28.3 Content
28.3.1 Newton’s Law of Universal Gravitation
Every mass attracts every other mass with a force:
\[F = \frac{GMm}{r^2}\]
where: - \(F\) = gravitational force (N) - \(G = 6.67 \times 10^{-11}\ \text{N·m}^2/\text{kg}^2\) (gravitational constant) - \(M, m\) = masses (kg) - \(r\) = separation between centres of mass (m)
- Gravity is always attractive
- Force acts along the line joining the centres
- Force follows the inverse square law
28.3.2 Interactive: Gravitational Force
Explore how gravitational force varies with distance:
28.3.3 Gravitational Field Strength
Gravitational field strength (\(g\)) is the force per unit mass at a point:
\[g = \frac{F}{m} = \frac{GM}{r^2}\]
At Earth’s surface (\(r = R_E = 6.37 \times 10^6\ \text{m}\)): \[g = 9.8\ \text{m/s}^2\]
As altitude increases, \(r\) increases and \(g\) decreases. At twice the Earth’s radius, \(g\) is reduced to one-quarter.
28.3.4 Interactive: Gravitational Field
28.3.5 Orbital Motion
For a circular orbit, gravitational force provides the centripetal force:
\[\frac{GMm}{r^2} = \frac{mv^2}{r}\]
This gives orbital speed: \[v = \sqrt{\frac{GM}{r}}\]
Orbital speed depends only on the central mass \(M\) and orbital radius \(r\)—not on the orbiting mass \(m\).
28.3.6 Orbital Period
Using \(v = 2\pi r/T\):
\[T = 2\pi\sqrt{\frac{r^3}{GM}}\]
This leads to Kepler’s Third Law: \[\frac{r^3}{T^2} = \frac{GM}{4\pi^2}\]
28.3.7 Interactive: Orbital Mechanics
28.3.8 Geostationary Orbits
A geostationary satellite: - Has period \(T = 24\ \text{hours} = 86400\ \text{s}\) - Orbits above the equator - Moves in the same direction as Earth’s rotation - Appears stationary from Earth’s surface
Geostationary radius: \(r \approx 4.2 \times 10^7\ \text{m}\) (about 36,000 km altitude)
28.3.9 Kepler’s Laws
First Law: Planets move in elliptical orbits with the Sun at one focus.
Second Law: A line from planet to Sun sweeps equal areas in equal times.
Third Law: \(r^3/T^2 = \text{constant}\) for all objects orbiting the same central mass.
28.3.10 Gravitational Potential Energy
In a gravitational field:
\[U = -\frac{GMm}{r}\]
Gravitational potential energy is negative because: - Zero reference is at infinite separation - Energy is released when objects fall together - Work must be done to separate them
28.3.11 Escape Velocity
The minimum speed needed to escape a gravitational field (reach infinity with zero speed):
\[v_{esc} = \sqrt{\frac{2GM}{r}}\]
At Earth’s surface: \(v_{esc} = 11.2\ \text{km/s}\)
28.3.12 Total Energy in Orbit
For a circular orbit:
Kinetic energy: \(K = \frac{1}{2}mv^2 = \frac{GMm}{2r}\)
Total energy: \(E = K + U = -\frac{GMm}{2r}\)
Total energy is negative for bound orbits. The satellite cannot escape unless energy is added.