D.1 Gravitational fields

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.

Worked example — Earth and a book

For $m=1.0\,\mathrm{kg}$, $M=6.0\times10^{24}\,\mathrm{kg}$ and $r=6.4\times10^6\,\mathrm{m}$, $F=(6.67\times10^{-11})(1.0)(6.0\times10^{24})/(6.4\times10^6)^2=9.8\,\mathrm{N}$. This is the book’s weight; the book attracts Earth with the same force magnitude.

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.

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.

Worked example — surface field

For $M=4.87\times10^{24}\,\mathrm{kg}$ and $r=6.05\times10^6\,\mathrm{m}$, $g=(6.67\times10^{-11})(4.87\times10^{24})/(6.05\times10^6)^2=8.87\,\mathrm{N\,kg^{-1}}$. The result is the force per kilogram at the surface, directed inward.

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 $W=mg$.

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.