recall and use g = GM / r

Gravitational Field – The Relationship \(g = \dfrac{GM}{r^{2}}\)

Learning Objectives (Cambridge AO)

• AO1 – Knowledge: Recall the definition of a gravitational field and the formula \(g = GM/r^{2}\).
• AO2 – Application: Use the formula to calculate field strength, orbital speed and period, and to estimate the change of \(g\) with altitude.
• AO3 – Analysis: Derive the field of a point mass and of a spherically symmetric body; explain the linear variation of \(g\) inside a uniform solid sphere.
• AO4 – Evaluation (A‑Level): Discuss the assumptions (point mass, spherical symmetry, neglect of rotation) and the limits of the formula.

1. Key Definitions

• Gravitational field (\(\mathbf g\)): The force per unit test mass at a point in space.
  \[ \mathbf g = \frac{\mathbf F}{m_{\text{test}}} \]
• Field‑line representation: Lines drawn radially inward toward the attracting mass; the density of lines is proportional to \(|\mathbf g|\).

2. Newton’s Law of Universal Gravitation

The vector form of the attractive force between two point masses \(M\) and \(m\) separated by a distance \(r\) is

\[ \mathbf F = -\,\frac{GMm}{r^{2}}\;\hat{\mathbf r} \] where

• \(G = 6.674\times10^{-11}\ \text{N·m}^{2}\text{kg}^{-2}\) (universal gravitational constant),
• \(\hat{\mathbf r}\) is a unit vector directed **away** from the test mass (so the force points toward \(M\)).

3. Gravitational Field of a Point Mass

1. Start from Newton’s law: \(\mathbf F = -GMm/r^{2}\,\hat{\mathbf r}\).
2. Divide by the test mass \(m\): \[ \boxed{\mathbf g = -\,\frac{GM}{r^{2}}\;\hat{\mathbf r}} \] The magnitude is \(g = GM/r^{2}\); the direction is radially inward.

4. Extension to Spherically Symmetric Bodies

If the mass distribution is spherically symmetric (e.g. a planet or a uniform solid sphere), the field outside the body is identical to that of a point mass placed at its centre (Shell Theorem). Hence for any point at distance \(r\ge R\) (where \(R\) is the radius of the body)

\[ g = \frac{GM}{r^{2}} . \]

4.1 Inside a Uniform Solid Sphere

For a point at distance \(r \[ g_{\text{inside}} = \frac{G M_{\text{enc}}}{r^{2}} = \frac{GM}{R^{3}}\,r . \]

This shows a **linear increase** of \(g\) from zero at the centre to the surface value \(GM/R^{2}\).

5. Near‑Surface Approximation

When the height \(h\) above the surface is much smaller than the planetary radius (\(h\ll R\)), we can write

\[ g(h) = \frac{GM}{(R+h)^{2}} \approx g_{0}\!\left(1-\frac{2h}{R}\right), \] where \(g_{0}=GM/R^{2}\) is the surface value (≈ 9.81 m s\(^{-2}\) for Earth). This approximation justifies treating \(g\) as constant for most everyday problems.

6. Orbital‑Motion Applications

• Centripetal condition: For a circular orbit of radius \(r\), \[ \frac{GMm}{r^{2}} = \frac{mv^{2}}{r}. \]
• Orbital speed: \[ \boxed{v = \sqrt{\frac{GM}{r}}} \]
• Orbital period: \[ \boxed{T = 2\pi\sqrt{\frac{r^{3}}{GM}}} \]
• Acceleration in orbit: \[ a_{\text{c}} = \frac{v^{2}}{r}= \frac{GM}{r^{2}} = g(r). \]

7. Practical Skills (IGCSE/A‑Level)

• Using a calibrated spring balance or a simple pendulum to measure the local value of \(g\) and compare with the theoretical value \(GM/R^{2}\).
• Applying a scientific calculator correctly (significant figures, scientific notation, square‑root function).
• Drawing accurate field‑line diagrams and labeling vectors.
• Checking units at every step – \([G]=\text{N·m}^{2}\text{kg}^{-2}\), \([M]=\text{kg}\), \([r]=\text{m}\) → \([g]=\text{m s}^{-2}\).

8. Common Mistakes to Avoid

• Using \(r\) as the height above the surface instead of the distance from the centre.
• Omitting the square on \(r\) in the denominator.
• Including the test mass \(m\) in the expression for \(g\); it cancels out.
• Confusing the orbital‑speed formula \(v^{2}=GM/r\) with the field formula \(g=GM/r^{2}\).
• Assuming the formula is valid for non‑spherical bodies or for points inside a non‑uniform sphere without justification.

9. Summary Table of Symbols

Symbol	Quantity	Units	Typical Earth Value
G	Universal gravitational constant	N·m²·kg⁻²	6.674 × 10⁻¹¹
M	Mass of attracting body	kg	5.97 × 10²⁴
R	Radius of the body (centre → surface)	m	6.37 × 10⁶
r	Distance from centre of mass (general)	m	—
h	Height above the surface	m	—
g	Gravitational field strength	m s⁻²	9.81 (surface)
v	Orbital speed (circular)	m s⁻¹	—
T	Orbital period	s	—

10. Worked Example

Problem: Calculate the gravitational field strength at a height of 300 km above the Earth’s surface.

1. Given: \(R_{\oplus}=6.37\times10^{6}\ \text{m}\), \(h=3.0\times10^{5}\ \text{m}\), \(M_{\oplus}=5.97\times10^{24}\ \text{kg}\).
2. Distance from centre: \(r = R_{\oplus}+h = 6.37\times10^{6}+3.0\times10^{5}=6.67\times10^{6}\ \text{m}\).
3. Use \(g = GM/r^{2}\): \[ g = \frac{(6.674\times10^{-11})(5.97\times10^{24})}{(6.67\times10^{6})^{2}} = \frac{3.986\times10^{14}}{4.45\times10^{13}} \approx 8.96\ \text{m s}^{-2}. \]
4. Result: At 300 km altitude, \(g \approx 8.96\ \text{m s}^{-2}\) (≈ 9 % lower than at the surface).

11. Practice Questions

1. (2 marks) A satellite orbits a planet of mass \(8.0\times10^{24}\ \text{kg}\) at a radius of \(2.0\times10^{7}\ \text{m}\). Calculate its orbital speed using \(v=\sqrt{GM/r}\).
   Answer: \(v = 5.8\times10^{3}\ \text{m s}^{-1}\).
2. (3 marks) Determine the altitude above Earth where the gravitational field strength is 80 % of the surface value.
   Solution outline: Set \(\frac{GM}{(R+h)^{2}} = 0.8\,\frac{GM}{R^{2}}\) → \((R+h) = R/\sqrt{0.8}\) → \(h = R\left(\frac{1}{\sqrt{0.8}}-1\right) \approx 1.1\times10^{5}\ \text{m}\) (≈ 110 km).
3. (4 marks) Explain why the field inside a uniform solid sphere varies linearly with distance from the centre, and derive the expression \(g = (GM/R^{3})\,r\).
   Key points:* Use the Shell Theorem, consider only the mass enclosed within radius \(r\); substitute \(M_{\text{enc}} = M(r^{3}/R^{3})\) into \(g = GM_{\text{enc}}/r^{2}\).
4. (2 marks) A pendulum of length 1.00 m has a period of 2.01 s. Use the period formula \(T=2\pi\sqrt{L/g}\) to find the experimental value of \(g\) and compare with the theoretical \(9.81\ \text{m s}^{-2}\).
   Answer: \(g_{\text{exp}} = 9.77\ \text{m s}^{-2}\); the difference is 0.4 %.

12. Diagram – Gravitational Field of a Point Mass

13. Further Reading & Extensions

• Derivation of the Shell Theorem (Gauss’s law for gravity).
• Kepler’s three laws and their connection to \(g = GM/r^{2}\).
• Gravitational potential energy, escape velocity, and binding energy of planets.
• Effect of planetary rotation on the effective value of \(g\) (centrifugal reduction).
• Non‑spherical bodies: how the field deviates from the simple \(GM/r^{2}\) form.