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

This guide teaches the standard level and higher level part of sub-topic D.1, Gravitational fields, from the IB Diploma Programme Physics guide, first assessment 2025. It covers Kepler’s three laws, Newton’s law of gravitation, extended bodies treated as point masses, gravitational field strength and field lines. Everything here is for both SL and HL. The potential-based content in D.1 (potential energy, potential, escape speed and orbital speed) is HL only and is taught separately.

Use the IB Physics hub to see every sub-topic and the printable checklist to tick off each statement. Companion pages: the revision notes and the practice questions. You will also need circular motion and Newton’s laws from the forces and momentum guide. To find your gaps first, take the free 10-minute diagnostics.

Use G = 6.67 × 10⁻¹¹ N m² kg⁻² throughout.

What this unit covers

Guide statement (D.1)	What you must be able to do	Tier
Kepler’s three laws of orbital motion	State each law, apply the second and third to numbers	SL and HL
Newton’s universal law, F = G m₁m₂/r²	Calculate the force between point masses and see how it scales	SL and HL
Extended bodies as point masses	Say when it is valid and measure r to the centre	SL and HL
Field strength, g = F/m = GM/r²	Define g, calculate it, link it to weight	SL and HL
Gravitational field lines	Sketch and interpret radial and uniform fields	SL and HL

The guide says orbits are treated as circular in calculations, and that the resultant field strength is only needed at points along a line joining two bodies.

Kepler’s three laws

Kepler found these laws by fitting observations of the planets. They describe the motion without explaining it. Newton’s law of gravitation supplies the explanation.

• First law. Each planet moves on an ellipse, with the Sun at one focus.
• Second law. In equal intervals of time, the straight line from the Sun to a planet covers equal areas.
• Third law. The square of the orbital period is proportional to the cube of the mean orbital radius: T² ∝ r³. For an ellipse, r is the semi-major axis. For a circle, it is the radius.

The laws hold for any set of bodies orbiting one central mass, such as moons around a planet. Each system has its own constant of proportionality.

Using the second law

An orbiting body moves fastest when it is closest to the central mass and slowest when furthest away. At the two points where the velocity is at right angles to the line to the central mass, the area swept in a short time Δt is about ½ r v Δt. Equal areas then give:

r₁ v₁ = r₂ v₂

Worked example 1. An asteroid is 2.2 × 10¹¹ m from the Sun at its closest and moving at 2.8 × 10⁴ m s⁻¹. At its furthest it is 5.5 × 10¹¹ m away. Find its speed there.

v₂ = r₁ v₁ / r₂ = (2.2e11 × 2.8e4) / 5.5e11
   = 1.12 × 10⁴ m s⁻¹

The speed falls by the same factor as the distance rises (2.5 times), so the answer is sensible.

Using the third law

Because T²/r³ is constant for one system, compare two orbits without knowing the central mass:

T₂/T₁ = (r₂/r₁)^(3/2)

Worked example 2. These data are supplied for two moons of Saturn.

Moon	Orbital radius / m	Period / days
Titan	1.22 × 10⁹	15.9
Rhea	5.27 × 10⁸	4.52

Show that the data support Kepler’s third law.

Convert days to seconds (× 86 400), then find T²/r³ for each.

Titan: (15.9 × 86400)² / (1.22e9)³ = 1.04 × 10⁻¹⁵ s² m⁻³
Rhea:  (4.52 × 86400)² / (5.27e8)³ = 1.04 × 10⁻¹⁵ s² m⁻³

The two values agree to three significant figures, so T² ∝ r³ for both moons.

A shortcut for a planet orbiting at four times another’s distance: T scales as 4^(3/2) = 8, so its period is eight times longer.

Newton’s universal law of gravitation

Every pair of point masses attracts with equal and opposite forces of magnitude:

F = G m₁ m₂ / r²

Here r is the separation of the two points. Key features:

• The force is always attractive.
• It is an inverse square law: double the separation and F falls to a quarter.
• F is proportional to each mass, so doubling one mass doubles F.
• The two forces form a Newton’s third law pair. The small body pulls the large body with the same size of force that the large body pulls the small one.

Worked example 3. Two uniform spheres of mass 250 kg and 40 kg have their centres 0.80 m apart. Find the force between them, then the force if the separation becomes 1.6 m.

F = 6.67e-11 × 250 × 40 / 0.80²
  = 1.04 × 10⁻⁶ N

Doubling r divides F by 2² = 4, so the new force is 2.61 × 10⁻⁷ N.

The forces are tiny because G is tiny. Gravity is only noticeable when at least one mass is planet-sized.

Extended bodies as point masses

The law is written for point masses. A real body can be treated as a point mass at its centre when:

• it is a sphere (or has spherical symmetry) with uniform density, so its mass acts as if concentrated at its centre (the guide asks for this for spherical masses of uniform density); or
• the separation is much larger than the size of the bodies, so differences in distance across each body do not matter.

In every case, r is measured centre to centre, never from the surface.

Worked example 4. A satellite of mass 1200 kg orbits 600 km above the surface of the Earth. Take the Earth’s mass as 5.97 × 10²⁴ kg and its radius as 6.37 × 10⁶ m. Find the gravitational force on the satellite.

r = 6.37e6 + 0.60e6 = 6.97 × 10⁶ m
F = 6.67e-11 × 5.97e24 × 1200 / (6.97e6)²
  = 9.84 × 10³ N

Using 6.37 × 10⁶ m for r would be the most common slip here.

Gravitational field strength

A gravitational field is a region where a mass experiences a force because of another mass. At any point, the gravitational field strength g is defined as the gravitational force felt by a small point mass there, divided by that mass:

g = F / m

Its unit is N kg⁻¹, and it is a vector pointing in the direction of the force, towards the source. Keep the test mass tiny, otherwise it would alter the field you are trying to measure.

Put F = GMm/r² into the definition. The test mass cancels:

g = GM / r²

Here M is the mass of the source (a point mass or a uniform sphere) and r is the distance from its centre. The field is independent of the test mass and falls as 1/r².

Near a planet’s surface, the force on a mass m is its weight, W = mg. A freely falling object has acceleration a = F/m = g, so N kg⁻¹ and m s⁻² describe the same quantity.

Worked example 5. A Mars lander problem gives the planet’s radius as 3.39 × 10⁶ m and its mass as 6.42 × 10²³ kg. Find g at its surface, the weight of an 85 kg astronaut, and the height above the surface where g falls to 1.00 N kg⁻¹.

g = GM/R² = 6.67e-11 × 6.42e23 / (3.39e6)² = 3.73 N kg⁻¹
W = mg = 85 × 3.726 = 317 N

For g = 1.00, find r first:

r² = GM/g = 6.67e-11 × 6.42e23 / 1.00 = 4.282e13 m²
r  = 6.54 × 10⁶ m
h  = r − R = 6.54e6 − 3.39e6 = 3.15 × 10⁶ m

Notice that r is measured from the centre, so you must subtract R at the end.

Two bodies along a line

When two bodies act, add the two field strengths as vectors. On the line joining them, that means adding or subtracting. Between the bodies the fields point in opposite directions. Outside the pair, both point towards the bodies, so they add.

Worked example 6. Bodies A and B have masses 3.6 × 10²² kg and 9.0 × 10²¹ kg, with their centres 4.8 × 10⁸ m apart. Find (a) the point on the line between them where the resultant field is zero, and (b) the resultant field 1.6 × 10⁸ m from A, on the line between them.

(a) At the null point, gA = gB. With x measured from A:

M_A / x² = M_B / (d − x)²
(d − x)/x = √(M_B/M_A) = √(9.0e21/3.6e22) = 0.5
x = d/1.5 = 3.2 × 10⁸ m from A

The null point is closer to the smaller mass, which is what you expect.

(b) At 1.6 × 10⁸ m from A, the distance from B is 3.2 × 10⁸ m.

g_A = 6.67e-11 × 3.6e22 / (1.6e8)² = 9.38 × 10⁻⁵ N kg⁻¹ towards A
g_B = 6.67e-11 × 9.0e21 / (3.2e8)² = 5.86 × 10⁻⁶ N kg⁻¹ towards B
resultant = 9.38e-5 − 5.86e-6 = 8.79 × 10⁻⁵ N kg⁻¹ towards A

Always state the direction. A resultant field strength is a vector, so the number alone is incomplete.

Gravitational field lines

Field lines show the direction a small mass would be pushed. The guide requires you to sketch and interpret them.

• Radial field (point mass or sphere). Straight lines point inwards to the centre, spread evenly in all directions. They crowd together near the mass, where the field is strong, and spread out further away.
• Uniform field. Close to a planet’s surface, the lines are parallel, equally spaced and pointing down. The guide asks you to treat the field there as uniform.
• Rules. Arrows show the direction of the force on a mass. Line spacing shows strength: closer lines mean a stronger field. Lines never cross, because the field has one direction at each point. Between two masses, the lines curve around a null point where the field is zero.

Worked example 7. Show that treating g as uniform is reasonable over a height of 8.0 km above the Earth’s surface (Earth data: 5.97 × 10²⁴ kg and 6.37 × 10⁶ m).

g at surface   = GM/R²         = 9.813 N kg⁻¹
g at 8.0 km    = GM/(R+8000)²  = 9.789 N kg⁻¹
decrease       = 0.25 %

A quarter of one percent is small. Over heights much smaller than R, treating g as constant is a good approximation.

Common errors

• Measuring r from the surface instead of from the centre.
• Forgetting to square r, or squaring G.
• Adding two field strengths between the bodies instead of subtracting.
• Saying g depends on the test mass. It does not.
• Drawing field lines that cross or that start on a planet’s surface instead of pointing at its centre.
• Using Kepler’s third law with the radius to the surface rather than the orbital radius.

Next steps

Work through the practice questions and use the revision notes for the final weeks.

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. The additional higher level content of D.1 is not covered here.