Welcome to Gravitational Fields!
Have you ever wondered what keeps the Moon in orbit around the Earth, or why you stay firmly grounded on the floor? The answer is gravity. In this chapter of your A2 Physics course, we explore gravitational fields — invisible regions of space where masses exert attractive forces on one another. Don't worry if field theories sound a little abstract at first! We will break down every concept step-by-step, using clear analogies and straightforward mathematics.
1. Newton's Law of Universal Gravitation
Sir Isaac Newton realized that every single mass in the universe attracts every other mass. Whether it is an apple falling from a tree or planets orbiting the Sun, the fundamental rule is the same.
The Statement
Newton's Law of Universal Gravitation states that the attractive gravitational force \(F\) between two point masses is:
• Directly proportional to the product of their masses (\(m_1 \times m_2\))
• Inversely proportional to the square of the distance between their centers (\(r^2\))
The Mathematical Formula
\(F = \frac{G m_1 m_2}{r^2}\)
Where:
• \(F\) = Gravitational force between the two masses (measured in Newtons, \(\text{N}\))
• \(G\) = Universal Gravitational Constant (\(6.67 \times 10^{-11} \text{ N m}^2 \text{ kg}^{-2}\))
• \(m_1\) and \(m_2\) = The two interacting masses (measured in kilograms, \(\text{kg}\))
• \(r\) = Distance between the centers of mass of the two objects (measured in meters, \(\text{m}\))
Point Masses and Spherical Bodies
Newton's equation strictly applies to point masses (masses concentrated at a single point). However, for uniform spherical bodies like planets and stars, we can treat them as point masses with all their mass concentrated at their center.
Common Mistake Alert: Always measure \(r\) from the center of one object to the center of the other, NOT from their surfaces!
Did You Know?
The gravitational constant \(G\) is extremely small (\(6.67 \times 10^{-11}\)). This explains why you don't physically feel yourself being pulled towards your desk or your textbook — you need enormous masses (like the Earth) for gravitational force to become noticeably large!
Key Takeaway: Gravitational force is always attractive and follows an inverse-square law: if you double the distance between two masses, the force drops to one-quarter (\(\frac{1}{4}\)) of its original value.
2. Gravitational Field Strength (\(g\))
A gravitational field is a region of space where any object with mass experiences an attractive gravitational force.
Definition of Gravitational Field Strength
Gravitational field strength, \(g\), at a point is defined as the force per unit mass acting on a small test mass placed at that point.
\(g = \frac{F}{m}\)
Units for \(g\): Newtons per kilogram (\(\text{N kg}^{-1}\)) or meters per second squared (\(\text{m s}^{-2}\)).
Radial Field of a Point Mass (or Planet)
If we substitute Newton's Law of Gravitation \(F = \frac{G M m}{r^2}\) into the definition \(g = \frac{F}{m}\), the small test mass \(m\) cancels out:
\(g = \frac{G M}{r^2}\)
Where \(M\) is the mass creating the field, and \(r\) is the distance from the center of that mass.
Uniform Fields vs. Radial Fields
• Radial Fields: Around isolated spheres or planets, field lines point inward towards the center. The lines spread out as you move further away, showing that \(g\) decreases with distance (\(g \propto \frac{1}{r^2}\)).
• Uniform Fields: Close to the surface of the Earth (over small distances and heights), the curvature of the Earth is negligible. The field lines are parallel and equally spaced, meaning \(g\) is constant (\(\approx 9.81 \text{ N kg}^{-1}\)).
Key Takeaway: Gravitational field strength depends only on the mass of the object creating the field and the distance from its center — not on the mass of the object placed in it.
3. Gravitational Potential (\(V\)) and Potential Energy (\(E_p\))
What is Gravitational Potential?
The gravitational potential, \(V\), at a point in a gravitational field is defined as the work done per unit mass in bringing a small test mass from infinity to that point.
\(V = -\frac{G M}{r}\)
Units for \(V\): Joules per kilogram (\(\text{J kg}^{-1}\)).
Why is Gravitational Potential Always Negative?
This is one of the most common questions students ask! Here is the step-by-step logic:
1. We define the gravitational potential at infinity (infinitely far away from all masses) to be zero (\(V = 0\)).
2. Because gravity is always attractive, the field naturally pulls a mass towards the planet. No external work is needed to pull it in; instead, the field does work on the mass.
3. Since moving towards the mass releases energy, the potential energy must decrease below zero.
4. Therefore, anywhere closer than infinity has a negative potential (\(V < 0\)).
Analogy: Imagine being at the top of a deep well (ground level = \(0 \text{ m}\)). As you climb down into the well, your position becomes negative (\(-10 \text{ m}\), \(-20 \text{ m}\)). To climb back out to ground level (infinity), you must do work!
Gravitational Potential Energy (\(E_p\))
If a body of mass \(m\) is placed at a point where the gravitational potential is \(V\), its gravitational potential energy is:
\(E_p = m V = -\frac{G M m}{r}\)
When an object moves between two points with a potential difference \(\Delta V\), the change in gravitational potential energy is:
\(\Delta E_p = m \Delta V\)
Equipotential Surfaces
An equipotential surface is a surface where every point has the exact same gravitational potential. Around a spherical planet, equipotentials are concentric spheres. Moving along an equipotential surface requires zero work because \(\Delta V = 0\).
Key Takeaway: Gravitational potential is zero at infinity and becomes increasingly negative as you get closer to a mass.
4. Planetary and Satellite Orbits
When a satellite or planet moves in a circular orbit, gravity provides the required centripetal force to keep it moving in a circle.
Deriving Kepler's Third Law (\(T^2 \propto r^3\))
Let's follow the derivation step-by-step:
Step 1: Equate gravitational force to centripetal force:
\(F_{\text{gravity}} = F_{\text{centripetal}}\)
\(\frac{G M m}{r^2} = \frac{m v^2}{r}\)
Step 2: Simplify to find the orbital speed \(v\):
\(v^2 = \frac{G M}{r} \implies v = \sqrt{\frac{G M}{r}}\)
Notice that the mass of the satellite \(m\) cancels out! All satellites at the same orbital radius travel at the same speed.
Step 3: Express orbital speed in terms of orbital period \(T\):
\(v = \frac{\text{distance}}{\text{time}} = \frac{2 \pi r}{T}\)
Step 4: Substitute \(v\) into our equation from Step 2:
\(\left(\frac{2 \pi r}{T}\right)^2 = \frac{G M}{r}\)
\(\frac{4 \pi^2 r^2}{T^2} = \frac{G M}{r}\)
Step 5: Rearrange for \(T^2\):
\(T^2 = \left(\frac{4 \pi^2}{G M}\right) r^3\)
Since \(\frac{4 \pi^2}{G M}\) is constant for a given central body, we find that \(T^2 \propto r^3\). This is Kepler's Third Law.
Geostationary Satellites
A geostationary satellite appears stationary above a fixed point on the Earth's equator. To do this, it must satisfy four key conditions:
• Orbit period = exactly 24 hours (equal to Earth's rotational period)
• Orbit directly above the equator
• Travel in the same direction as Earth's rotation (West to East)
• Be at a specific fixed height above the Earth's surface (\(\approx 36,000 \text{ km}\))
Uses: Telecommunications, satellite TV, and weather monitoring.
Key Takeaway: The farther a satellite is from Earth, the longer its orbital period and the slower its orbital speed.
5. Escape Velocity
Escape velocity (\(v_{\text{esc}}\)) is the minimum speed an object must be launched with from a planet's surface to completely escape the planet's gravitational pull and reach infinity.
Step-by-Step Derivation
To reach infinity where total energy is zero, the initial kinetic energy of the launched object must equal the gravitational potential energy needed to overcome the field:
\(\text{Kinetic Energy gained} = \text{Work needed to reach infinity}\)
\(\frac{1}{2} m v_{\text{esc}}^2 = \frac{G M m}{r}\)
Cancel the mass \(m\) of the object from both sides:
\(\frac{1}{2} v_{\text{esc}}^2 = \frac{G M}{r}\)
\(v_{\text{esc}} = \sqrt{\frac{2 G M}{r}}\)
For Earth, the escape velocity from the surface is approximately \(11.2 \text{ km s}^{-1}\).
Key Takeaway: Escape velocity does not depend on the mass of the projectile — a rocket and a pebble need the same escape velocity to leave Earth.
6. Quick Revision Summary & Formulas
Here is a handy summary of the core equations for your exam revision:
• Newton's Law of Gravitation: \(F = \frac{G m_1 m_2}{r^2}\)
• Gravitational Field Strength (definition): \(g = \frac{F}{m}\)
• Gravitational Field Strength (radial field): \(g = \frac{G M}{r^2}\)
• Gravitational Potential: \(V = -\frac{G M}{r}\)
• Gravitational Potential Energy: \(E_p = -\frac{G M m}{r}\)
• Kepler's Third Law Relationship: \(T^2 = \left(\frac{4 \pi^2}{G M}\right) r^3\)
• Orbital Speed: \(v = \sqrt{\frac{G M}{r}}\)
• Escape Velocity: \(v_{\text{esc}} = \sqrt{\frac{2 G M}{r}}\)
Top Exam Tips
• Check your radius values: If an exam question gives you an altitude above the surface (\(h\)), you MUST add the planet's radius (\(R\)) so that \(r = R + h\).
• Remember units: Distances must always be converted to meters (\(\text{m}\)), and periods must be converted to seconds (\(\text{s}\)).
• Sign of potential: Never forget the minus sign for gravitational potential \(V\) and potential energy \(E_p\)!