Welcome to Gravitation
Welcome to one of the most exciting topics in AS 2 Section B: Mechanics 2! Up until now, you have probably treated gravity as a constant force pulling objects downwards with an acceleration of \(g \approx 9.8 \text{ m s}^{-2}\). While that works brilliantly near the surface of the Earth, the universe is much bigger than our classroom.
In this chapter, you will discover how gravity works across vast cosmic distances. We will explore how gravity holds the Moon in orbit, why satellites stay in space, and how to calculate the exact speed and time needed for planetary orbits. Don't worry if this seems a bit daunting at first—we will break every single idea down into simple, step-by-step pieces!
1. Newton's Law of Universal Gravitation
Sir Isaac Newton realised that the same invisible pull that makes an apple fall from a tree also keeps planets orbiting the Sun. His law states that every particle of matter in the universe attracts every other particle with a force directly proportional to the product of their masses and inversely proportional to the square of the distance between their centres.
The Master Formula
\(F = \frac{G M m}{r^2}\)
Let's break down each symbol in this formula:
• \(F\) is the magnitude of the gravitational force of attraction between the two bodies, measured in Newtons (\(\text{N}\)).
• \(G\) is the Universal Gravitational Constant, which has the value \(G \approx 6.67 \times 10^{-11} \text{ N m}^2 \text{kg}^{-2}\). (This is a constant of nature and will always be provided in your data sheet).
• \(M\) and \(m\) are the masses of the two interacting objects, measured in kilograms (\(\text{kg}\)).
• \(r\) is the distance between the centres of mass of the two objects, measured in metres (\(\text{m}\)).
Understanding the Inverse-Square Law
Notice the \(r^2\) in the denominator? This means gravity follows an inverse-square law:
• If you double the distance (\(2r\)), the gravitational force becomes \(\frac{1}{2^2} = \frac{1}{4}\) of its original strength.
• If you triple the distance (\(3r\)), the force drops to \(\frac{1}{3^2} = \frac{1}{9}\) of its original strength.
Common Pitfall Alert
Always measure from the centre! A very common mistake in exam questions is using the height above a planet's surface as \(r\). If an object is at a height \(h\) above a planet of radius \(R\), the total distance from the centre is \(r = R + h\).
Key Takeaway
Gravitational force acts along the line joining the centres of two bodies. It increases if either mass increases, and it decreases rapidly as the distance between them increases: \(F = \frac{G M m}{r^2}\).
2. Gravitational Field Strength and Acceleration due to Gravity (\(g\))
What is the link between Newton's Universal Law and our familiar weight formula \(W = mg\)?
Consider an object of mass \(m\) sitting on the surface of a planet with mass \(M\) and radius \(R\). The gravitational force on the object is its weight:
\(F = m g\)
According to Newton's Law of Gravitation, this force is also:
\(F = \frac{G M m}{R^2}\)
Setting these two equal to each other:
\(m g = \frac{G M m}{R^2}\)
Dividing both sides by the small mass \(m\) gives:
\(g = \frac{G M}{R^2}\)
Value of \(g\) at a Height Above the Surface
If you travel to an altitude \(h\) above the planet's surface, the distance from the centre becomes \(r = R + h\). The acceleration due to gravity at this height is:
\(g_h = \frac{G M}{(R + h)^2}\)
Because \(G M = g_0 R^2\) (where \(g_0\) is the surface value), we can also express the gravitational acceleration at height \(h\) as:
\(g_h = g_0 \left(\frac{R}{R + h}\right)^2\)
Did You Know?
Because the Earth is slightly flattened at the poles and bulges at the equator, the radius at the poles is smaller than at the equator. Consequently, you weigh slightly more at the North Pole than at the equator!
Key Takeaway
The acceleration due to gravity on any spherical body depends only on the mass of the body and the distance from its centre: \(g = \frac{G M}{r^2}\). The mass of the falling object does not affect \(g\).
3. Circular Orbits and Satellite Motion
How does a satellite stay in space without falling down? The answer is that it is constantly falling—but it is moving sideways so fast that the Earth curves away beneath it at the exact same rate!
For a satellite of mass \(m\) moving in a circular orbit of radius \(r\) around a planet of mass \(M\) with speed \(v\), the gravitational attraction provides the required centripetal force.
Step-by-Step Derivation of Orbital Speed
Step 1: Write down the centripetal force equation from circular motion:
\(F_{\text{centripetal}} = \frac{m v^2}{r}\)
Step 2: Equate this to Newton's gravitational force:
\(\frac{G M m}{r^2} = \frac{m v^2}{r}\)
Step 3: Cancel the satellite mass \(m\) from both sides and multiply by \(r\):
\(v^2 = \frac{G M}{r}\)
Step 4: Take the square root:
\(v = \sqrt{\frac{G M}{r}}\)
Important Observations
• Mass Independence: The orbital speed \(v\) does not depend on the mass \(m\) of the satellite. A tiny screw and a massive space station in the same orbit move at the exact same speed.
• Speed vs Altitude: As orbital radius \(r\) increases, the speed \(v\) decreases. Satellites closer to Earth must travel much faster than satellites further away.
Key Takeaway
For any circular orbit, equating gravitational force to centripetal force gives the orbital speed: \(v = \sqrt{\frac{G M}{r}}\).
4. Orbital Period and Kepler's Third Law
The orbital period \(T\) is the time it takes for a satellite or planet to complete one full revolution around its orbit.
Deriving the Period Equation
We know that for circular motion at constant speed \(v\):
\(v = \frac{\text{distance}}{\text{time}} = \frac{2 \pi r}{T}\)
We also know from circular dynamics that:
\(\frac{G M m}{r^2} = m r \omega^2\)
Since angular speed is \(\omega = \frac{2\pi}{T}\), substituting this gives:
\(\frac{G M}{r^2} = r \left(\frac{2\pi}{T}\right)^2 = \frac{4 \pi^2 r}{T^2}\)
Rearranging this equation to make \(T^2\) the subject:
\(T^2 = \left(\frac{4 \pi^2}{G M}\right) r^3\)
Kepler's Third Law
Since \(\frac{4 \pi^2}{G M}\) is a constant for all bodies orbiting the same central mass \(M\), this gives us Kepler's Third Law of Planetary Motion:
\(T^2 \propto r^3\)
The square of the orbital period is directly proportional to the cube of the radius of the orbit.
Ratio Shortcut for Exams
If you are comparing two satellites or planets orbiting the same central body (for example, two planets orbiting the Sun):
\(\frac{T_1^2}{T_2^2} = \frac{r_1^3}{r_2^3}\)
This ratio relationship can save you a lot of time in calculation questions!
Key Takeaway
The time for one orbit is related to the orbital radius by \(T^2 = \left(\frac{4 \pi^2}{G M}\right) r^3\). The further a satellite is from the planet, the longer it takes to complete an orbit.
5. Geostationary Satellites
A geostationary satellite is an artificial satellite that remains directly above the exact same point on the Earth's equator at all times. They are essential for telecommunications, satellite television, and weather forecasting.
Requirements for a Geostationary Orbit
To stay above the exact same spot on Earth, a satellite must meet three strict conditions:
1. Orbital Period: Its period must be exactly equal to Earth's rotational period, which is \(T = 24 \text{ hours} = 86400 \text{ s}\).
2. Plane of Orbit: The orbit must lie directly in the equatorial plane.
3. Direction of Motion: It must orbit from west to east, in the same direction as the Earth rotates.
Calculating the Geostationary Altitude
By substituting \(T = 86400 \text{ s}\), \(G = 6.67 \times 10^{-11} \text{ N m}^2 \text{kg}^{-2}\), and Earth's mass \(M \approx 5.97 \times 10^{24} \text{ kg}\) into Kepler's equation:
\(r^3 = \frac{G M T^2}{4 \pi^2}\)
\(r \approx 4.22 \times 10^7 \text{ m} \approx 42200 \text{ km}\)
Subtracting Earth's radius (\(R \approx 6400 \text{ km}\)) gives an altitude of approximately \(35800 \text{ km}\) above the Earth's surface.
Key Takeaway
There is only one unique orbital radius where a satellite can be geostationary. All geostationary satellites orbit at this exact altitude above the equator.
6. Summary of Key Formulas and Exam Tips
Core Formulas Reference
• Newton's Gravitational Law: \(F = \frac{G M m}{r^2}\)
• Gravitational Field Strength: \(g = \frac{G M}{r^2}\)
• Surface Relationship: \(g_0 = \frac{G M}{R^2} \implies G M = g_0 R^2\)
• Orbital Speed: \(v = \sqrt{\frac{G M}{r}}\)
• Orbital Period & Kepler's 3rd Law: \(T^2 = \left(\frac{4 \pi^2}{G M}\right) r^3\)
Exam Checklist & Pitfalls to Avoid
• Check Your Distance (\(r\)): Read carefully whether the question gives the radius of orbit \(r\) or the altitude \(h\) above the surface. Remember \(r = R + h\).
• Unit Conversions: Always convert kilometres to metres (\(1 \text{ km} = 1000 \text{ m}\)) and time to seconds (\(1 \text{ hour} = 3600 \text{ s}\), \(1 \text{ day} = 86400 \text{ s}\)) before calculating.
• Mass Substitution: When calculating orbital speed or period, \(M\) is the mass of the central body being orbited (e.g. Earth or Sun), not the orbiting satellite.
• Clear Algebraic Steps: In "show that" questions, always write out the initial balance of forces (\(\frac{GMm}{r^2} = \frac{mv^2}{r}\) or \(\frac{GMm}{r^2} = mr\omega^2\)) before rearranging.