D.1 Gravitational fields

State the three laws

1. A planet follows an elliptical orbit with the star at one focus.
2. The line from the star to the planet sweeps out equal areas in equal time intervals.
3. For bodies orbiting the same central star, the square of the orbital period is proportional to the cube of the semi-major axis: $T^2\propto a^3$.

Interpret the geometry

The semi-major axis $a$ is half the longest diameter of the ellipse. In an elliptical orbit the planet is closer to the star at one focus-side end and farther away at the other. Equal swept areas mean the planet moves faster when it is closer to the star and slower when it is farther away.

Compare orbital systems

For two planets around the same star,
$$\frac{T_Y}{T_X}=\left(\frac{a_Y}{a_X}\right)^{3/2}$$
Use the semi-major axes, not automatically the instantaneous distance from the star. The third law comparison assumes the same central mass.

Common trap

The star is at a focus, not generally at the centre of the ellipse. Also, the second law refers to equal swept areas, not equal arc lengths or equal distances travelled.

Core idea

Any two point masses attract one another with force magnitude
$$F=G\frac{m_1m_2}{r^2}$$
where $r$ is the separation between their centres and $G$ is the universal gravitational constant. The force acts along the line joining the masses and is attractive.

Read the scaling

Doubling either mass doubles the force. Doubling the separation reduces the force to one quarter. The two masses exert equal-magnitude forces on each other in opposite directions; the equation gives the interaction force, not two independent forces on the same object.

Choose the model

Use the point-mass equation when the bodies can be treated as point masses, or when a spherically symmetric body is outside its surface and the centre-to-centre separation is used. For extended irregular bodies, the simple centre-to-centre model may not be valid.

Common trap

Do not use the radius of one body as $r$ unless the other mass is effectively at its centre. Convert kilometres to metres before substituting, and remember that gravitational force is attractive rather than repulsive.

Core idea

The point-mass model replaces an extended body by a single mass at a representative point, usually its centre of mass. It is suitable when the body’s size is negligible compared with the separation involved, or when the body is spherically symmetric and the point of interest is outside it.

Use spherical symmetry

A satellite orbiting a spherical planet of uniform density can be modelled as if the planet’s entire mass were concentrated at its centre. The gravitational force then depends on the centre-to-centre distance. The same approximation can be used for two spherically symmetric bodies when their separation is measured between centres.

Check the limit

The approximation is not automatically valid for nearby irregular bodies, for points inside an extended body, or when the object’s size is comparable with the separation. In those cases different parts of the mass are at significantly different distances and their contributions cannot be represented by one point without further justification.

Common trap

Do not justify the model only by saying that the planet is “large”. The relevant reasons are small satellite-to-planet size ratio and/or spherical symmetry with an external point of interest.

Define the field

Gravitational field strength at a point is the gravitational force per unit mass on a small test mass placed at that point:
$$g=\frac{F}{m}$$
It is a vector quantity with SI unit $\mathrm{N\,kg^{-1}}$, numerically equivalent to $\mathrm{m\,s^{-2}}$.

Calculate the field of a point mass

Combining $F=GMm/r^2$ with $g=F/m$ gives
$$g=\frac{GM}{r^2}$$
where $M$ is the mass creating the field and $r$ is the distance from its centre. The test mass cancels because field strength describes the source field, not the weight of one particular object.

Read the scaling

At a fixed distance, $g$ is proportional to $M$. At a fixed source mass, doubling $r$ reduces $g$ to one quarter. The field direction is toward the source mass; the scalar expression gives the magnitude. Near a surface, the weight of a mass $m$ is $W=mg$.

Common trap

Do not use the object’s own mass in $g=GM/r^2$ as $M$, and do not use altitude alone for $r$: use distance from the source centre.

Interpret a field line

A gravitational field line is a drawn line whose tangent gives the direction of the gravitational field at each point. Arrows point toward the mass creating the field because gravity is attractive. Field lines are a representation of the vector field, not physical paths followed by test masses.

Read the pattern

Around an isolated point mass or spherical mass, field lines are radial and point inward. Where lines are closer together, the field is stronger; where they are farther apart, it is weaker. This matches the inverse-square decrease of field strength with distance.

Combine sources

For more than one source mass, the net field is the vector sum of the individual fields. At a point between two stars, draw each contribution along the line joining that star to the point and point each arrow toward its source; the resultant can then be found by vector addition.

Common trap

Do not draw field lines as zigzags, make them cross, or point them away from a positive-looking “source” label: gravitational field arrows always point toward mass. Line density indicates relative strength, not a separate force on each line.