Welcome to Gravitation!
Ever wondered why the Moon doesn’t crash into Earth, or how GPS satellites stay in perfect sync to help you navigate? The answer lies in one of the fundamental forces of the universe: Gravitation.
In this chapter of AS 2 Section B: Mechanics 2, we will bridge the gap between simple terrestrial gravity (where we used a constant \(g = 9.8\text{ m s}^{-2}\)) and the cosmic mechanics that govern planets and satellites. Don't worry if this sounds intimidating at first—once you see how gravity pairs up with circular motion, everything clicks into place!
1. Newton's Law of Universal Gravitation
Back in earlier mechanics units, we assumed gravity was constant. But as you move hundreds or thousands of kilometres above the Earth, gravity weakens. Sir Isaac Newton realised that every particle of matter attracts every other particle with a force that depends on their masses and the distance between them.
The Law Defined
Newton's Law of Universal Gravitation states that 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 centres.
Mathematically, we write this as:
\(F = \frac{G M m}{r^2}\)
Where:
• \(F\) = Gravitational force of attraction (measured in Newtons, \(\text{N}\))
• \(G\) = Universal Gravitational Constant (\(G \approx 6.67 \times 10^{-11}\text{ N m}^2\text{ kg}^{-2}\))
• \(M\) and \(m\) = The masses of the two interacting bodies (measured in \(\text{kg}\))
• \(r\) = Distance between the centres of the two masses (measured in \(\text{m}\))
Important Properties to Remember
1. 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, it drops to \(\frac{1}{9}\).
2. Spherical Bodies: Uniform spherical bodies (like planets and moons) act gravitationally as if all their mass were concentrated at their geometric centre.
Common Mistake to Avoid: Always measure \(r\) from the centre of the planet, not its surface! If a satellite is at height \(h\) above a planet of radius \(R\), then \(r = R + h\).
Key Takeaway
Gravity is an attractive force pulling along the line joining the centres of two masses. It gets stronger with larger masses and drops off rapidly as the distance between them increases.
2. Gravitational Field Strength (\(g\))
What actually is "\(g\)"? We know \(g \approx 9.8\text{ m s}^{-2}\) at the surface of the Earth, but where does that number come from?
Definition of Field Strength
The gravitational field strength, \(g\), at any point in a field is defined as the gravitational force per unit mass exerted on a small test mass placed at that point:
\(g = \frac{F}{m}\)
Substituting Newton's law of gravitation (\(F = \frac{GMm}{r^2}\)) into this definition gives:
\(g = \frac{\left(\frac{G M m}{r^2}\right)}{m} \implies g = \frac{GM}{r^2}\)
Gravity at the Earth's Surface vs. at Height \(h\)
• At the Earth's surface (\(r = R\), where \(R\) is the radius of the Earth):
\(g_0 = \frac{GM}{R^2}\)
This gives us the incredibly handy algebraic substitution: \(GM = g_0 R^2\). You will use this trick often when the exam doesn't give you \(G\) or \(M\) directly!
• At a height \(h\) above the surface (\(r = R + h\)):
\(g_h = \frac{GM}{(R + h)^2} = g_0 \left(\frac{R}{R + h}\right)^2\)
Key Takeaway
Gravitational field strength \(g\) is the acceleration due to gravity at a specific distance \(r\) from the centre of mass \(M\). As you rise above the planet, \(g\) decreases according to the inverse-square law.
3. Circular Orbits of Satellites and Planets
How does a satellite stay in orbit without falling down to Earth? It is actually in constant free-fall! The satellite moves forward so fast that as gravity pulls it down, the surface of the Earth curves away underneath it at the exact same rate.
The Fundamental Orbit Equation
For any circular orbit, the gravitational attraction provides the centripetal force required to keep the body moving in a circle.
\(F_{\text{gravity}} = F_{\text{centripetal}}\)
\(\frac{GMm}{r^2} = \frac{m v^2}{r}\)
Notice that the mass of the satellite (\(m\)) cancels out on both sides!
\(\frac{GM}{r} = v^2 \implies v = \sqrt{\frac{GM}{r}}\)
Did you know? Because the satellite's mass cancels out, an astronaut floating outside the International Space Station travels at the exact same orbital speed as the 400-tonne station itself!
Key Relationship: Speed and Altitude
Because \(v = \sqrt{\frac{GM}{r}}\), as the orbital radius \(r\) increases, the orbital speed \(v\) decreases. Satellites closer to Earth must travel significantly faster to stay in orbit than satellites further away.
Key Takeaway
To solve almost any satellite problem, start with: \(\text{Gravitational Force} = \text{Centripetal Force}\).
4. Kepler's Third Law of Planetary Motion
Johannes Kepler discovered empirically that the time a planet takes to complete an orbit is linked to its distance from the Sun. Using Newton's laws, we can prove this mathematically!
Deriving Kepler's Third Law
Recall the relationship between linear velocity \(v\), orbital radius \(r\), and orbital period \(T\) (the time for one complete orbit):
\(v = \frac{\text{distance}}{\text{time}} = \frac{2\pi r}{T}\)
Now, substitute this into the orbital speed formula \(v^2 = \frac{GM}{r}\):
\(\left(\frac{2\pi r}{T}\right)^2 = \frac{GM}{r}\)
\(\frac{4\pi^2 r^2}{T^2} = \frac{GM}{r}\)
Rearranging for \(T^2\) gives:
\(T^2 = \left(\frac{4\pi^2}{GM}\right) r^3\)
Since \(G\), \(M\), and \(\pi\) are all constants for a given central body:
\(T^2 \propto r^3\)
This is Kepler's Third Law: The square of the orbital period is directly proportional to the cube of the radius of the orbit.
Step-by-Step Worked Example
Question: A satellite orbits Earth at a distance of \(r_1 = 7.0 \times 10^6\text{ m}\) with a period of \(T_1 = 5800\text{ s}\). Find the period \(T_2\) of a second satellite orbiting at a distance of \(r_2 = 1.4 \times 10^7\text{ m}\).
Step 1: Set up the ratio using Kepler's Third Law:
\(\frac{T_2^2}{T_1^2} = \frac{r_2^3}{r_1^3}\)
Step 2: Substitute the given values:
\(\frac{T_2^2}{5800^2} = \left(\frac{1.4 \times 10^7}{7.0 \times 10^6}\right)^3 = (2)^3 = 8\)
Step 3: Solve for \(T_2\):
\(T_2^2 = 8 \times 5800^2 = 8 \times 3.364 \times 10^7 = 2.6912 \times 10^8\)
\(T_2 = \sqrt{2.6912 \times 10^8} \approx 16405\text{ s} \approx 4.56\text{ hours}\)
Key Takeaway
For any two satellites orbiting the same central body, \(\frac{T^2}{r^3} = \text{constant}\). If distance doubles, the period increases by a factor of \(\sqrt{8} \approx 2.83\).
5. Geostationary Orbits
A geostationary satellite is a special type of satellite that appears completely motionless from the perspective of an observer on the ground. These satellites are essential for telecommunications, weather forecasting, and satellite television (which is why satellite dishes on houses always point in a fixed direction!).
The Three Key Conditions for a Geostationary Orbit
For a satellite to remain fixed over the same point on Earth's surface, it must satisfy three strict criteria:
1. Orbital Period: Its period must be exactly equal to the Earth's rotational period (\(T = 24\text{ hours} = 86400\text{ s}\)).
2. Orbital Plane: It must orbit directly above the Equator.
3. Direction of Motion: It must travel from West to East (the same direction as the Earth rotates).
Calculating the Altitude of a Geostationary Orbit
Using Kepler's Third Law: \(r^3 = \frac{GM T^2}{4\pi^2}\)
Using Earth's parameters (\(M = 5.97 \times 10^{24}\text{ kg}\), \(T = 86400\text{ s}\), \(G = 6.67 \times 10^{-11}\text{ N m}^2\text{ kg}^{-2}\)):
\(r^3 = \frac{(6.67 \times 10^{-11})(5.97 \times 10^{24})(86400)^2}{4\pi^2} \approx 7.54 \times 10^{22}\text{ m}^3\)
\(r = \sqrt[3]{7.54 \times 10^{22}} \approx 4.22 \times 10^7\text{ m} = 42200\text{ km}\)
To find the height \(h\) above the Earth's surface (radius \(R \approx 6400\text{ km}\)):
\(h = r - R = 42200\text{ km} - 6400\text{ km} \approx 35800\text{ km}\)
Key Takeaway
There is only one unique radius (approx. \(42200\text{ km}\) from the centre of Earth) at which a satellite can achieve a geostationary orbit.
Chapter Summary & Revision Checklist
Before heading into past paper questions, ensure you are comfortable with these core relationships:
• Newton's Law of Gravitation: \(F = \frac{GMm}{r^2}\)
• Gravitational Field Strength: \(g = \frac{GM}{r^2}\)
• Surface Gravity Connection: \(GM = g_0 R^2\)
• Orbital Speed: \(v = \sqrt{\frac{GM}{r}}\)
• Kepler's Third Law: \(T^2 = \frac{4\pi^2}{GM} r^3\)
• Geostationary Orbits: Orbit above equator, West to East, \(T = 24\text{ hours}\).