derive, from Newton’s law of gravitation and the definition of gravitational field, the equation g = GM / r 2 for the gravitational field strength due to a point mass

Deriving the Gravitational Field Strength \(g = \dfrac{GM}{r^{2}}\)

1. Syllabus Coverage (Cambridge AS & A‑Level 9702)

2. Learning Objectives (Cambridge Assessment Objectives)

• AO1 – Knowledge & understanding: State Newton’s law of universal gravitation, define a gravitational field, and describe field‑line behaviour.
• AO2 – Application: Derive \(g = GM/r^{2}\); calculate field strength, work done, orbital speed, period, escape velocity and tidal acceleration for given data.
• AO3 – Analysis & evaluation: Interpret free‑fall or pendulum data, assess uncertainties, and discuss the limits of the point‑mass model (e.g., Earth’s non‑spherical shape, tidal effects).

3. Symbols, Quantities & Units

4. Vector Refresher (Topic 1.4)

• Vectors are written in bold (\(\mathbf{A}\)) or with an arrow (\(\vec{A}\)).
• Magnitude: \(|\mathbf{A}|\) or simply \(A\).
• Unit vector: \(\hat{\mathbf{r}}\) with \(|\hat{\mathbf{r}}|=1\).
• Vector addition: \(\mathbf{A}+\mathbf{B}\) is performed tip‑to‑tail; the resultant’s magnitude follows from the cosine rule.
• Important for gravitation: \(\mathbf{F}\) and \(\mathbf{g}\) are collinear with \(\hat{\mathbf{r}}\) (radial direction).

5. Newton’s Law of Universal Gravitation (Topic 3.1)

For two point masses \(M\) and \(m\) separated by a distance \(r\):

\[

\mathbf{F}= -\,\frac{G M m}{r^{2}}\;\hat{\mathbf{r}}

\]

• \(\hat{\mathbf{r}}\) points from the source mass \(M\) toward the test mass \(m\).
• The minus sign indicates the attractive nature of the force.

6. Definition of the Gravitational Field (Topic 13.1)

The gravitational field (or field strength) at a point is the force experienced per unit test mass:

\[

\mathbf{g}= \frac{\mathbf{F}}{m}\qquad\text{(units m s⁻²)}

\]

7. Derivation of the Field Strength

1. Insert the expression for \(\mathbf{F}\) from Newton’s law:

   \[

   \mathbf{g}= \frac{-\,G M m}{r^{2}}\;\frac{\hat{\mathbf{r}}}{m}

   = -\,\frac{G M}{r^{2}}\;\hat{\mathbf{r}}.

   \]

2. Take the magnitude to obtain the scalar field strength:

   \[

   g = |\mathbf{g}| = \frac{G M}{r^{2}}.

   \]

3. Direction: \(\mathbf{g}\) points radially inward toward the source mass.

8. Gravitational Field Lines (Visualising the Vector Field)

• Field lines are drawn such that the tangent at any point gives the direction of \(\mathbf{g}\).
• For a point mass they are straight lines converging on the mass; the density of lines is proportional to \(|\mathbf{g}|\) (i.e., to \(1/r^{2}\)).

9. Work Done in a Gravitational Field (Topic 5.2 & 13.3)

Work is the dot product of force and displacement. Moving a test mass \(m\) from radius \(r{1}\) to \(r{2}\) (with \(r{2}>r{1}\)) gives:

\[

W = \int{r{1}}^{r_{2}} \mathbf{F}\cdot d\mathbf{l}

= -\int{r{1}}^{r_{2}} \frac{G M m}{r^{2}}\,dr

= -G M m\left(\frac{1}{r{2}}-\frac{1}{r{1}}\right).

\]

Because the force is attractive, the work is negative when the mass moves away from the source.

Worked Example

Problem: A 2 kg satellite is raised from a low‑Earth orbit of radius \(r{1}=6.7\times10^{6}\) m to a geostationary orbit of radius \(r{2}=4.22\times10^{7}\) m. Calculate the work required.

Solution:

\[

W = -G M{\oplus} m\left(\frac{1}{r{2}}-\frac{1}{r_{1}}\right)

= -\bigl(6.674\times10^{-11}\bigr)(5.97\times10^{24})(2)

\left(\frac{1}{4.22\times10^{7}}-\frac{1}{6.7\times10^{6}}\right)

\approx +5.1\times10^{10}\;\text{J}.

\]

(The positive value indicates energy that must be supplied by rockets.)

10. Gravitational Potential and Potential Energy

• Potential \(\phi(r)\) is the work done per unit mass in bringing a test mass from infinity to \(r\):

  \[

  \phi(r)= -\int_{\infty}^{r}\mathbf{g}\cdot d\mathbf{l}

  = -\int_{\infty}^{r}\frac{GM}{l^{2}}\,dl

  = -\frac{GM}{r}.

  \]

• Potential energy of the test mass:

  \[

  U = m\phi = -\frac{GMm}{r}.

  \]

• Zero of potential is conventionally chosen at infinity; therefore \(U\) is negative close to the mass.

11. Orbital Motion (Topic 13.1‑13.5)

11.1 Circular Orbits

For a mass \(m\) in a circular orbit of radius \(r\) around a point mass \(M\):

\[

\frac{mv^{2}}{r}= \frac{GMm}{r^{2}} \;\Longrightarrow\; v=\sqrt{\frac{GM}{r}}.

\]

Orbital period:

\[

T = \frac{2\pi r}{v}=2\pi\sqrt{\frac{r^{3}}{GM}}.

\]

11.2 Elliptical (Keplerian) Orbits

• Specific mechanical energy (energy per unit mass):

  \[

  \epsilon = \frac{v^{2}}{2}-\frac{GM}{r}= -\frac{GM}{2a},

  \]

  where \(a\) is the semi‑major axis.

• Angular momentum per unit mass \(h = r v_{\perp}\) is constant, leading to the conic‑section equation

  \[

  r = \frac{h^{2}/GM}{1+e\cos\theta},

  \]

  with eccentricity \(e\).

11.3 Kepler’s Laws (Topic 25 – Astronomy)

1. Law 1 (Ellipses): Planets move in ellipses with the Sun at one focus.
2. Law 2 (Equal Areas): A line joining a planet to the Sun sweeps out equal areas in equal times (consequence of constant angular momentum).
3. Law 3 (Period‑Semi‑major axis): \(T^{2}= \dfrac{4\pi^{2}}{GM_{\odot}}\,a^{3}\). For circular orbits this reduces to the period formula above.

11.4 Tidal Acceleration (Advanced Topic)

Because the gravitational field varies with \(r\), a body of finite size experiences a differential force (tidal force). For a satellite of radius \(R_{s}\) at distance \(r\) from Earth:

\[

a{\text{tidal}} \approx 2\,\frac{GM{\oplus}}{r^{3}}\,R_{s}.

\]

This explains ocean tides and the Roche limit for satellite disruption.

11.5 Escape Velocity

Escape occurs when kinetic energy equals the magnitude of the gravitational potential energy at the launch point:

\[

\frac{1}{2}mv_{\text{esc}}^{2}= \frac{GMm}{r}

\;\Longrightarrow\;

v_{\text{esc}} = \sqrt{\frac{2GM}{r}}.

\]

12. Example Calculations

12.1 Gravitational field at the Earth’s surface

\[

g_{\oplus}= \frac{(6.674\times10^{-11})(5.97\times10^{24})}{(6.37\times10^{6})^{2}}

\approx 9.81\;\text{m s}^{-2}.

\]

12.2 Low‑Earth satellite orbital speed ( \(r=6.78\times10^{6}\) m )

\[

v = \sqrt{\frac{GM_{\oplus}}{r}}

= \sqrt{\frac{6.674\times10^{-11}\times5.97\times10^{24}}{6.78\times10^{6}}}

\approx 7.67\;\text{km s}^{-1}.

\]

12.3 Escape velocity from the Moon ( \(M=7.35\times10^{22}\) kg, \(R=1.74\times10^{6}\) m )

\[

v_{\text{esc}} = \sqrt{\frac{2GM}{R}}

\approx 2.38\;\text{km s}^{-1}.

\]

12.4 Tidal acceleration on a 1 m object at Earth’s surface

\[

a{\text{tidal}} \approx 2\,\frac{GM{\oplus}}{R{\oplus}^{3}}\,R{s}

= 2\,\frac{6.674\times10^{-11}\times5.97\times10^{24}}{(6.37\times10^{6})^{3}}\times1

\approx 3.1\times10^{-7}\;\text{m s}^{-2}.

\]

13. Practical Investigation (Paper 3 & 5 Skill)

14. Common Misconceptions & How to Address Them

• “Gravity only exists near Earth.” – The inverse‑square law applies to any point mass; Earth is just a convenient example.
• “\(g\) is constant everywhere.” – Emphasise the \(1/r^{2}\) dependence; over small height ranges \(g\) appears constant, but it decreases with altitude.
• “The field points outward.” – The vector form \(\mathbf{g}= -GM/r^{2}\,\hat{\mathbf{r}}\) makes the inward direction explicit.
• “Potential energy is always positive.” – Because the zero is set at infinity, \(U=-GMm/r\) is negative near the mass.
• “All orbits are circular.” – Show the derivation of elliptical orbits and Kepler’s second law to illustrate varying speed and distance.

15. Summary

• Starting from Newton’s law of universal gravitation and the definition \(\mathbf{g}= \mathbf{F}/m\), we derived the vector field \(\mathbf{g}= -GM/r^{2}\,\hat{\mathbf{r}}\) and its magnitude \(g = GM/r^{2}\).
• Field lines are radial and their density follows the inverse‑square law.
• Integrating the field gives the gravitational potential \(\phi = -GM/r\) and potential energy \(U = -GMm/r\).
• Applications include circular and elliptical orbital motion, Kepler’s laws, escape velocity, and tidal accelerations.
• A simple pendulum provides a practical route to measuring \(g\) and developing AO3 skills.
• Understanding the limits of the point‑mass model (non‑spherical Earth, tidal forces) prepares students for higher‑level astrophysics topics.