understand that a satellite in a geostationary orbit remains at the same point above the Earth’s surface, with an orbital period of 24 hours, orbiting from west to east, directly above the Equator

1. Gravitational Field

1.1 Definition (Cambridge 9702 §13.1)

The gravitational field strength \( \mathbf{g} \) at a distance \(r\) from a point mass \(M\) is the force that would act on a test mass \(m_{\text{test}}\) placed at that point, per unit test mass:

\[

\mathbf{g}(r)=\frac{\mathbf{F}}{m_{\text{test}}}= -\,\frac{GM}{r^{2}}\;\hat{\mathbf{r}}

\]

• g – magnitude of the field (N kg⁻¹ = m s⁻²)
• G – universal gravitational constant, \(6.674\times10^{-11}\ \text{N m}^{2}\text{kg}^{-2}\)
• M – mass that creates the field (kg)
• r – distance from the centre of the mass (m)
• Field lines point radially inward, showing that the field is always attractive.

1.2 Worked Example – Gravitational Force (Cambridge 9702 §13.2)

Calculate the magnitude of the gravitational force between the Earth (\(M{\oplus}=5.972\times10^{24}\ \text{kg}\)) and the Moon (\(m{\text{Moon}}=7.35\times10^{22}\ \text{kg}\)), whose centre‑to‑centre distance is \(r=3.84\times10^{8}\ \text{m}\).

\[

F = G\frac{M{\oplus}m{\text{Moon}}}{r^{2}}

= 6.674\times10^{-11}\frac{(5.972\times10^{24})(7.35\times10^{22})}{(3.84\times10^{8})^{2}}

\approx 1.99\times10^{20}\ \text{N}

\]

This value can be used in an AO2 question that asks for the force or the corresponding field strength at the Moon’s position.

1.3 Derivation of \(g = GM/r^{2}\) (Cambridge 9702 §13.3)

Starting from Newton’s law of universal gravitation for two point masses,

\[

F = G\frac{M\,m_{\text{test}}}{r^{2}},

\]

divide both sides by the test mass \(m_{\text{test}}\) to obtain the field strength:

\[

g = \frac{F}{m_{\text{test}}}= \frac{GM}{r^{2}}.

\]

2. Gravitational Potential (Cambridge 9702 §13.4)

2.1 Definition

The gravitational potential \(\phi\) at distance \(r\) from a point mass \(M\) is the work done per unit mass in bringing a test mass from infinity to that point:

\[

\phi(r)= -\,\frac{GM}{r}\qquad\bigl[\text{J kg}^{-1}\bigr]

\]

2.2 Potential Energy of a Mass

\[

E_{p}=m\,\phi = -\,\frac{GMm}{r}

\]

2.3 Numerical Example

Find the gravitational potential at a height of 10 000 km above Earth’s surface.

• Distance from Earth’s centre: \(r = R_{\oplus}+10\,000\text{ km}=6.371\times10^{6}+1.0\times10^{7}=1.6371\times10^{7}\ \text{m}\)
• \(\phi = -\dfrac{GM_{\oplus}}{r}= -\dfrac{6.674\times10^{-11}\times5.972\times10^{24}}{1.6371\times10^{7}}

  \approx -2.44\times10^{7}\ \text{J kg}^{-1}\)

The negative sign indicates that work must be done against the attractive field to move the mass to infinity.

3. Motion in a Circle (Cambridge 9702 §12)

3.1 Key Relationships

• Linear (centripetal) acceleration: \(\displaystyle a_{\text{c}}=\frac{v^{2}}{r}\)
• Angular speed: \(\displaystyle \omega=\frac{v}{r}\)
• Period: \(\displaystyle T=\frac{2\pi}{\omega}=\frac{2\pi r}{v}\)

3.2 Satellite in a Circular Orbit

For a satellite of mass \(m\) in a circular orbit of radius \(r\) the required centripetal acceleration is supplied entirely by the Earth’s gravitational field:

\[

\frac{v^{2}}{r}=g(r)=\frac{GM_{\oplus}}{r^{2}}

\]

Hence

\[

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

\omega=\sqrt{\frac{GM_{\oplus}}{r^{3}}},\qquad

T=2\pi\sqrt{\frac{r^{3}}{GM_{\oplus}}}.

\]

4. Geostationary Orbit (Cambridge 9702 §13.5)

4.1 Syllabus Conditions

1. Period: \(T = 24\ \text{h}=86\,400\ \text{s}\) – exactly the Earth’s rotational period.
2. Plane: Orbit must lie in the equatorial plane (\(i=0^{\circ}\)).
3. Direction: Satellite must travel west‑to‑east, the same sense as Earth’s rotation.

4.2 Deriving the Geostationary Radius

Set the orbital period formula equal to 24 h and solve for \(r\):

\[

r{\text{geo}}=\left(\frac{GM{\oplus}T^{2}}{4\pi^{2}}\right)^{1/3}

\]

4.3 Numerical Result

Altitude is obtained by \(h{\text{geo}} = r{\text{geo}}-R_{\oplus}\).

4.4 Sample AO2 Calculation – Verify the Period

Using the derived radius, compute the orbital period and compare it with the required 24 h.

\[

T{\text{calc}} = 2\pi\sqrt{\frac{r{\text{geo}}^{3}}{GM_{\oplus}}}

= 2\pi\sqrt{\frac{(4.2164\times10^{7})^{3}}{6.674\times10^{-11}\times5.972\times10^{24}}}

\approx 86\,400\ \text{s} = 24\ \text{h}

\]

Thus the radius is consistent with the geostationary condition.

4.5 Propagation of Uncertainty (AO2)

Given typical relative uncertainties:

• \(\frac{\Delta G}{G}=0.001\) (0.1 %)
• \(\frac{\Delta M{\oplus}}{M{\oplus}}=0.001\) (0.1 %)
• \(\frac{\Delta T}{T}=0\) (the period is defined exactly)

For \(r{\text{geo}} = \bigl(GM{\oplus}T^{2}/4\pi^{2}\bigr)^{1/3}\) the relative uncertainty is

\[

\frac{\Delta r}{r}= \frac{1}{3}\left(\frac{\Delta G}{G}+\frac{\Delta M{\oplus}}{M{\oplus}}\right)

\approx \frac{1}{3}(0.001+0.001)=0.00067\;(0.067\%).

\]

Hence the altitude is known to better than 0.1 % – a useful figure for an AO2 “uncertainty” question.

5. Experimental / Modelling Skills (AO3)

5.1 Objective

Model Kepler’s 3^rd law (\(T^{2}\propto r^{3}\)) with a simple rotating “satellite” system and analyse the data.

5.2 Materials

• Motor‑driven platform (acts as the satellite)
• Four interchangeable circular rails of known radii (e.g., 0.10 m, 0.15 m, 0.20 m, 0.25 m)
• Stopwatch (0.01 s resolution) or a digital timer
• Ruler or caliper for measuring rail radii
• Safety goggles and a guard to prevent accidental contact with rotating parts

5.3 Method (linked to syllabus AO3)

1. Secure a rail of radius \(r_{1}\) on a stable base.
2. Start the motor at a low, constant speed and allow the platform to reach steady rotation.
3. Using the stopwatch, time 10 complete revolutions; record \(t{1}\) and compute the period \(T{1}=t_{1}/10\).
4. Repeat steps 1‑3 for the remaining radii (\(r{2}, r{3}, r_{4}\)).
5. Calculate \(T^{2}\) and \(r^{3}\) for each run and plot \(T^{2}\) (y‑axis) against \(r^{3}\) (x‑axis). The points should lie on a straight line through the origin.
6. Determine the gradient of the best‑fit line; compare it with the theoretical value \(\displaystyle \frac{4\pi^{2}}{g{\text{eff}}}\) where \(g{\text{eff}}\) is the centripetal “force” supplied by the motor.

5.4 Evaluation Checklist

• Systematic errors: friction in the bearings, non‑uniform motor torque, mis‑measurement of radii.
• Random errors: reaction time when using a manual stopwatch.
• Uncertainty analysis: propagate uncertainties in \(t\) (±0.02 s) and \(r\) (±0.5 mm) to obtain uncertainties in \(T^{2}\) and \(r^{3}\).
• Model relevance: Discuss how the motor’s constant torque replaces the gravitational force and why the experiment still demonstrates the \(T^{2}\propto r^{3}\) relationship.
• Safety: Keep hands away from rotating parts, wear goggles, ensure the motor is securely mounted.

6. Summary of Key Points (AO1)

• Gravitational field strength: \(g = GM/r^{2}\) (force per unit test mass).
• Gravitational potential: \(\phi = -GM/r\); potential energy \(E_{p}=m\phi\).
• For circular motion: \(a_{\text{c}}=v^{2}/r\), \(\omega=v/r\), and \(T=2\pi\sqrt{r^{3}/GM}\).
• Geostationary orbit must satisfy three syllabus conditions – 24 h period, equatorial plane, eastward motion.
• Derived geostationary radius ≈ 42 164 km from Earth’s centre (altitude ≈ 35 786 km).
• Uncertainty propagation shows the altitude is known to better than 0.1 %.
• The tabletop experiment provides hands‑on practice of AO2 calculations and AO3 experimental skills.