic D.1, Gravitational fields, from the IB Diploma Programme Physics guide, first assessment 2025. The study guide has the full explanations and longer examples. Try the practice questions. The potential-based, HL only content of D.1 is not in these notes.

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Use G = 6.67 × 10⁻¹¹ N m² kg⁻².

Kepler’s laws at a glance

Law	Statement	What it lets you do
First	Each orbit is an ellipse, with the Sun at a focus	Describe the shape of a planetary orbit
Second	The Sun-planet line covers equal areas in equal times	Compare speeds at different distances
Third	T² ∝ r³	Compare periods and radii of bodies orbiting the same mass

In calculations the guide treats orbits as circular, so r is the orbital radius.

Method: second law at the closest and furthest points

1. Identify the two points where the velocity is at right angles to the line to the central mass.
2. Write r₁v₁ = r₂v₂.
3. Rearrange for the unknown speed.
4. Check that the speed is higher at the smaller distance.

Method: third law between two orbits

1. Confirm both bodies orbit the same central mass.
2. Write T₂/T₁ = (r₂/r₁)^(3/2).
3. Keep the same units for both periods and both radii. No conversion is needed.
4. Check the sign: a larger orbit gives a longer period.

Worked reminder: an orbit nine times larger gives T 9^(3/2) = 27 times longer.

Newton’s law of gravitation

F = G m₁ m₂ / r²

Fact	Detail
Nature	Attractive only
Dependence	F ∝ m₁, F ∝ m₂, F ∝ 1/r²
Direction	Along the line joining the two masses
Pair	The two forces are equal and opposite (Newton’s third law)
Distance	r is centre to centre, never surface to centre

Scaling shortcut: work out the factor for each change and multiply them. Triple r gives 1/9, double one mass gives 2, so together F becomes 2/9 of its original value.

Extended bodies as point masses

Treat a body as a point mass at its centre when:

• it is spherical with uniform density (or spherically symmetric), or
• the separation is far bigger than the size of either body.

Do not take r from the surface.

Field strength

Quantity	Result	Unit
Definition	g = F/m, force divided by the mass of a small point mass	N kg⁻¹
Point mass or uniform sphere	g = GM/r²	N kg⁻¹
Weight	W = mg	N
Free fall	a = g	m s⁻²

• g is a vector that points towards the source.
• g does not depend on the test mass.
• Inverse square: at r = 2R, g is a quarter of its surface value.
• N kg⁻¹ and m s⁻² are equivalent units.

Method: g at a given height

1. Find r = R + h.
2. Substitute into g = GM/r².
3. Keep four digits in the working, then round to three significant figures.
4. State the unit and direction (towards the centre).

Method: two bodies on a line

1. Draw the line and mark the point you want.
2. Find each body’s g at that point using its own distance.
3. Between the bodies: subtract the field strengths. Outside: add them.
4. Give the direction of the larger one.
5. For a null point, solve M₁/x² = M₂/(d − x)². Taking square roots, x/(d − x) = √(M₁/M₂). The null point is nearer the smaller mass.

Worked reminder: if one mass is 16 times the other, √16 = 4, so the null point is 4/5 of the separation from the larger mass.

Field lines

Field	Pattern	Spacing
Radial (point mass, sphere)	Straight lines pointing to the centre	Closer near the mass
Uniform (near a planet’s surface)	Parallel, equally spaced, pointing down	Constant
Two masses	Lines curve between them, with a null point	Sparse at the null point

Rules to remember:

• The arrow shows the direction of the force on a mass.
• Closer lines mean a stronger field.
• Lines never cross.
• Lines end at the surface of a sphere, pointing at its centre.

Worked reminders

Field strength at a height. Find g at 500 km above a planet of mass 6.0 × 10²³ kg and radius 3.0 × 10⁶ m.

r = 3.0e6 + 0.5e6 = 3.5 × 10⁶ m
g = 6.67e-11 × 6.0e23 / (3.5e6)² = 3.27 N kg⁻¹

Third law with a ratio. A moon orbits at 3.0 × 10⁸ m with period 2.0 days. A second moon of the same planet orbits at 6.0 × 10⁸ m. The radius doubles, so T grows by 2^(3/2) = 2.83, giving 5.7 days.

Weight from field strength. A 12 kg instrument on a planet where g = 3.27 N kg⁻¹ weighs 12 × 3.27 = 39 N.

Units check

Quantity	Symbol	Unit
Gravitational constant	G	N m² kg⁻²
Field strength	g	N kg⁻¹
Period	T	s (or days, if you use a ratio)
Separation	r	m, measured centre to centre
Mass	M, m	kg

A quick check on any answer: the unit of GM/r² is N m² kg⁻² × kg / m² = N kg⁻¹, which confirms the formula is dimensionally consistent.

Must-know distinctions

• g and G. G is a universal constant. g is a field strength that changes with position.
• Mass and weight. Mass is in kg and does not change. Weight is a force in N and depends on g.
• Kepler and Newton. Kepler’s laws describe the orbits. Newton’s law of gravitation explains them.
• Radial and uniform. The field is radial in general and treated as uniform only over heights small compared with the planet’s radius.

Quick self-test

1. State Kepler’s second law in your own words.
2. A planet orbits a star at nine times the distance of a second planet. By what factor is its period longer?
3. Find the force between masses of 2.0 × 10³ kg and 3.0 × 10³ kg whose centres are 10 m apart.
4. The separation of two masses is tripled. By what factor does the force change?
5. The Earth pulls a satellite with a force of 800 N. What force does the satellite exert on the Earth?
6. At the surface g = 9.81 N kg⁻¹. What is g at a distance of 2R from the planet’s centre?
7. Find g at the surface of the Moon, whose mass is 7.35 × 10²² kg and whose radius is 1.74 × 10⁶ m.
8. Find the weight of a 60 kg person on the Moon, using your answer to question 7.
9. A comet is 1.0 × 10¹⁰ m from a star moving at 6.0 × 10⁴ m s⁻¹. What is its speed when it is 4.0 × 10¹⁰ m from the star, at a point where the velocity is again perpendicular to the line to the star?
10. Two masses differ by a factor of 16 and are separated by 5.0 m. How far from the heavier mass is the null point?

Answers

1. The line from the Sun to a planet covers equal areas in equal intervals of time.
2. 9^(3/2) = 27 times longer.
3. F = 6.67 × 10⁻¹¹ × 2.0 × 10³ × 3.0 × 10³ / 10² = 4.00 × 10⁻⁶ N.
4. 1/3² = one ninth.
5. 800 N, towards the satellite (Newton’s third law).
6. g is a quarter: 9.81/4 = 2.45 N kg⁻¹.
7. g = 6.67 × 10⁻¹¹ × 7.35 × 10²² / (1.74 × 10⁶)² = 1.62 N kg⁻¹.
8. W = 60 × 1.619 = 97.2 N.
9. v₂ = 6.0 × 10⁴ × 1.0 × 10¹⁰ / 4.0 × 10¹⁰ = 1.5 × 10⁴ m s⁻¹.
10. √16 = 4, so the null point is 4/5 of 5.0 m from the heavier mass: 4.0 m.

Where marks are usually lost

• Using the height above the surface as r, instead of R + h.
• Giving g in N, or force in N kg⁻¹. Check the unit against the quantity.
• Writing “g is the force on a mass” without “per unit mass” in a definition.
• Stating that g depends on the mass of the test object.
• Treating Kepler’s laws as the cause of orbits rather than a description of them.
• Missing the direction on a resultant field strength.
• Subtracting field strengths outside the pair of bodies, where they add.
• Drawing field lines for a sphere that do not point at the centre, or that cross.
• Mixing days and seconds when working out T²/r³ for a single body, though not when taking a ratio.

Official syllabus

This page follows sub-topic D.1 (Gravitational fields), standard level and higher level content, of the IB Diploma Programme Physics guide, first assessment 2025, published by the International Baccalaureate Organization.