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.

Define the reference

Gravitational potential energy is defined as the work done to assemble the masses of a system from infinite separation. Set the potential energy at infinite separation to zero. Because gravity is attractive, bringing masses together releases energy, so the gravitational potential energy of a bound system is negative.

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.

Common trap

Do not make gravitational potential energy positive simply because the mass is high above a planet. With zero at infinity, every finite point in the isolated attractive field has negative potential energy.

Use the two-body expression

For two point masses, or two spherically symmetric bodies modelled at their centres, the gravitational potential energy is
$$E_p=-G\frac{m_1m_2}{r}$$
where $r$ is the centre-to-centre separation. The reference is $E_p=0$ at infinite separation.

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.

Common trap

Do not omit the minus sign or use $r^2$ in the potential-energy expression. $r^2$ belongs to force and field-strength laws; potential energy varies as $1/r$.

Define potential at a point

Gravitational potential $V_g$ at a point is the work done per unit mass in bringing a small test mass from infinity to that point. Set $V_g=0$ at infinity. Its SI unit is $\mathrm{J\,kg^{-1}}$, and it is a scalar quantity.

Use the point-source expression

For a point mass or spherical source of mass $M$,
$$V_g=-\frac{GM}{r}$$
where $r$ is the distance from the source centre. The test mass does not appear: potential describes the field at the point, while potential energy for a test mass is $E_p=mV_g$.

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: $W=m\Delta V_g$.

Common trap

Do not confuse potential $V_g$ in J kg⁻¹ with potential energy $E_p$ in joules, and do not use $r^2$: potential follows $1/r$, whereas field strength follows $1/r^2$.

Use the gradient relationship

Gravitational field strength is the negative gradient of gravitational potential:
$$g=-\frac{\Delta V_g}{\Delta r}$$
For a smooth graph, use the gradient of the tangent at the point. The negative sign means the field points toward decreasing potential.

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.

Common trap

Do not use the average slope over a wide curved section as the field at one point unless the question’s graph is effectively linear there. Do not drop the negative sign without stating whether you are reporting a vector component or a magnitude.

Use the potential difference

When a mass $m$ moves between two points, the work done on the mass by an external agent under the potential convention is
$$W_{\text{on}}=m\Delta V_g=m(V_{g,2}-V_{g,1})$$
where point 1 is the initial position and point 2 is the final position. Gravitational potential is a scalar, so the result depends only on the endpoints.

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.

Common trap

Check whether the question asks for work by the field or work done on the mass. Reversing the order of the potential values changes the sign.

Define an equipotential

An equipotential surface is a surface on which every point has the same gravitational potential $V_g$. Moving a mass along one equipotential gives $\Delta V_g=0$, so no work is done by the field and no external work is required for quasistatic motion along the surface.

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.

Common trap

Do not claim that every move between nearby-looking surfaces requires equal work. Read the potential labels and the starting and ending surfaces; motion along one surface has zero potential difference.

Use the perpendicular relationship

Gravitational field lines cross equipotential surfaces at right angles. The field points in the direction of decreasing gravitational potential, so the field-line arrow is normal to the equipotential and toward lower $V_g$.

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.

Common trap

Do not draw field lines along equipotentials. Moving along an equipotential has zero potential difference, while the gravitational field points across it, toward lower potential.

Define escape speed

Escape speed is the minimum speed an object must have at a point—usually the surface of a planet—to reach infinity with zero remaining speed, assuming no air resistance and no other significant gravitational fields. It is not the speed needed to enter a circular orbit.

Derive the model

At the limiting case, the initial kinetic energy supplies the increase in gravitational potential energy from $V_g=-GM/r$ to zero at infinity:
$$\frac12mv_{\rm esc}^2=\frac{GMm}{r}$$
so
$$v_{\rm esc}=\sqrt{\frac{2GM}{r}}$$
The escaping object’s mass cancels.

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 $r$.

Common trap

Do not use the circular-orbit speed $\sqrt{GM/r}$ or say that escape means “overcoming gravity” at a finite boundary. The limiting condition is reaching infinity with zero final speed.

Set the circular-orbit model

For a small mass $m$ in a circular orbit of radius $r$ around a much larger mass $M$, gravity supplies the centripetal force:
$$\frac{GMm}{r^2}=\frac{mv^2}{r}$$

Solve for speed

After cancelling $m$ and rearranging,
$$v_{\rm orbital}=\sqrt{\frac{GM}{r}}$$
The radius is measured from the centre of the central mass. The orbit is assumed circular in this syllabus objective.

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.

Common trap

Do not use escape speed for a bound circular orbit: $v_{\rm esc}=\sqrt2\,v_{\rm orbital}$ at the same radius. Also add the planet’s radius to altitude before using $r$.

Start with energy loss

Atmospheric drag opposes the satellite’s motion and removes mechanical energy from the orbit. The total orbital energy becomes more negative, so the satellite moves to a lower orbit. It does not simply slow while remaining at the same radius.

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.

Common trap

Do not conclude that drag causes the final orbital speed to decrease merely because drag opposes motion. Distinguish the instantaneous force from the speed of the new lower circular orbit.