recall and use Newton’s law of gravitation F = Gm1m2 / r2 for the force between two point masses

Newton’s Law of Universal Gravitation

For any two point masses \(m_{1}\) and \(m_{2}\) separated by a centre‑to‑centre distance \(r\), the magnitude of the attractive force is

\[ F = \frac{G\,m_{1}m_{2}}{r^{2}} \]

• \(G\) = \(6.674\times10^{-11}\;\text{N m}^{2}\text{kg}^{-2}\) – universal gravitational constant.
• \(F\) is a vector directed along the line joining the two masses (each mass feels a force toward the other).

Key Variables and Units

Symbol	Quantity	SI Unit	Typical Value (if useful)
\(F\)	Gravitational force	newton (N)	–
\(G\)	Universal gravitational constant	m³ kg⁻¹ s⁻²	\(6.674\times10^{-11}\)
\(m_{1},\,m_{2}\)	Masses of the interacting bodies	kilogram (kg)	–
\(r\)	Separation of the centres	metre (m)	–

13 Gravitational Fields (Cambridge Syllabus)

13.1 Gravitational Field and Field Lines

• Definition: The gravitational field \(\mathbf{g}\) at a point is the force per unit test mass placed at that point \[ \mathbf{g} = \frac{\mathbf{F}}{m_{\text{test}}}\qquad\bigl[\mathbf{g}\bigr]=\text{N kg}^{-1}=\text{m s}^{-2} \]
• Direction of \(\mathbf{g}\) is **toward** the source mass (field lines point inward).
• Field‑line diagrams are a visual aid – the density of lines indicates the magnitude \(|\mathbf{g}|\).

Activity: Sketch a field‑line diagram for a single point mass. Show that the lines are radial, point toward the mass, and become less dense as the distance increases.

13.2 Gravitational Force Between Point Masses

Combining the definition of \(\mathbf{g}\) with Newton’s law gives the field produced by a point mass \(M\):

\[ \mathbf{g} = -\,\frac{GM}{r^{2}}\;\hat{\mathbf{r}} \]

The minus sign indicates that the field points toward the mass.

13.3 Gravitational Field of a Point Mass

• Magnitude: \(\displaystyle g = \frac{GM}{r^{2}}\).
• At the surface of the Earth (\(M_{E}=5.97\times10^{24}\,\text{kg}\), \(R_{E}=6.371\times10^{6}\,\text{m}\)): \[ g_{\text{Earth}} = \frac{GM_{E}}{R_{E}^{2}} \approx 9.81\;\text{m s}^{-2}

• Because \(g\propto 1/r^{2}\), the field falls rapidly with height; e.g. at \(r=2R_{E}\) the field is one‑quarter of its surface value.

13.4 Gravitational Potential and Potential Energy

• Gravitational potential \(\phi\) (potential energy per unit mass): \[ \phi = -\,\frac{GM}{r}\qquad\bigl[\phi\bigr]=\text{J kg}^{-1} \]
• Gravitational potential energy \(U\) of a mass \(m\) in the field of \(M\): \[ U = m\phi = -\,\frac{GMm}{r} \]
• The negative sign reflects that work must be done against the attractive force to separate the masses to infinity, where \(U=0\).

Weight and the Local \(g\)‑value

• Weight of an object of mass \(m\) near a planet’s surface: \[ W = mg\qquad(g\approx9.81\;\text{m s}^{-2}\text{ on Earth}) \]
• Variation of \(g\) with altitude \(h\): \[ g(h) = \frac{GM}{\bigl(R+h\bigr)^{2}} \] Hence weight decreases with height.

Orbital Motion – Circular Orbits

For a satellite of mass \(m\) in a circular orbit of radius \(r\) around a planet of mass \(M\), gravity provides the required centripetal force:

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

The orbital period follows from \(T = 2\pi r/v\):

\[ T = 2\pi\sqrt{\frac{r^{3}}{GM}} \]

Superposition of Gravitational Fields

• The net gravitational field at a point is the vector sum of the fields due to all individual masses: \[ \mathbf{g}_{\text{net}} = \sum_{i}\mathbf{g}_{i} \]
• This principle is used for calculating the field inside the Earth, on irregular bodies, or in multi‑body problems.

Experimental Determination – Cavendish (Torsion‑Balance) Experiment

• Two small lead spheres are attached to a horizontal rod suspended by a thin wire.
• Large stationary spheres are placed nearby; their mutual attraction twists the wire.
• Measuring the torsion angle gives the tiny force \(F\). With the known masses and separation \(r\), the universal constant is obtained from \[ F = \frac{G M_{1}M_{2}}{r^{2}} \;\Longrightarrow\; G = \frac{Fr^{2}}{M_{1}M_{2}} \]
• This is the only practical laboratory method for determining \(G\) (relevant to AO3 – “experimental techniques”).

Worked Examples

Example 1 – Force between two 1 kg masses 1 m apart

\[ F = \frac{(6.674\times10^{-11})(1)(1)}{(1)^{2}} = 6.7\times10^{-11}\;\text{N} \]

Order‑of‑magnitude check: the force is essentially negligible compared with everyday forces.

Example 2 – Gravitational field 500 km above Earth’s surface

\[ r = R_{E}+h = 6.371\times10^{6}\,\text{m}+5.0\times10^{5}\,\text{m}=6.871\times10^{6}\,\text{m} \] \[ g = \frac{GM_{E}}{r^{2}} = \frac{6.674\times10^{-11}\times5.97\times10^{24}}{(6.871\times10^{6})^{2}} \approx 8.7\;\text{m s}^{-2} \]

Example 3 – Orbital speed of a low‑Earth satellite (altitude = 400 km)

\[ r = 6.371\times10^{6}\,\text{m}+4.0\times10^{5}\,\text{m}=6.771\times10^{6}\,\text{m} \] \[ v = \sqrt{\frac{GM_{E}}{r}} = \sqrt{\frac{6.674\times10^{-11}\times5.97\times10^{24}}{6.771\times10^{6}}} \approx 7.67\times10^{3}\;\text{m s}^{-1} \]

Example 4 – Gravitational potential energy of a 2 kg mass at Earth’s surface

\[ U = -\frac{GM_{E}m}{R_{E}} = -\frac{6.674\times10^{-11}\times5.97\times10^{24}\times2}{6.371\times10^{6}} \approx -1.25\times10^{8}\;\text{J} \]

Common Misconceptions

• “Gravitational force is proportional to distance.” It varies as \(1/r^{2}\), not linearly.
• “Only large bodies attract each other.” Any two masses attract; the force is simply very small for everyday objects.
• “Weight is a property of the object.” Weight \(W=mg\) depends on the local gravitational field, not just on the object's mass.
• “The field is a contact force.” Gravitational fields act at a distance; they are described by vectors, not by physical contact.
• “Potential energy is always positive.” Gravitational potential energy is defined as negative, with zero at infinite separation.

Practice Questions

1. Two 2 kg masses are placed 0.5 m apart. Calculate the magnitude of the gravitational force between them.
2. A planet of mass \(8.0\times10^{24}\,\text{kg}\) exerts a force of \(3.2\times10^{22}\,\text{N}\) on a spacecraft of mass \(2.0\times10^{3}\,\text{kg}\). Find the distance between the planet’s centre and the spacecraft.
3. Determine the gravitational field \(g\) at a distance of \(3R_{E}\) from the centre of the Earth.
4. Derive the expression for the orbital speed of a satellite in a circular orbit of radius \(r\) using Newton’s law of gravitation and the centripetal‑force condition.
5. Explain why the gravitational attraction between two electrons (\(m_{e}=9.11\times10^{-31}\,\text{kg}\), charge \(e=1.60\times10^{-19}\,\text{C}\)) is negligible compared with the electrostatic force. (Use \(G=6.674\times10^{-11}\,\text{N m}^{2}\text{kg}^{-2}\) and \(k_{e}=8.988\times10^{9}\,\text{N m}^{2}\text{C}^{-2}\).)

Suggested Answers (brief)

1. \(F = \dfrac{Gm_{1}m_{2}}{r^{2}} = 6.674\times10^{-11}\times\frac{(2)(2)}{(0.5)^{2}} = 1.07\times10^{-9}\;\text{N}\).
2. Rearrange \(F = GMm/r^{2}\) → \(r = \sqrt{GMm/F}\) \[ r = \sqrt{\frac{(6.674\times10^{-11})(8.0\times10^{24})(2.0\times10^{3})}{3.2\times10^{22}}} \approx 1.0\times10^{7}\;\text{m} \]
3. \(g = \dfrac{GM_{E}}{(3R_{E})^{2}} = \dfrac{g_{\text{surface}}}{9} \approx \dfrac{9.81}{9}=1.09\;\text{m s}^{-2}\).
4. Set \(mv^{2}/r = GMm/r^{2}\) → \(v = \sqrt{GM/r}\). This is the required orbital‑speed formula.
5. Gravitational force: \(F_{g}=Gm_{e}^{2}/r^{2}\). Electrostatic force: \(F_{e}=k_{e}e^{2}/r^{2}\). Ratio \(\displaystyle\frac{F_{e}}{F_{g}} = \frac{k_{e}e^{2}}{Gm_{e}^{2}} \approx 10^{42}\). Hence the electrostatic attraction is enormously larger than the gravitational attraction.

Summary of Key Formulae

Concept	Formula	What it gives
Gravitational force	\(F = \dfrac{GM_{1}M_{2}}{r^{2}}\)	Magnitude of the attractive force between two point masses.
Gravitational field (point mass)	\(\mathbf{g} = -\dfrac{GM}{r^{2}}\hat{\mathbf{r}}\)	Force per unit test mass; direction toward the source.
Gravitational potential	\(\phi = -\dfrac{GM}{r}\)	Potential energy per unit mass.
Potential energy	\(U = -\dfrac{GMm}{r}\)	Energy of a mass \(m\) in the field of \(M\).
Weight near a surface	\(W = mg\)	Force exerted by gravity on a body of mass \(m\).
Orbital speed (circular)	\(v = \sqrt{\dfrac{GM}{r}}\)	Speed required for a circular orbit of radius \(r\).
Orbital period (circular)	\(T = 2\pi\sqrt{\dfrac{r^{3}}{GM}}\)	Time taken for one complete orbit.
Superposition of fields	\(\mathbf{g}_{\text{net}} = \sum_i \mathbf{g}_i\)	Resultant field from many masses.

Syllabus Checklist – Topic 13 Gravitational Fields

Syllabus Requirement	Covered in Notes	Action Needed
13.1 Definition & field‑line diagrams	Definition given; field‑line description included.	Add the “Sketch a field‑line diagram” activity (see above).
13.2 Gravitational force between point masses	Newton’s law stated; derivation of \(\mathbf{g}\) from it shown.	None – fully satisfied.
13.3 Gravitational field of a point mass	Magnitude, direction, and Earth‑surface example provided.	None.
13.4 Gravitational potential & potential energy	Formulas and physical meaning explained.	None.
Weight and variation of \(g\) with altitude	Weight formula and \(g(h)\) expression included.	None.
Orbital motion (circular orbits)	Derivation of speed and period, plus worked example.	None.
Superposition principle	Vector‑addition statement and brief comment.	None.
Experimental determination of \(G\) (Cavendish)	Full description of the torsion‑balance method.	None.

Final Remarks

• Newton’s law provides the quantitative basis for all gravitational calculations in the Cambridge IGCSE/A‑Level syllabus.
• Understanding the relationship between force, field, potential, and energy enables you to move fluidly between vector‑based and scalar‑based problem‑solving approaches.
• Master the ability to sketch field‑line diagrams, apply superposition, and use the Cavendish experiment as a concrete link between theory and measurement.
• These notes, examples, and practice questions are designed to give you a solid foundation for the full range of exam tasks on gravitational fields, orbital motion, and weight.