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.

Interpret the sign

Moving a mass farther from an attracting body increases gravitational potential energy toward zero and requires positive external work if done slowly. Moving it inward makes the potential energy more negative; the gravitational field can do positive work and transfer potential energy into kinetic energy.

Use the field model

Gravitational force is conservative, so the work between two fixed positions depends only on the endpoints, not the path. Near Earth’s surface, where $g$ is approximately constant, changes can be approximated by $\Delta E_p=mg\Delta h$; for large distances use the field-based potential model rather than a constant-g approximation.

Worked example — 1.0 kg at Earth’s surface

Using $M=6.0\times10^{24}\,\mathrm{kg}$, $m=1.0\,\mathrm{kg}$ and $r=6.4\times10^6\,\mathrm{m}$, $E_p=-(6.67\times10^{-11})(6.0\times10^{24})(1.0)/(6.4\times10^6)=-6.3\times10^7\,\mathrm{J}$. The negative result means energy must be supplied to separate the pair to infinity.

Interpret the negative sign

At every finite separation, $E_p<0$ because the masses form a bound configuration relative to infinity. Increasing $r$ makes $E_p$ less negative; decreasing $r$ makes it more negative. The change in potential energy is what matters when comparing two positions.

Connect to a circular orbit

For a satellite in a circular orbit, the gravitational potential energy is still $-GMm/r$. If the orbital relation gives $K=GMm/(2r)$, then the total mechanical energy is $E_T=K+E_p=-GMm/(2r)$. This orbit result is a consequence of the circular-orbit model, not a replacement for the general two-body potential-energy equation.

Worked example — orbital potential

At $r=7.9\times10^6\,\mathrm{m}$ from Earth’s centre, with $M=6.0\times10^{24}\,\mathrm{kg}$, $V_g=-(6.67\times10^{-11})(6.0\times10^{24})/(7.9\times10^6)=-5.1\times10^7\,\mathrm{J\,kg^{-1}}$. The negative value is potential energy per kilogram relative to zero at infinity.

Interpret the sign

At finite distance the potential is negative because the field does work as an attracting mass is brought inward from infinity. Moving outward increases $V_g$ toward zero; moving inward makes it more negative. The potential difference between two points is what determines work: $W=m\Delta V_g$.

Worked example — graph gradient

If a tangent changes by $3.8\times10^8\,\mathrm{J\,kg^{-1}}$ over $4.2\times10^7\,\mathrm{m}$, then $|g|=(3.8\times10^8)/(4.2\times10^7)=9.0\,\mathrm{J\,kg^{-1}m^{-1}}=9.0\,\mathrm{N\,kg^{-1}}$. Direction comes from the negative gradient.

Read a potential–distance graph

The gradient is $\Delta V_g/\Delta r$, with units $\mathrm{J\,kg^{-1}m^{-1}}=\mathrm{N\,kg^{-1}}$. A negative slope gives a positive outward radial magnitude only after the vector direction and sign convention are interpreted. Near a source, the potential changes more rapidly with distance, so the field is stronger.

Connect to work

For a mass $m$ moved between two points, $W=m\Delta V_g$ is the work done on the mass by the external agent under the stated sign convention. The field strength relation is local; potential difference and work compare endpoints.

Worked example — changing orbit

For $m=850\,\mathrm{kg}$, $V_{g,1}=-5.07\times10^7\,\mathrm{J\,kg^{-1}}$ and $V_{g,2}=-5.40\times10^7\,\mathrm{J\,kg^{-1}}$, $W_{\mathrm{on}}=850[(-5.40)-(-5.07)]\times10^7=-2.8\times10^9\,\mathrm{J}$. The negative result means the satellite must lose energy to enter the lower orbit.

Distinguish the work agent

The work done by the gravitational field is the negative of the work done on the mass by an external agent when the motion is quasistatic:
$$W_{\text{field}}=-m\Delta V_g=m(V_{g,1}-V_{g,2})$$
Moving outward raises potential toward zero, so the field does negative work; moving inward lowers potential, so the field does positive work.

Read a graph

If a potential–distance graph gives $V_{g,1}$ and $V_{g,2}$, use their difference, not the area under the graph. An area under a force–distance graph can represent work, but a potential graph already gives work per unit mass through its vertical difference.

Read the geometry

Around an isolated spherical mass, equipotential surfaces are concentric spheres; in a two-dimensional diagram they appear as concentric circles. For multiple masses, the shape is distorted by the scalar sum of the individual potentials. The numerical spacing of drawn surfaces is a choice, so use labelled potential values rather than assuming equal physical spacing means equal potential difference.

Compare movements

The external work needed to move a mass slowly between surfaces is $W_{\text{on}}=m\Delta V_g$. The greatest work for a fixed mass occurs for the largest potential difference, not automatically for the longest geometric path. Equipotentials help identify where the potential changes and where the field is strong.

Apply it to a radial field

Around an isolated spherical mass, equipotential surfaces are concentric spheres and field lines are radial. In a two-dimensional sketch, draw concentric equipotential circles and radial field arrows crossing them normally toward the mass.

Read field strength

If equal potential intervals are drawn, closer equipotential lines mean a larger potential gradient and therefore a stronger field. Farther spacing indicates a weaker field. The line/surface geometry gives direction and relative strength; the potential labels give the quantitative difference.

Derive the model

At the limiting escape condition, initial kinetic energy supplies the increase in potential energy from $-GMm/r$ to zero at infinity. The escaping object’s mass cancels.

Worked example — Earth

With $M=6.0\times10^{24}\,\mathrm{kg}$ and $r=6.4\times10^6\,\mathrm{m}$, $v_{\mathrm{esc}}=\sqrt{2(6.67\times10^{-11})(6.0\times10^{24})/(6.4\times10^6)}=1.1\times10^4\,\mathrm{m\,s^{-1}}$, about $11\,\mathrm{km\,s^{-1}}$. This is the minimum no-drag speed for zero speed at infinity.

Read the scaling

Escape speed increases with the square root of source mass and decreases with the square root of distance from its centre. For bodies with the same density, $M\propto R^3$, so at the surface $v_{\rm esc}\propto R$. Always use the source centre-to-point distance $r$.

Worked example — lunar orbit

At $100\,\mathrm{km}$ above the Moon, $r=1.737\times10^6+0.100\times10^6=1.837\times10^6\,\mathrm{m}$. With $M=7.35\times10^{22}\,\mathrm{kg}$, $v=\sqrt{(6.67\times10^{-11})(7.35\times10^{22})/(1.837\times10^6)}=1.63\times10^3\,\mathrm{m\,s^{-1}}$.

Compare orbits

At the same central mass, orbital speed decreases as $r^{-1/2}$. A satellite at a smaller circular-orbit radius moves faster. The satellite mass does not affect the required speed in this ideal model.

Explain the speed increase

For a circular orbit, $v=\sqrt{GM/r}$. As drag lowers $r$, the new circular orbital speed is larger, so the satellite speeds up as it spirals inward even though drag is an opposing force at every instant.

Follow the feedback

At lower altitude the atmosphere is generally denser, and the higher orbital speed can increase the drag effect. Continued energy loss can therefore make the orbit decay further, eventually leading to re-entry, burning or impact depending on the body and conditions.