Welcome to Gravitational Fields: Understanding the Universe's Glue!
Welcome to Topic 13! This chapter moves beyond simple forces near the Earth's surface (\(F=mg\)) and dives into how gravity works across vast cosmic distances. Understanding gravitational fields is essential because it explains why planets orbit stars, why satellites stay in space, and fundamentally, how the universe holds itself together.
Don't worry if the formulae look intimidating at first. We will break down each concept step-by-step, starting with definitions and moving towards powerful equations that govern astronomical motion.
13.1 The Concept of a Gravitational Field
What is a Field of Force?
A field of force is simply a region in which an object experiences a non-contact force. The gravitational field is the specific region around a mass where another mass will experience a force due to gravity.
It is crucial to define the strength of this field:
Definition: Gravitational Field Strength (\(g\))
The 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.
- Equation: \(g = \frac{F}{m}\)
- Units: Newtons per kilogram (\(\text{N kg}^{-1}\)).
- Note: Since \(F=ma\), the units \(\text{N kg}^{-1}\) are equivalent to \(\text{m s}^{-2}\) (acceleration of free fall).
Key Takeaway: Gravitational field strength is a vector quantity. Its direction is always the direction of the force experienced by the mass (i.e., towards the centre of the attracting mass).
Representing Fields: Field Lines
We represent gravitational fields using field lines (also called lines of force):
- The direction of the field line shows the direction of the force acting on a mass (always inwards, towards the attracting mass).
- The density (closeness) of the field lines shows the strength of the field. Where lines are closer, the field is stronger.
For an isolated point mass (like the Earth, viewed from far away), the field lines are radial (pointing directly inward to the centre).
Gravitational fields are always attractive (masses only pull, never push), and they are radial around point masses.
13.2 Gravitational Force Between Point Masses: Newton's Law
The Universal Law of Gravitation
Sir Isaac Newton deduced the rule that governs the attractive force between any two masses in the universe.
Newton's Law of Gravitation
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 the separation (\(r\)) between their centres.
The mathematical relationship is: \[ F = \frac{G m_1 m_2}{r^2} \]
Where:
- \(F\) is the gravitational force (N).
- \(m_1\) and \(m_2\) are the two masses (kg).
- \(r\) is the distance between the centres of the masses (m).
- \(G\) is the Universal Gravitational Constant (\(6.67 \times 10^{-11} \text{ N m}^2 \text{ kg}^{-2}\)).
Important Simplification (Uniform Spheres)
When dealing with large, uniform spherical objects (like planets), we treat them as if all their mass is concentrated at a single point at their centre.
For any point outside a uniform sphere, the mass of the sphere may be considered to be a point mass at its centre.
Analogy: This is why it doesn't matter where you stand on the Earth's surface—the pull of gravity always seems to come straight down from the centre.
The Inverse Square Law
The dependence on \(1/r^2\) means gravity is an inverse square law force.
- If you double the distance \(r\), the force \(F\) decreases by a factor of \(2^2 = 4\).
- If you triple the distance \(r\), the force \(F\) decreases by a factor of \(3^2 = 9\).
This decrease is very rapid, which is why we don't feel the gravitational pull of distant planets in our daily lives.
The Universal Gravitational Constant \(G\) is one of the hardest physical constants to measure accurately! It tells us that gravity is an incredibly weak force compared to the electric force (which we will study later).
13.3 Gravitational Field Strength Due to a Point Mass
We can now combine the definition of field strength (\(g = F/m\)) with Newton's Law (\(F = G M m / r^2\)) to find the formula for the gravitational field strength \(g\) created by a single mass \(M\).
Derivation of \(g = \frac{G M}{r^2}\)
- Start with Newton's Law, where \(M\) is the central mass (e.g., Earth) and \(m\) is the test mass: \[ F = \frac{G M m}{r^2} \]
- Substitute this expression for \(F\) into the definition of gravitational field strength, \(g = F/m\): \[ g = \frac{\left( \frac{G M m}{r^2} \right)}{m} \]
- The test mass \(m\) cancels out, leaving the final expression: \[ g = \frac{G M}{r^2} \]
This formula shows that the gravitational field strength \(g\) at any point depends only on the mass \(M\) creating the field and the distance \(r\) from its centre.
Why is \(g\) Approximately Constant Near Earth's Surface?
We often use \(g \approx 9.81 \text{ N kg}^{-1}\) for objects near the Earth. This approximation holds because:
- The radius of the Earth (\(R\)) is about \(6.4 \times 10^6 \text{ m}\).
- For objects on Earth's surface or even high mountains, the change in distance \(\Delta r\) is very small compared to the radius \(R\).
- Since \(g \propto 1/r^2\), if \(r\) barely changes, \(g\) barely changes.
However, if we look at a satellite orbiting hundreds of kilometers up, the change in \(r\) is significant, and we must use the full formula \(g = G M / r^2\).
\(F\) (Force/Weight) depends on BOTH the central mass \(M\) and the object's mass \(m\).
\(g\) (Field Strength) depends ONLY on the central mass \(M\) and the distance \(r\).
13.4 Gravitational Potential (\(\phi\))
Gravitational fields are conservative, meaning the work done moving a mass between two points is independent of the path taken. This allows us to define gravitational potential energy and gravitational potential.
Defining Gravitational Potential (\(\phi\))
Just as gravitational field strength is force per unit mass, Gravitational Potential (\(\phi\)) is potential energy per unit mass.
Definition: Gravitational Potential (\(\phi\))
Gravitational potential (\(\phi\)) at a point is the work done per unit mass in bringing a small test mass from infinity to that point.
- Units: Joules per kilogram (\(\text{J kg}^{-1}\)).
The Significance of Infinity and Negative Signs
In gravitational theory:
- Infinity: Infinity is chosen as the point of zero potential (\(\phi=0\)) because the gravitational force effectively becomes zero there.
- Negative Potential: Since gravity is an attractive force, work is done by the gravitational field when you move a mass from infinity (zero potential) closer to the central mass. If the field does the work, the potential energy must decrease. Therefore, gravitational potential is always negative.
The formula for the gravitational potential due to a point mass \(M\) at distance \(r\) is: \[ \phi = - \frac{G M}{r} \]
Think of it like being in debt: the potential is lowest (most negative) when you are close to the mass, representing the maximum amount of work needed to escape to zero potential (infinity).
Gravitational Potential Energy (\(E_p\))
The gravitational potential energy (\(E_p\)) of two point masses, \(M\) and \(m\), separated by distance \(r\), is simply the potential \(\phi\) multiplied by the mass \(m\).
- Equation: \(E_p = m \phi\)
- Resulting formula: \[ E_p = - \frac{G M m}{r} \]
This formula represents the work required to separate the two masses to infinity, or equivalently, the energy released when the two masses come together from infinity.
Potential (\(\phi\)) is a scalar quantity (just a magnitude, always negative). It uses \(1/r\).
Field Strength (\(g\)) is a vector quantity (magnitude and direction). It uses \(1/r^2\).
13.5 Application: Circular Orbits and Satellites (13.2)
Gravitational fields are the driving mechanism behind orbital motion. For a satellite of mass \(m\) orbiting a much larger central mass \(M\) (like Earth) in a circular orbit of radius \(r\), the gravitational force is the sole provider of the centripetal force required for circular motion.
Orbital Analysis
We equate the gravitational force (\(F_{\text{G}}\)) to the centripetal force (\(F_{\text{C}}\)):
\[ F_{\text{G}} = F_{\text{C}} \] \[ \frac{G M m}{r^2} = \frac{m v^2}{r} \quad (\text{using } F_C = mv^2/r) \]
We can simplify this relationship to find the orbital speed \(v\):
- Cancel \(m\) (mass of the satellite) from both sides (meaning orbital speed is independent of satellite mass).
- Cancel one \(r\) from both sides. \[ \frac{G M}{r} = v^2 \]
- Orbital Speed: \[ v = \sqrt{\frac{G M}{r}} \]
This equation is powerful: it shows that the closer the satellite is to the central mass (smaller \(r\)), the faster it must move to stay in orbit.
The Geostationary Orbit (A Special Case)
A geostationary satellite is an essential application of orbital mechanics. It is a satellite that remains fixed above the same point on the Earth's surface.
For an orbit to be geostationary, it must satisfy three very strict conditions:
- Period (\(T\)) must be 24 hours (86,400 seconds). This ensures the satellite completes one orbit in exactly the same time the Earth completes one rotation.
- It must orbit directly above the Equator. Any inclination would cause the satellite to drift north and south relative to a ground point.
- It must orbit from West to East (in the same direction as Earth's rotation).
Application: Geostationary satellites are vital for fixed communication dishes (e.g., satellite TV) because the dish never has to move—it always points to the same spot in the sky.
Calculating Geostationary Radius:
Since \(v = r \omega\) and \(\omega = 2\pi / T\), we can substitute the orbital speed equation:
\[
v^2 = \frac{G M}{r} \quad \implies \quad \left( \frac{2\pi r}{T} \right)^2 = \frac{G M}{r}
\]
Solving for \(r\) (the radius of the orbit) allows scientists to calculate the exact height required for a geostationary orbit (approximately 42,000 km from the centre of the Earth).
Common Pitfall
When calculating orbital radius or speed, remember that \(r\) must always be the distance from the centre of the planet, not the height above the surface. If you are given the height, you must add the radius of the planet to find \(r\).
\(r_{\text{orbit}} = R_{\text{planet}} + h_{\text{altitude}}\)