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Here is a comprehensive, meticulously structured set of revision notes tailored for a “Top in World” (A*) level of understanding for the Cambridge A-Level Physics (9702) syllabus.

These notes are formatted specifically for Obsidian (.md), utilizing callouts, LaTeX mathematics, and a logical progression that builds conceptual depth while perfectly aligning with the 2025-2027 syllabus requirements.

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🌌 Topic 13: Gravitational Fields

Tags: 9702 Gravitational-Fields A-Star-Notes Macro-Physics Related Topics: Circular Motion, Electric Fields (for analogies)

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13.1 The Concept of a Gravitational Field

Definition: Gravitational Field

A gravitational field is an example of a field of force. It is a region of space where a mass experiences a force. It is formally defined as the gravitational force per unit mass at that point.

• Nature of the Force: Gravitational forces are strictly attractive (unlike electric/magnetic fields which can repel). The force always acts towards the centre of the mass creating the field.
• Representing Fields:
  • Radial Fields: Field lines point towards the center of a point mass or uniform sphere. As distance increases, field lines get further apart, showing decreasing field strength.
  • Uniform Fields: Represented by parallel, equally spaced lines. Near the Earth’s surface, the field is effectively uniform over small changes in height.

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13.2 Newton’s Law of Gravitation & Point Masses

Definition: Newton's Law of Gravitation

The gravitational force of attraction between two point masses is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers.

$F=\frac{G{m}_1{m}_2}{r^2}$

• $F$ = Gravitational force (N)
• $G$ = Universal Gravitational Constant ($6.67\times 10^{-11}{\text{ N m}}^2{\text{ kg}}^{-2}$)
• $m_1,m_2$ = The interacting masses (kg)
• $r$ = Distance between the centres of the masses (m)

The Point Mass Approximation

For a point outside a uniform sphere (like a planet), the mass of the sphere can be considered to be completely concentrated as a point mass at its centre. Condition: This is only valid if the sphere has uniform density (or spherical symmetry) and the distance $r$ is greater than the radius of the sphere.

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13.3 Gravitational Field Strength ($g$)

Definition: Gravitational Field Strength ( $g$)

The gravitational force exerted per unit mass on a small object placed at that point. $g=\frac{F}{m}$ Units: ${\text{N kg}}^{-1}$ (equivalent to ${\text{m s}}^{-2}$)

Deriving $g$ for a Point Mass

(Syllabus Requirement 13.3.1)

1. Start with Newton’s Law for a test mass $m$ in the field of a large mass $M$: $F=\frac{GMm}{r^2}$
2. Use the definition of field strength: $g=\frac{F}{m}$
3. Substitute $F$: $g=\frac{\frac{GMm}{r^2}}{m}$
4. Final Equation: $g=\frac{GM}{r^2}$ (Note: $g$ obeys an inverse-square law with distance $r$.)

Why is $g$ constant near the Earth’s surface?

• The Earth’s radius $R\approx 6400\text{ km}$.
• For small changes in height $h$ (e.g., $h=100\text{ m}$), the distance from the center is $R+h$.
• Because $R\gg h$, the value of $(R+h{)}^2\approx R^2$. Therefore, $g$ changes by a negligible amount and is approximated as a constant $9.81{\text{ N kg}}^{-1}$.

Examiner Pitfall: $r$ vs Altitude

A classic trap! Orbital altitude / height ($h$) is measured from the surface. To use the equations, you must find $r$ by adding the radius of the planet ($R$). $r=R+h$

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13.4 Circular Orbits & Satellites

In orbit, the gravitational force provides the exact centripetal force required to keep the object in circular motion.

$F_G=F_C$ $\frac{GMm}{r^2}=\frac{m{v}^2}{r}=mr{\omega }^2$

From this, you can derive several critical relationships:

1. Orbital Speed ($v$): $v=\sqrt{\frac{GM}{r}}$ (Notice: mass of the satellite $m$ cancels out. All objects at radius $r$ orbit at the exact same speed).
2. Kepler’s Third Law ($T^2\propto r^3$): Using $\omega =\frac{2\pi }{T}$: $\frac{GM}{r^2}=r{\left(\frac{2\pi }{T}\right)}^2⟹T^2=\left(\frac{4{\pi }^2}{GM}\right)r^3$

Top-in-World Application: Binary Star Systems

In binary star questions (two stars orbiting a common center), both stars share the same angular velocity ($\omega$) and time period ($T$). The gravitational force between them provides the centripetal force for each star. $F=\frac{G{m}_1{m}_2}{(r_1+r_2{)}^2}=m_1{r}_1{\omega }^2=m_2{r}_2{\omega }^2$

Geostationary Orbits

A geostationary satellite remains at the exact same point above the Earth’s surface. To do this, it must satisfy three strict conditions:

1. It must have an orbital period of exactly 24 hours.
2. It must orbit from West to East (the same direction the Earth rotates).
3. It must orbit directly above the Equator. (Main use: Telecommunications and satellite TV, because satellite dishes on Earth do not need to move to track it).

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13.5 Gravitational Potential ($\varphi$)

Definition: Gravitational Potential ( $\varphi$)

The work done per unit mass in bringing a small test mass from infinity to that point in the field. $\varphi =-\frac{GM}{r}$ Units: ${\text{J kg}}^{-1}$ (Scalar quantity)

Why is Potential ALWAYS Negative?

This is a guaranteed exam question. Memorize this logic:

1. Gravitational potential is strictly defined as zero at infinity.
2. Gravitational forces are always attractive.
3. Therefore, as a mass moves from infinity towards a planet, the field does work on the mass (it accelerates).
4. To move it back to infinity, external work must be done against the field. Thus, the potential everywhere closer than infinity must be less than zero (negative).

Equipotential Surfaces

• Lines or surfaces where the gravitational potential $\varphi$ is identical.
• Moving a mass along an equipotential surface requires zero work.
• Equipotentials are always perpendicular to gravitational field lines.

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13.6 Gravitational Potential Energy ($E_p$)

Gravitational Potential Energy of a two-mass system is the work done bringing the masses from infinity to a separation $r$. Because $E_p=m\varphi$, we get:

$E_p=-\frac{GMm}{r}$

Calculating Changes in Energy ($\Delta E_p$)

To find the work done moving a satellite from radius $r_1$ to radius $r_2$: $\text{Work Done}=\Delta E_p=E_{p(\text{final})}-E_{p(\text{initial})}$ $\Delta E_p=\left(-\frac{GMm}{r_2}\right)-\left(-\frac{GMm}{r_1}\right)=GMm\left(\frac{1}{r_1}-\frac{1}{r_2}\right)$

$mgh$ vs $-\frac{GMm}{r}$

Never use $\Delta E_p=mgh$ for satellites in space! $mgh$ is an approximation that only works near the Earth’s surface where $g$ is constant. For large distances (orbits), $g$ changes, so you must use the $E_p=-\frac{GMm}{r}$ formula.

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🏆 A* / Top in World Examiner Insights

1. Big $G$ vs Little $g$: Never confuse them in writing. $G$ is the Universal Gravitational Constant (a fundamental property of the universe). $g$ is local field strength (depends on where you are and the mass of the nearby planet).
2. Logarithmic Graphs (Paper 5 Focus): You are often given $T^2\propto r^3$ and asked to verify it.
   • Taking logs: $\log (T)=\frac{3}{2}\log (r)+\text{constant}$.
   • A graph of $\log (T)$ vs $\log (r)$ will yield a straight line with a gradient of exactly 1.5.
3. Resultant Field Strength: $g$ is a vector. If a question asks for the point between Earth and the Moon where the net field is zero (neutral point), you must equate the magnitudes: $g_{earth}=g_{moon}$ $⟹\frac{G{M}_E}{r_1^2}=\frac{G{M}_M}{r_2^2}$.
4. Work Done calculation signs: If an object falls towards a planet, it loses $E_p$ (potential becomes more negative) and gains Kinetic Energy ($E_k$). If launched away, it gains $E_p$ (potential becomes less negative, closer to zero) and loses $E_k$. Keep strict track of your minus signs!
5. Definitions are absolute: When defining fields or potentials, the exact syllabus wording is required. Missing “per unit mass” or “from infinity” will instantly drop the mark.