Gravitational Field – Cambridge International AS & A Level Physics (9702)
1. Definition and Vector Nature (AO1)
2. Derivation from Newton’s Law of Gravitation (AO2)
- Newton’s law for the force between two point masses is
\[
\mathbf{F}= -\,\frac{G M m}{r^{2}}\;\hat{\mathbf r}
\]
where \(\hat{\mathbf r}\) is the outward radial unit vector and the minus sign shows the force is attractive.
- Dividing by the test mass \(m\) gives the field produced by the source mass \(M\):
\[
\boxed{\;\mathbf{g}= -\,\frac{G M}{r^{2}}\;\hat{\mathbf r}\;}
\]
Hence the magnitude is \(g=\dfrac{G M}{r^{2}}\) and the direction is radially inward.
- Constant: \(G = 6.67\times10^{-11}\;\text{N·m}^{2}\text{kg}^{-2}\).
3. Gravitational Potential and Its Relation to the Field (AO2)
4. Gravitational Field Inside a Uniform Solid Sphere (AO1)
5. Representing a Gravitational Field with Field Lines (AO1)
Field lines are a visual aid that convey the direction and *relative* strength of a vector field.
| Rule | Meaning for a Gravitational Field |
|---|
| Origin and termination |
Lines start at infinity and terminate on the mass that creates the field; they never begin or end in empty space. |
| Direction |
Arrows on the lines point **inward**, reflecting the attractive nature of gravity. |
| Density |
The density of lines (number per unit area) is **qualitatively** proportional to the magnitude of \(\mathbf{g}\). Closer to the source the lines are more densely packed. |
| Crossing |
Lines never cross; a crossing would imply two different directions at the same point, which is impossible for a vector field. |
| Superposition |
For several masses the resultant field is obtained by vector addition of the individual fields; the combined line pattern reflects this addition. |
6. Sketches of Field‑Line Diagrams for Common Configurations (AO1)
- Isolated point mass – Radial lines converge on the mass. The spacing decreases as \(r\) decreases, illustrating the \(1/r^{2}\) dependence.
- Two equal masses (binary system) – Each mass emits its own set of lines. Mid‑way a neutral (saddle) point occurs where the fields cancel; the lines are equally spaced on either side of this point.
- Two unequal masses – The neutral point is nearer to the smaller mass. The denser set of lines belongs to the larger mass.
- Earth’s surface (or any small region of a large sphere) – Over a limited area the field may be approximated as uniform; parallel lines represent \(g\approx9.81\;\text{m s}^{-2}\).
- Uniform spherical shell – No field lines enter the interior; the field inside the shell is zero.
- Uniform solid sphere (inside) – Lines emerge from the centre with a spacing that increases linearly with radius, reflecting the linear increase of \(g\) with \(r\) (section 4).
Diagram placeholders (to be drawn by the student):
- Figure 1 – Radial lines for a single point mass.
- Figure 2 – Lines for two equal masses showing the neutral point.
- Figure 3 – Lines for two unequal masses with the shifted neutral point.
- Figure 4 – Parallel lines near Earth’s surface.
- Figure 5 – No lines inside a spherical shell.
- Figure 6 – Linear‑density lines inside a uniform solid sphere.
7. Gravitational Gauss’s Law (AO2)
- The total gravitational flux through any closed surface that encloses a point mass \(M\) is
\[
\boxed{\;\displaystyle\oint \mathbf{g}\!\cdot\!d\mathbf{A}=4\pi G M\;}
\]
This result is independent of the shape or size of the surface.
- For a spherical surface of radius \(r\):
\[
\Phi = g\,(4\pi r^{2}) = \frac{G M}{r^{2}}\,(4\pi r^{2}) = 4\pi G M .
\]
- In the syllabus the emphasis is on the *qualitative* link: a greater number of field lines crossing a surface indicates a larger flux; the exact “flux per line” is not required.
8. Superposition of Gravitational Fields (AO1 & AO2)
9. Summary Table (Quick Revision)
| Aspect |
Mathematical description |
Field‑line representation |
| Source |
Mass \(M\) (point, sphere, shell) |
Lines terminate on \(M\); start at infinity. |
| Direction |
\(\mathbf{g}\) points toward the source. |
Arrows on lines point inward. |
| Magnitude (point mass) |
\(g = \dfrac{G M}{r^{2}}\) |
Line density ∝ \(g\); denser nearer the mass. |
| Inside uniform solid sphere |
\(\mathbf{g}(r)= -\dfrac{G M}{R^{3}}\,r\;\hat{\mathbf r}\) |
Lines emerge from centre with spacing increasing linearly with \(r\). |
| Potential |
\(\Phi = -\dfrac{G M}{r}\) |
Equipotential surfaces are concentric spheres; field lines are perpendicular to them. |
| Superposition |
\(\mathbf{g}_{\text{total}} = \sum_i \mathbf{g}_i\) |
Resultant pattern is the vector sum of the individual line sets. |
| Gauss’s law (flux) |
\(\displaystyle\oint \mathbf{g}\!\cdot\!d\mathbf{A}=4\pi G M\) |
Same total number of lines cross any closed surface surrounding the mass. |
10. Common Misconceptions (AO1)
- Field lines are physical objects. They are a pictorial aid; the gravitational field exists everywhere, whether or not we draw lines.
- Number of drawn lines matters. Only the *relative* density conveys information about field strength; the absolute number is arbitrary.
- Field inside a uniform solid sphere. Many think the field is strongest at the centre; in fact it is zero at the centre and increases linearly with radius.
- Neutral point in a binary system. It lies closer to the smaller mass, not exactly midway.
- Equipotential surfaces are the same as field lines. Equipotentials are perpendicular to field lines; they are not the same thing.
11. Practice Questions (AO1 – AO3)
- Diagramming (AO1) – Draw the field‑line diagram for two masses \(M_{1}=5\;\text{kg}\) and \(M_{2}=10\;\text{kg}\) separated by \(0.20\;\text{m}\). Mark the point where the net field is zero and explain why it is nearer to the smaller mass.
- Gauss’s law calculation (AO2) – Using gravitational Gauss’s law, find the flux through a spherical surface of radius \(0.50\;\text{m}\) that encloses a point mass of \(2\;\text{kg}\). (Answer: \(\Phi = 4\pi G(2\;\text{kg})\)).
- Field‑line density with altitude (AO1) – Explain qualitatively how the spacing of the lines changes when moving from the Earth’s surface to an altitude of \(400\;\text{km}\) (ISS height). Use the \(1/r^{2}\) dependence to justify the change.
- Superposition problem (AO2) – Two equal masses are placed \(0.10\;\text{m}\) apart. Determine the magnitude and direction of the gravitational field at a point midway between them using vector addition, then illustrate the result with field lines.
- Potential and field relation (AO2) – A point mass \(M=3\;\text{kg}\) is at the origin. Write the expression for the gravitational potential \(\Phi(r)\) and show, by differentiating, that \(\mathbf{g}= -abla\Phi\) yields the field formula of section 2.
- Orbital motion application (AO3) – An artificial satellite moves in a circular orbit of radius \(r=7.0\times10^{6}\;\text{m}\) around Earth (mass \(M_{\earth}=5.97\times10^{24}\;\text{kg}\)). Using \(g = v^{2}/r\) and the expression for \(\mathbf{g}\), calculate the orbital speed \(v\). Comment on how the field‑line picture (parallel lines near the surface) relates to the uniform‑field approximation used in low‑orbit calculations.
12. Further Reading (AO1)
- Cambridge International AS & A Level Physics (9702) – Syllabus sections 5.1–5.4 (Gravitation).
- Textbook: Physics for Cambridge International A Level, chapters on Newton’s law of gravitation, gravitational potential, and Gauss’s law for gravity.
- Online:
- Cambridge Assessment International Education – “Gravitational fields and potentials” tutorial videos.
- Khan Academy – “Gravitational field and potential” playlists (useful for visualising field lines).