Physics | Gravitational fields

Hello Future IB Physicists! Welcome to Gravitational Fields (D.1)

Welcome to the "Fields" section of Physics! This chapter, Gravitational fields, is where we stop thinking about forces that push or pull things directly, and start thinking about the invisible "influence" objects have on the space around them.

Why is this important? Because gravitational fields govern the behavior of everything in the universe—from why you stay glued to the ground to how satellites orbit Earth. Understanding fields is a major conceptual leap in physics, providing the necessary foundation for studying Electric and Magnetic Fields later on.

Don't worry if the formulas look scary! We will break down the concepts, especially the role of the negative sign, step-by-step.

1. Newton's Law of Universal Gravitation

1.1 The Fundamental Force

The foundation of this entire chapter is Newton’s realization that the force keeping the Moon in orbit is the same force that causes an apple to fall to the ground. This is the Universal Gravitation Law.

What it states: Every point mass in the universe attracts every other point mass with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers.

The Formula:

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

Where:
\( F \) is the gravitational force (vector quantity, always attractive).
\( M \) and \( m \) are the two interacting masses.
\( r \) is the distance between the centers of the two masses.
\( G \) is the Universal Gravitational Constant, a very tiny number:
$$ G \approx 6.67 \times 10^{-11} \, \text{N m}^2 \text{ kg}^{-2} $$

Analogy: Think of this force like a really quiet friend. For everyday objects (like you and your textbook), the gravitational force is so tiny we don't notice it because G is so small. But when one of the masses is planetary (like Earth), the force becomes enormous!

Key Concept: The Inverse Square Law

Notice the \( r^2 \) in the denominator? This means gravity follows an Inverse Square Law:

• If you double the distance (\( r \times 2 \)), the force decreases by a factor of four (\( F / 4 \)).
• If you triple the distance (\( r \times 3 \)), the force decreases by a factor of nine (\( F / 9 \)).

2. Gravitational Field Strength ($g$)

In field theory, we want to know about the space around a mass, even if there isn't another mass there yet. A gravitational field is the region of space around an object in which another object experiences a gravitational force.

2.1 Definition and Calculation

Gravitational Field Strength (\( g \)) is defined as the force experienced per unit mass placed at that point in the field.

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

Units for \( g \) are \( \text{N kg}^{-1} \). Notice that this is dimensionally equivalent to the standard acceleration due to gravity unit, \( \text{m s}^{-2} \).

Calculating Field Strength (The creating mass \( M \)):

By substituting Newton’s Universal Law (\( F = G \frac{M m}{r^2} \)) into the definition of field strength (\( g = \frac{F}{m} \)):

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

Where \( M \) is the mass creating the field (e.g., the Earth).

Important: The field strength \( g \) does not depend on the mass \( m \) of the object placed in the field. It only depends on the source mass \( M \) and the distance \( r \).

2.2 Field Lines

We represent gravitational fields using field lines:

• Direction: Field lines always point inwards, toward the center of the mass \( M \), because gravity is always attractive.
• Density: The closer the lines are together, the stronger the field.
• Shape: For a spherical mass (like a planet), the field lines are radial.

3. Gravitational Potential Energy ($E_p$)

When dealing with fields over very large distances (where \( g \) is not constant), we must use a more rigorous definition of gravitational potential energy than the simple \( mgh \) you learned previously.

3.1 Defining Zero Potential Energy

In gravitational field calculations, we define the zero reference point for potential energy as being at infinity (\( r = \infty \)).

Why infinity? At an infinite distance, the gravitational force \( F \) between masses is zero. It requires zero work to move the mass further away.

3.2 The Gravitational Potential Energy Formula

Gravitational potential energy is the work done to move a mass \( m \) from infinity to a point \( r \) in the field of mass \( M \). Since gravity is attractive, you do negative work to bring the mass *in*, or positive work to move it *out* (away from the attraction).

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

Understanding the Negative Sign (Crucial Concept!)

Since \( E_p = 0 \) at infinity, and energy decreases as masses get closer (because the force is attractive and they are doing work on each other), all finite distances must have negative potential energy.

Memory Aid: A mass is "trapped" in a potential well. You must ADD positive energy to it to free it and bring its total energy up to zero (the state of being free at infinity).

Key Takeaway: Potential Energy increases (becomes less negative) as the mass moves away from the source mass \( M \).

4. Gravitational Potential ($V_g$)

Just as gravitational field strength (\( g \)) is force per unit mass, Gravitational Potential (\( V_g \)) is gravitational potential energy per unit mass.

4.1 Definition and Calculation

Potential is a scalar quantity, which makes calculations involving multiple masses much simpler than dealing with vector field strengths!

$$ V_g = \frac{E_p}{m} $$

Using the formula for \( E_p \):

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

Units for \( V_g \) are \( \text{J kg}^{-1} \) (Joules per kilogram).

Relationship to \( E_p \): If you know the potential \( V_g \) at a point, the potential energy \( E_p \) of any mass \( m \) placed there is simply \( E_p = m V_g \).

4.2 Movement in a Field

Masses naturally move from a region of higher potential (less negative, closer to zero) to a region of lower potential (more negative).

Example: If you drop a ball, it moves from the potential near the top of a building (e.g., \( -60 \, \text{MJ kg}^{-1} \)) down to the potential at ground level (e.g., \( -62 \, \text{MJ kg}^{-1} \)). It moves toward the more negative (lower) value.

4.3 Equipotential Surfaces (HL Focus)

An equipotential surface is a surface on which every point has the same gravitational potential \( V_g \).

• For a spherical mass, equipotential surfaces are concentric spheres.
• Since potential is constant on these surfaces, no work is done moving a mass along an equipotential surface.
• Crucial Link: Equipotential surfaces are always perpendicular to the gravitational field lines.

5. Applications of Gravitation

5.1 Orbital Motion

When a satellite (mass \( m \)) orbits a planet (mass \( M \)) in a stable circular path, the gravitational force provides the necessary centripetal force.

$$ F_{centripetal} = F_{gravity} $$ $$ \frac{m v^2}{r} = G \frac{M m}{r^2} $$

We can solve this for the orbital speed \( v \):

$$ v = \sqrt{\frac{G M}{r}} $$

Did you know? The speed required to maintain a stable orbit depends only on the mass of the central body (\( M \)) and the orbital radius (\( r \)). It does not depend on the mass of the satellite (\( m \))!

Total Energy of an Orbiting Satellite (HL Extension)

The total mechanical energy \( E_{total} \) of a satellite in a circular orbit is the sum of its Kinetic Energy (KE) and Gravitational Potential Energy (PE).

1. Kinetic Energy: \( KE = \frac{1}{2} m v^2 \). Using the orbital speed formula, \( KE = \frac{1}{2} m (\frac{G M}{r}) = \frac{G M m}{2r} \).

2. Potential Energy: \( PE = -G \frac{M m}{r} \).

3. Total Energy: $$ E_{total} = KE + PE = \frac{G M m}{2r} - G \frac{M m}{r} $$ $$ E_{total} = -\frac{G M m}{2r} $$

Since \( E_{total} \) is negative, the satellite is bound to the central mass (it won't drift away on its own).

5.2 Escape Speed (\( v_{esc} \))

Escape speed is the minimum speed an object must have on the surface of a mass \( M \) (radius \( R \)) to escape the gravitational influence completely, reaching infinity with zero kinetic energy.

This is calculated using the Principle of Conservation of Energy. We set the final total energy (at infinity) to zero:

$$ E_{initial} = E_{final} = 0 $$ $$ (KE + PE)_{surface} = 0 $$ $$ \frac{1}{2} m v_{esc}^2 + (-G \frac{M m}{R}) = 0 $$

Solving for \( v_{esc} \):

$$ v_{esc} = \sqrt{\frac{2 G M}{R}} $$

Analogy: Imagine throwing a ball up. If you throw it slowly, gravity wins, and it comes back down. If you throw it at escape speed, you have given it exactly enough kinetic energy to overcome its negative potential energy, allowing it to "coast" to infinity.

6. Kepler’s Laws and Field Derivations (HL ONLY)

This section dives deeper into orbital mechanics, connecting Newton’s laws to Kepler’s empirical observations.

6.1 Kepler's Third Law (The Law of Periods)

Kepler's Third Law states that for objects orbiting the same central mass, the square of the orbital period (\( T \)) is directly proportional to the cube of the average orbital radius (\( r \)).

Step-by-Step Derivation:

1. Start with the centripetal force derived from gravity (Section 5.1): $$ G \frac{M m}{r^2} = F_{centripetal} $$
2. Express the centripetal force using angular speed or period (\( T \)). Recall that \( v = \frac{2 \pi r}{T} \). $$ F_{centripetal} = m \omega^2 r = m (\frac{2 \pi}{T})^2 r $$
3. Set the forces equal: $$ G \frac{M m}{r^2} = m (\frac{4 \pi^2}{T^2}) r $$
4. Cancel \( m \) and rearrange to isolate \( T^2 \): $$ T^2 = \frac{4 \pi^2 r^3}{G M} $$

Since \( 4 \pi^2 \), \( G \), and the central mass \( M \) are all constants, we confirm Kepler’s Third Law: $$ T^2 \propto r^3 $$

This equation is incredibly powerful, allowing us to determine the mass \( M \) of a central body (like the Sun or Jupiter) merely by observing the orbital period and radius of one of its satellites.

6.2 Relating Potential and Field Strength (HL Calculus Concept)

For HL students, it is vital to know the formal mathematical relationship between the scalar potential \( V_g \) and the vector field strength \( g \).

The gravitational field strength \( g \) is the negative derivative (the negative gradient) of the gravitational potential \( V_g \) with respect to distance \( r \).

$$ g = - \frac{\Delta V_g}{\Delta r} \quad \text{or} \quad g = - \frac{d V_g}{d r} $$

Interpretation: The field strength vector points in the direction of the steepest decrease in potential. Since potential decreases (gets more negative) as you move closer to the mass, the field \( g \) points inwards, confirming our earlier definition.