Physics (9702) | Gravitational field

Gravitational Fields (9702 A Level Physics) Study Notes

Hello future Physicist! Welcome to one of the most exciting topics in A Level Physics: Gravitational Fields. This chapter explains how large masses, like planets and stars, interact across vast distances. Don't worry if the formulas look intimidating—we will break down the concepts of force, field strength, and potential step-by-step. Understanding fields is essential for studying satellites, space travel, and even mapping the universe!

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13.1 The Gravitational Field Concept

Before we dive into calculations, we need to understand what a field actually is.

What is a Field of Force?

In physics, a field of force is a region of space where an object experiences a non-contact force. You learned about weight ($W = mg$) in AS Physics; now we define the cause of that weight—the gravitational field.

• A gravitational field is created by any object possessing mass.
• It is an example of an attractive field of force.

Defining Gravitational Field Strength (g)

The key definition you must memorize is:

Definition: The gravitational field strength ($g$) at a point is the force per unit mass exerted on a small test mass placed at that point.

This links directly to our familiar weight equation:

$$g = \frac{F}{m}$$

where $F$ is the gravitational force acting on mass $m$.

• Unit: Since $g = F/m$, the SI unit for gravitational field strength is Newtons per kilogram ($\text{N}\text{kg}^{-1}$).
• Vector Quantity: Gravitational field strength is a vector. It always points in the direction of the force—which is always towards the mass creating the field.

Representing Fields with Field Lines

We visualize fields using gravitational field lines (or lines of force):

1. Field lines show the direction of the force acting on a small test mass (always inwards, towards the attracting mass).
2. The density (closeness) of the field lines shows the magnitude of the field strength ($g$). Where lines are closer together, the field is stronger.

Analogy: Imagine rain pouring onto Earth. The closer the drops, the stronger the field. All "drops" fall directly toward the center of the Earth.

Key Takeaway (13.1): A gravitational field is defined as the force experienced per unit mass. Field lines are always directed inwards towards the massive object.

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13.2 Gravitational Force between Point Masses

Newton’s Law of Gravitation

This fundamental law allows us to calculate the attractive force between any two masses.

Newton’s Law of Gravitation states: The gravitational force ($F$) between two point masses ($m_1$ and $m_2$) is directly proportional to the product of their masses and inversely proportional to the square of their separation ($r$).

$$F = G \frac{m_1 m_2}{r^2}$$

Key Components and Jargon Busting:

• $F$: The gravitational force (in N). This force is equal and opposite on both masses (Newton’s Third Law).
• $m_1, m_2$: The interacting masses (in kg).
• $r$: The separation distance between the centres of the masses (in m).
• $G$: The Universal Gravitational Constant. This is a constant value that you will find in your data booklet: \(G \approx 6.67 \times 10^{-11} \, \text{N}\text{m}^2\text{kg}^{-2}\).

The Inverse Square Law

The term \(1/r^2\) means gravity follows an inverse square law:

• If you double the distance ($r \to 2r$), the force decreases by a factor of $2^2 = 4$ ($F \to F/4$).
• If you triple the distance ($r \to 3r$), the force decreases by a factor of $3^2 = 9$ ($F \to F/9$).

Memory Aid: "Distance Kills Force Squared."

Point Masses and Uniform Spheres

For large objects like planets, it's difficult to measure $r$ accurately unless we treat them simply. The syllabus requires you to understand that:

Assumption: For a point outside a uniform sphere (like Earth), the entire mass of the sphere ($M$) may be considered to be concentrated as a point mass at its centre.

This simplifies all gravitational calculations for planets, moons, and stars—$r$ is measured from the center of the sphere to the center of the other mass (or point of interest).

Key Takeaway (13.2): Gravitational force depends strongly on distance (inverse square law) and mass. Always measure distance $r$ from the centre of the masses.

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13.3 Gravitational Field Strength (g) due to a Point Mass

We can now combine the definition of field strength ($g=F/m$) with Newton’s Law ($F = G M m/r^2$).

Derivation of $g = GM/r^2$ (Syllabus Requirement)

This derivation shows how the gravitational field strength depends only on the mass creating the field and the distance from it.

1. Start with Newton’s Law for a mass $M$ creating the field and a small test mass $m$: $$F = G \frac{M m}{r^2}$$
2. Recall the definition of gravitational field strength: $$g = \frac{F}{m}$$
3. Substitute (1) into (2): $$g = \frac{G \frac{M m}{r^2}}{m}$$
4. The test mass $m$ cancels out: $$g = \frac{G M}{r^2}$$

This is a powerful result! The gravitational field strength ($g$) at any point around a massive object ($M$) does not depend on the mass of the object experiencing the field ($m$).

Gravitational Field Strength near Earth’s Surface

For A Level, you must understand why the acceleration of free fall $g$ is often treated as constant near the surface of the Earth (\(g \approx 9.81 \, \text{m}\text{s}^{-2}\) or $\text{N}\text{kg}^{-1}$).

The formula \(g = \frac{G M}{r^2}\) tells us that $g$ changes with distance $r$.

• $M$ is the mass of the Earth.
• $r$ is the distance from the centre of the Earth.
• The radius of the Earth ($R_E$) is about $6400 \, \text{km}$.

If you climb a mountain (say, $1 \, \text{km}$ high), the distance $r$ changes from $6400 \, \text{km}$ to $6401 \, \text{km}$. This change in $r$ is tiny (less than 0.02%) compared to the total distance. Therefore, $g$ is approximately constant for small changes in height near the Earth’s surface.

However, when dealing with satellites orbiting hundreds or thousands of kilometres away, $r$ changes significantly, and you must use the full formula \(g = \frac{G M}{r^2}\).

Key Takeaway (13.3): Field strength ($g = GM/r^2$) is the vector equivalent of force per unit mass. $g$ is constant near Earth because small changes in height are negligible compared to the Earth’s radius.

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13.4 Gravitational Potential ($\phi$) and Potential Energy ($E_p$)

Gravitational field strength ($g$) is a vector. To simplify things, especially in energy problems, we use a scalar concept called Gravitational Potential.

Defining Gravitational Potential ($\phi$)

This definition is crucial and is very similar to electric potential, but involves mass instead of charge.

Definition: The gravitational potential ($\phi$) at a point is the work done per unit mass in bringing a small test mass from infinity to that point.

$$ \phi = \frac{W}{m} $$

Why the Negative Sign?

For potential due to a point mass $M$, the formula is:

$$ \phi = -\frac{G M}{r} $$

This negative sign is often confusing but makes perfect sense:

1. Reference Point: By convention, gravitational potential is defined as zero at infinity ($r = \infty$).
2. Gravity is Attractive: Since gravity is attractive, the field does work *on* the mass as it moves closer to $M$.
3. Work Done: If the field does the work, we don't need to put energy in (or we must apply an opposing force). Therefore, the potential must decrease from zero, becoming negative.

The closer you get to a mass, the more negative the potential becomes (e.g., $-10 \, \text{J}\text{kg}^{-1}$ is a lower potential than $-5 \, \text{J}\text{kg}^{-1}$).

• Unit: Since $\phi = W/m$, the SI unit is Joules per kilogram ($\text{J}\text{kg}^{-1}$).

Common Mistake Alert: Students often forget the negative sign. Remember, you fall into a potential well; potential is always negative in an attractive field!

Gravitational Potential Energy ($E_p$)

Just as potential is work per unit mass, Gravitational Potential Energy ($E_p$) is the total work done bringing the mass $m$ from infinity.

$$E_p = m \phi$$

For two point masses, $M$ and $m$, separated by $r$, the formula for the potential energy of the system is:

$$ E_p = -\frac{G M m}{r} $$

This is the energy needed to separate the two masses completely (to infinity).

Relationship between $g$ and $\phi$:

The gravitational field strength ($g$) is the negative of the potential gradient (\(d\phi/dr\)).

$$g = - \frac{\Delta \phi}{\Delta r}$$

In simple terms, $g$ tells you how steeply the potential ($\phi$) changes. Since $g$ is a vector pointing inwards (negative direction of $r$), it must be the negative of the scalar potential gradient.

Key Takeaway (13.4): Gravitational potential ($\phi$) and potential energy ($E_p$) are scalar quantities and are defined as zero at infinity. They are always negative in an attractive field.

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Applications: Orbits and Satellites

Circular Orbits

When a satellite of mass $m$ orbits a planet of mass $M$ in a circular path of radius $r$, the physics is straightforward: the gravitational attraction provides the necessary centripetal force.

$$ \text{Gravitational Force} = \text{Centripetal Force} $$

$$ F_G = F_C $$

Using the formulae for $F_G$ and $F_C$ (from Topic 12, Motion in a Circle):

$$ G \frac{M m}{r^2} = \frac{m v^2}{r} \quad \text{or} \quad G \frac{M m}{r^2} = m r \omega^2 $$

We can rearrange this relationship to find the required orbital speed ($v$) or angular speed ($\omega$):

$$ v^2 = \frac{G M}{r} $$

Notice that the mass of the satellite ($m$) cancels out. This means that a feather and a spacecraft would orbit at the same speed if placed in the same orbit $r$!

Geostationary Satellites

A geostationary orbit is a specific type of orbit used primarily for communication satellites, as they remain fixed relative to a point on Earth.

To be geostationary, the satellite must meet three strict criteria (which are all required knowledge):

1. Orbital Period (T): The satellite must have a period of 24 hours (equal to the rotation period of the Earth).
2. Direction of Orbit: It must orbit from West to East (the same direction as Earth’s rotation).
3. Location: It must orbit directly above the Equator.

If these conditions are met, the satellite will appear fixed in the sky relative to an observer on Earth, allowing ground antennae to remain pointed at it permanently.

Key Takeaway (Applications): Satellite motion is fundamentally a balance between gravitational force and centripetal force. Remember the three conditions for a geostationary orbit!

Congratulations! You've navigated the tricky waters of Gravitational Fields. Keep practicing those negative signs and inverse square calculations, and you'll master this topic!