13 Gravitational Fields and Forces (Cambridge A‑Level 9702)
13.1 Gravitational field (field strength \(g\))
- Definition: The gravitational field strength \(\mathbf g\) at a point is the force that would act on a test mass of 1 kg placed at that point.
\[
\mathbf g=\frac{\mathbf F}{m_{\text{test}}}\qquad\bigl[\mathbf g\bigr]=\mathrm{N\,kg^{-1}}=\mathrm{m\,s^{-2}}
- Vector form for a point mass \(M\) (or a sphere treated as a point mass):
\[
\boxed{\displaystyle \mathbf g = -\,\frac{GM}{r^{2}}\;\hat{\mathbf r}}
\]
\(\hat{\mathbf r}\) is a unit vector directed radially outward from the centre; the minus sign shows that \(\mathbf g\) points toward the mass.
- Scalar form (often used in calculations):
\[
g = \frac{GM}{r^{2}}
\]
(the direction is always toward the source).
- Field‑line picture: Radial lines pointing toward the mass; the density of lines is proportional to \(|\mathbf g|\). (Insert diagram of concentric arrows pointing inward.)
- Relation to weight: The weight of an object of mass \(m\) is
\[
W = mg = m|\mathbf g|
\]
Near the Earth’s surface \(g\approx9.8\;\text{m s}^{-2}\) and can be treated as constant for most A‑Level problems.
13.2 Gravitational force between point masses
Newton’s law of universal gravitation (vector form):
\[
\boxed{\displaystyle \mathbf F{12}= -\,\frac{G M{1}M_{2}}{r^{2}}\;\hat{\mathbf r}}
\]
where \(\mathbf F{12}\) is the force on mass \(M{2}\) due to mass \(M{1}\), \(r\) is the separation of their centres, and \(\hat{\mathbf r}\) points from \(M{1}\) to \(M_{2}\).
Scalar form (useful for magnitude calculations):
\[
F = \frac{G M{1}M{2}}{r^{2}}
\]
Link to the field:
- Force on a test mass \(m\) due to a source mass \(M\):
\(\displaystyle \mathbf F = -\frac{GMm}{r^{2}}\hat{\mathbf r}\).
- Dividing by the test mass gives the field:
\(\displaystyle \mathbf g = \frac{\mathbf F}{m}= -\frac{GM}{r^{2}}\hat{\mathbf r}\).
13.3 Shell theorem (uniform spherical masses)
The theorem has two parts that are directly usable in Cambridge problems.
- External point (\(r>R\)):
A spherically symmetric body of total mass \(M\) produces the same gravitational field at any point outside the body as if all its mass were concentrated at its centre.
\[
\boxed{\displaystyle \mathbf g_{\text{ext}} = -\frac{GM}{r^{2}}\hat{\mathbf r}},\qquad r\ge R
\]
- Internal point (\(r:
- Inside a thin spherical shell the net field is zero.
- Inside a solid sphere only the mass at radii \(
\[
\boxed{\displaystyle \mathbf g_{\text{int}} = -\frac{GM}{R^{3}}\,r\;\hat{\mathbf r}},\qquad r\le R
\]
Sketch of the proof (external point)
- Consider a thin spherical shell of radius \(R\) and total mass \(M\). Choose an external point \(P\) at distance \(r>R\) from the centre \(C\).
- Take two infinitesimal mass elements \(dm{1}\) and \(dm{2}\) that subtend the same solid angle \(\mathrm d\Omega\) at \(P\) and lie on opposite sides of the line \(CP\). Their distances from \(P\) are \(s{1}\) and \(s{2}\).
- The forces on a test mass \(m\) are
\[
\mathrm dF{1}= \frac{G\,dm{1}\,m}{s_{1}^{2}},\qquad
\mathrm dF{2}= \frac{G\,dm{2}\,m}{s_{2}^{2}}.
\]
The components perpendicular to \(CP\) cancel because \(\mathrm dF{1}\sin\theta{1}= \mathrm dF{2}\sin\theta{2}\).
- The radial components add, and after integrating over the whole shell one obtains
\[
F_{\text{net}} = \frac{GMm}{r^{2}},
\]
i.e. the same result as a point mass \(M\) at the centre.
Result for a solid sphere
A solid sphere can be regarded as a stack of concentric shells. Using the external‑point result for each shell and summing the contributions that lie inside radius \(r\) leads to the linear‑in‑\(r\) field shown above.
13.4 Gravitational potential energy (external points)
For a particle of mass \(m\) at a distance \(r\) from the centre of a uniform sphere (or any point mass) the gravitational potential energy is
\[
\boxed{U(r) = -\,\frac{GMm}{r}}\qquad (r\ge R)
\]
Derivation (work required to bring the particle from infinity to \(r\)):
\[
U = -\int_{\infty}^{r}\frac{GMm}{x^{2}}\,\mathrm dx
= -\frac{GMm}{r}.
\]
Quick‑reference formulae
| Quantity | Expression | Applicable region |
|---|
| Gravitational force (external) | \(F = \displaystyle\frac{GMm}{r^{2}}\) | \(r\ge R\) |
| Gravitational field (external) | \(\displaystyle\mathbf g = -\frac{GM}{r^{2}}\hat{\mathbf r}\) | \(r\ge R\) |
| Gravitational force (inside solid sphere) | \(F = \displaystyle\frac{GMm}{R^{3}}\,r\) | \(r\le R\) |
| Gravitational field (inside solid sphere) | \(\displaystyle\mathbf g = -\frac{GM}{R^{3}}\,r\;\hat{\mathbf r}\) | \(r\le R\) |
| Gravitational potential energy (external) | \(U = -\displaystyle\frac{GMm}{r}\) | \(r\ge R\) |
Worked example (using the shell theorem)
Problem: Find the weight of a 70 kg person standing on the surface of the Earth (\(M{\oplus}=5.97\times10^{24}\,\text{kg}\), \(R{\oplus}=6.37\times10^{6}\,\text{m}\)).
Solution:
- The person is at an external point, so the Earth can be treated as a point mass at its centre (Shell theorem).
- Apply \(F = GM_{\oplus}m / r^{2}\):
\[
F = 6.674\times10^{-11}\frac{(5.97\times10^{24})(70)}{(6.37\times10^{6})^{2}}
\approx 6.86\times10^{2}\ \text{N}.
\]
- Thus the weight \(W = mg \approx 686\ \text{N}\), which matches the familiar value \(W\approx70\times9.8\ \text{N}\).
Common misconceptions (and how to avoid them)
- “Mass distribution matters for the external field.” – For any spherically symmetric body the external field depends only on the total mass, not on how that mass is distributed.
- “Gravity inside a solid sphere is the same as on the surface.” – Inside, only the mass at radii smaller than the point contributes; the field falls linearly to zero at the centre.
- “A hollow sphere exerts the same force as a solid sphere everywhere.” – Outside they are identical, but inside a hollow shell the field is zero, whereas a solid sphere gives a non‑zero field.
- “The sign of \(U\) is arbitrary.” – Gravitational potential energy is defined to be zero at infinity; therefore it is always negative for bound systems.
Practice questions (Cambridge style)
- A uniform solid sphere of mass \(M=2.0\ \text{kg}\) and radius \(R=0.10\ \text{m}\) floats in space. Calculate the gravitational force on a particle of mass \(0.5\ \text{kg}\) located \(0.25\ \text{m}\) from the sphere’s centre.
- Show that the gravitational potential energy of a particle of mass \(m\) at a distance \(r\) from the centre of a uniform sphere (outside the sphere) is \(U=-GMm/r\). Include the integration step.
- A thin spherical shell of mass \(M=5.0\ \text{kg}\) and radius \(R=0.20\ \text{m}\) has a small hole through its centre. A \(0.1\ \text{kg}\) mass is placed at the centre of the hole. What is the net gravitational force on the mass? Explain using the shell theorem.
- Derive the expression for the gravitational field inside a uniform solid sphere and verify that at the surface (\(r=R\)) it reduces to \(g = GM/R^{2}\).
- Two identical solid spheres, each of mass \(M\) and radius \(R\), are placed a distance \(3R\) apart centre‑to‑centre. A small test mass \(m\) is located midway between them. Calculate the magnitude and direction of the net gravitational field at that point.
Summary
- The gravitational field strength is the force per unit mass: \(\mathbf g = -GM/r^{2}\,\hat{\mathbf r}\) for any point mass.
- By the shell theorem, a uniform sphere (solid or hollow) behaves like a point mass at its centre for all points outside the sphere.
- Inside a solid sphere the field varies linearly with distance, \(\mathbf g = -GM r / R^{3}\,\hat{\mathbf r}\); inside a thin shell it is zero.
- Gravitational potential energy for external points is \(U=-GMm/r\).
- These results allow rapid calculation of forces, fields, and energies for planets, stars, and laboratory spheres—exactly what the Cambridge 9702 syllabus expects.