Orbits of planets, moons and satellites (AQA A-Level Physics): Revision Notes

Orbits of planets, moons and satellites

Overview of orbital motion

The solar system contains numerous orbiting bodies. Most planets have natural satellites called moons orbiting them. For example, Earth has one moon, Mars has two, while gas giants like Jupiter and Saturn have over 60 known moons each. Understanding the physics of these orbits allows us to predict their motion and even determine the mass of celestial bodies.

Determining the mass of a planet from orbital data

When we know the orbital characteristics of a moon, we can calculate the mass of the planet it orbits. This technique relies on applying Newton's law of gravitation and the principles of circular motion.

Newton's law applied to orbits

A planet exerts a gravitational force on an orbiting moon according to Newton's law of gravitation. For a moon of mass $m_2$ orbiting a planet of mass $m_1$ at a distance $r$ from the planet's centre, the gravitational force is:

$F = \frac{Gm_1m_2}{r^2}$

where $G$ is the gravitational constant ($6.67 \times 10^{-11}$ N m² kg⁻²).

Centripetal force requirement

For the moon to maintain a circular orbit, it requires a centripetal force directed towards the planet's centre. This centripetal force is provided entirely by the gravitational force between the planet and moon.

The required centripetal force for circular motion is:

$F = m_2r\omega^2$

where $\omega$ is the angular speed of the moon in its orbit, measured in radians per second.

Deriving the mass formula

Since the gravitational force provides the centripetal force, we can equate these two expressions:

$\frac{Gm_1m_2}{r^2} = m_2r\omega^2$

By eliminating the moon's mass $m_2$ from both sides and rearranging, we obtain an expression for the planet's mass:

$m_1 = \frac{r^3\omega^2}{G}$

The angular speed $\omega$ relates to the orbital period $T$ through the relationship $\omega = \frac{2\pi}{T}$.

Kepler's third law

The German mathematician Johannes Kepler analyzed extensive observational data of planetary motion collected by Tycho Brahe. From this analysis, he discovered a mathematical relationship between orbital period and orbital radius that applies to all planets orbiting the Sun.

Statement of the law

Kepler's third law states that the square of the orbital period is proportional to the cube of the orbital radius:

$T^2 \propto r^3$

Derivation from Newton's law

Newton later demonstrated that Kepler's empirical observation could be derived mathematically from his law of universal gravitation. Starting with the relationship $\omega = \frac{2\pi}{T}$ for angular speed, and substituting this into the gravitational force equation:

$\frac{Gm_1}{r^2} = r\left(\frac{2\pi}{T}\right)^2$

Rearranging this expression gives:

$T^2 = \frac{4\pi^2}{Gm_1}r^3$

Since $\frac{4\pi^2}{Gm_1}$ is constant for a given central mass $m_1$, this proves that $T^2 \propto r^3$.

Satellites orbiting Earth

Artificial satellites serve numerous purposes in Earth orbit. Over 1200 operational satellites currently orbit Earth, with many more no longer functioning. These satellites occupy different orbital altitudes depending on their function.

Communication satellites orbit at altitudes exceeding 35,000 km, relaying television, telephone, and radio signals globally. Global Positioning System (GPS) satellites operate at altitudes over 20,000 km in medium Earth orbits, providing essential navigation data. Weather satellites and environmental monitoring satellites study atmospheric conditions and climate changes. Space observation satellites, positioned in low Earth orbits between 160 and 2000 km altitude, study astronomical phenomena. The International Space Station orbits at approximately 400 km altitude.

Geostationary orbits

Definition and characteristics

A geostationary orbit (GEO) is a special type of orbit where a satellite remains positioned above the same point on Earth's surface at all times. This occurs when the satellite orbits in the equatorial plane with an orbital period exactly equal to Earth's rotational period of 24 hours.

Requirements for geostationary orbits

A communication satellite in geostationary orbit maintains continuous contact with receiving dishes on Earth's surface. Since the satellite appears stationary from Earth's perspective, the receiving dishes can remain fixed pointing at the same spot in the sky.

Polar orbits

A satellite in a polar orbit passes over or very close to both the North and South Poles during each orbit. The orbital plane of a polar orbit is oriented at 90° to the equatorial plane, meaning the satellite travels in a north-south direction rather than east-west.

Polar orbiting satellites typically operate at much lower altitudes than geostationary satellites, traveling at higher orbital speeds and completing several orbits per day.

These satellites are used for mapping land features, monitoring polar ice caps, tracking environmental changes, and military surveillance purposes.

Synchronous orbits

A satellite following a synchronous orbit has an orbital period equal to the rotational period of the planet it orbits. For Earth, this means a 24-hour orbital period. A geostationary orbit is a specific type of synchronous orbit that additionally requires the satellite to orbit in the equatorial plane in the same direction as the planet's rotation.

Orbital speed

Relationship between speed and orbital radius

The orbital speed $v$ of a satellite depends on the mass $M$ of the planet it orbits and the orbital radius $r$. According to Newton's law, the gravitational force provides the centripetal force:

$\frac{GMm}{r^2} = \frac{mv^2}{r}$

where $m$ is the satellite's mass. Rearranging this equation to solve for orbital speed gives:

$v = \sqrt{\frac{GM}{r}}$

Energy of orbiting satellites

Gravitational potential energy

The gravitational potential energy of a satellite in orbit at distance $r$ from a planet's centre equals the work done by gravitational force in bringing the satellite from infinity to that position. For a satellite of mass $m$ orbiting a planet of mass $M$:

$E_p = -\frac{GMm}{r}$

Kinetic energy

The kinetic energy of an orbiting satellite can be expressed using its orbital speed. Since $v^2 = \frac{GM}{r}$ from the orbital speed equation, the kinetic energy is:

$E_k = \frac{1}{2}mv^2 = \frac{GMm}{2r}$

Note that the kinetic energy equals half the magnitude of the gravitational potential energy.

Total energy

The total energy of a satellite is the sum of its gravitational potential energy and kinetic energy:

$E_{total} = E_p + E_k = -\frac{GMm}{r} + \frac{GMm}{2r}$

$E_{total} = -\frac{GMm}{2r}$

As a satellite moves closer to Earth (decreasing $r$), its total energy becomes more negative. However, this does not mean the satellite slows down. In fact, the kinetic energy increases as the satellite loses altitude.

For satellites in low Earth orbit, atmospheric drag gradually removes energy from the system. As the total energy decreases, the satellite spirals inward, paradoxically speeding up as it loses energy. Eventually, friction with denser atmospheric layers generates intense heat, typically causing the satellite to burn up.

Escape velocity

Definition

The escape velocity is the minimum speed an object needs to completely escape from the gravitational field of a planet or other astronomical body. At escape velocity, an object has just enough kinetic energy to reach infinite distance from the planet, where the gravitational field has negligible effect.

Derivation

To derive escape velocity, we consider the energy required for an object of mass $m$ to escape from the gravitational field of a planet of mass $M$ and radius $R$. The work done against gravity to move from the planet's surface to infinity equals:

$\Delta W = m\Delta V = m\left(\frac{GM}{R}\right)$

If this work is provided by the object's initial kinetic energy, then:

$\frac{1}{2}mv^2 = \frac{mGM}{R}$

Solving for the escape velocity $v$:

$v = \sqrt{\frac{2GM}{R}}$