Kepler’s laws are observational rules that Newton’s gravitation later explains. The first law fixes the orbit geometry. The second law describes changing speed around an ellipse. The third law links orbital period to orbit size, but only when the orbiting bodies share the same central mass.

First law: an orbiting body moves in an elliptical orbit with the central mass at one focus.

Second law: a line from the central mass to the orbiting body sweeps out equal areas in equal time intervals.

The second law means the orbiting body moves faster when closer to the central mass and slower when farther away.

Third law: for bodies orbiting the same central mass, T^2 is proportional to r^3, where r is the mean orbital radius or semi-major axis.

For circular orbits around mass M, Newton’s law gives T^2 = 4π^2 r^3/(GM).

Match each Kepler law cue to the correct orbital statement.

Match

Reasons

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State Kepler’s second and third laws and explain why a comet moves faster when it is closer to the Sun.

Stating T^2 ∝ r^3 without saying that the same central mass is required.

State Kepler’s second and third laws and explain why a comet moves faster when it is closer to the Sun.

Choose

Newton’s law gives the magnitude of the attractive force between two masses. For spherical bodies, treat the mass as if concentrated at the centre when calculating the external field or force. The formula gives equal force magnitudes for both bodies; the acceleration can differ because acceleration depends on mass.

Newton’s law of gravitation is F = Gm1m2/r^2.

G = 6.67 × 10^-11 N m^2 kg^-2 is the gravitational constant.

r is the centre-to-centre distance between the two masses.

The force is always attractive and acts along the line joining the centres of the masses.

Each mass exerts the same magnitude force on the other, in opposite directions.

The inverse-square relationship means increasing r reduces F by the square of the scale factor.

Build Newton’s law of gravitation and state the distance and direction conventions.

Formula

Target formula F = G m1 m2 / r^2

F

gravitational force magnitude

N

G

gravitational constant

N m^2 kg^-2

m1

mass of first body

kg

m2

mass of second body

kg

r

centre-to-centre separation

m

1Identify both masses and the centre-to-centre separation.m1, m2, r

2Multiply the masses and G, then divide by the square of separation.F = G m1 m2 / r^2

3State the direction of the force.attractive, along the line joining centres

4Check inverse-square scaling.r doubled -> F/4

Build the formula

Two spherical masses attract each other gravitationally. State the formula for the force and explain what distance should be used for r.

Using surface separation instead of centre-to-centre separation or omitting that the force is attractive.

Two spherical masses attract each other gravitationally. State the formula for the force and explain what distance should be used for r.

Choose

The point-mass approximation is a modelling step before using inverse-square gravity. For planets and stars treated as spherical, external gravitational calculations use the centre as the location of the mass. This is why orbital radius is measured from the centre of the planet, not from the surface.

A point-mass approximation replaces an extended body by a single mass located at its centre of mass.

For a spherical body, the external gravitational field is the same as if all mass were concentrated at the centre.

In Newton’s law and g = GM/r^2, r is measured from the centre of the spherical body to the external point or the other body’s centre.

The approximation is best when the separation is much larger than the size of the bodies or when the body is spherical and the point is outside it.

Do not use the simple external point-mass formula inside a planet or near an irregular body unless a model has been justified.

Match each situation to whether the point-mass approximation is valid or needs caution.

Match

Reasons

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A satellite orbits 400 km above Earth’s surface. Explain why the orbital radius used in gravitational calculations is not 400 km.

Using altitude above the surface as r instead of measuring from Earth’s centre.

A satellite orbits 400 km above Earth’s surface. Explain why the orbital radius used in gravitational calculations is not 400 km.

Choose

A gravitational field describes what force a small test mass would experience at each point. Dividing force by the test mass gives g, so the field is a property of the source masses and position, not of the particular test mass. Around a spherical planet, g points radially inward and has magnitude GM/r^2 outside the planet.

Gravitational field strength g is defined as gravitational force per unit mass: g = F/m.

The unit N kg^-1 is equivalent to m s^-2.

For a point mass or outside a spherical mass M, the magnitude is g = GM/r^2.

The direction of g is the direction of the gravitational force on a small test mass, toward the attracting mass.

g decreases with inverse-square distance and fields from multiple masses combine by vector addition.

Build the gravitational field strength formulas and state the units and direction.

Formula

Target formula g = F/m; g = GM/r^2

g

gravitational field strength

N kg^-1 or m s^-2

F

gravitational force on a test mass

N

m

test mass experiencing the force

kg

G

gravitational constant

N m^2 kg^-2

M

source mass

kg

r

distance from centre of source mass to the field point

m

1Start with force per unit mass.g = F/m

2For a point or spherical source mass, substitute Newton’s law.g = GM/r^2

3State the units.N kg^-1 = m s^-2

4State the vector direction.toward the source mass

Build the formula

Define gravitational field strength and derive the expression for the field strength outside a spherical mass M.

Giving only g = GM/r^2 without defining g = F/m or stating the vector direction.

Define gravitational field strength and derive the expression for the field strength outside a spherical mass M.

Choose

A field-line diagram is a vector map. The arrow direction tells the direction a small mass would be pulled. Around an isolated spherical mass, all field lines point inward toward the centre. Near Earth’s surface over small distances, the radial lines are almost parallel, so the field is often drawn as uniform.

Field lines show the direction of gravitational force on a small test mass at each point.

For a point mass or spherical mass, field lines are radial and point toward the centre.

Line density represents field strength: closer lines mean larger g.

Near the surface of a large planet, the field can be approximated as uniform, with parallel equally spaced lines pointing downward.

Field lines should not cross because the field has only one direction at a point.

Label the gravitational field-line diagrams for a radial field and a uniform near-surface field.

Label

Labels

5

Sketch the gravitational field lines around an isolated spherical planet and explain how the diagram shows field direction and strength.

Drawing arrows away from the mass or failing to state that closer lines represent stronger field.

Sketch the gravitational field lines around an isolated spherical planet and explain how the diagram shows field direction and strength.

Choose

Core D.1 is secure when the student chooses the correct gravitational representation. Kepler describes observed orbit patterns. Newton explains the attractive inverse-square force. Field strength describes force per kilogram and field lines show direction and relative strength. Circular orbit questions use centripetal force, not escape-energy arguments.

Kepler’s laws describe elliptical orbits, equal areas in equal times, and T^2 proportional to r^3 for the same central mass.

Newton’s law of gravitation is F = Gm1m2/r^2, attractive and along the line joining centres.

For spherical bodies outside the mass, use the point-mass model and measure r from centre to centre.

Gravitational field strength is g = F/m = GM/r^2 and points toward the source mass; multiple fields add vectorially.

Field lines point in the force direction, are denser where the field is stronger, and are radial around a spherical mass.

For circular orbits, gravity provides centripetal force and v = sqrt(GM/r).

Match each core D.1 cue to the response it should trigger.

Match

Reasons

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IB Physics SL D.1 Gravitational fields Notes

Map Kepler’s three laws of orbital motion

Map Universal gravitation

Point-mass approximation

Map Gravitational field strength

Map Gravitational field lines

Retrieve the Core D.1 Gravitational fields Model