Know that the Milky Way is one of many billions of galaxies making up the Universe and that the diameter of the Milky Way is approximately 100000 light-years

6.1 The Earth and the Moon

Learning Objective

Students should be be able to:

• Describe the Earth’s rotation and its 23.5° axial tilt.
• Explain how rotation and tilt produce day‑night cycles, the variation of daylight length with latitude and season, and the four seasons.
• State the Moon’s orbital period, distinguish between the sidereal (27.3 d) and synodic (29.5 d) months, and explain the cause of lunar phases.
• Use the orbital‑speed formula \(\displaystyle v=\frac{2\pi r}{T}\) to calculate the speed of any planet (or moon) when the orbital radius and period are known.

Key Points – Earth

• Rotation

  • Period ≈ 24 h (one sidereal day).
  • Axis is tilted ≈ 23.5° to the plane of the Earth’s orbit (the ecliptic).
  • Rotation produces the daily alternation of daylight and darkness.

• Axial tilt and seasons

  • Because the tilt is constant while the Earth revolves around the Sun, different hemispheres receive different amounts of solar radiation at different times of the year.
  • Result: summer – higher solar elevation → longer days; winter – lower solar elevation → shorter days.
  • The length of daylight varies with latitude (e.g., 12 h at the equator year‑round, up to 24 h at the poles in summer).

• Revolution

  • Orbital period ≈ 365.25 days (one year).
  • Combined with the 23.5° tilt it gives the seasonal cycle.

Key Points – Moon

• Orbital periods

  • Sidereal month ≈ 27.3 days – time to complete one orbit relative to the stars.
  • Synodic month ≈ 29.5 days – time between identical phases (new → new). This is the value used in exam questions.

• Lunar phases

  • Phases arise from the changing Sun–Moon–Earth geometry.
  • When the Moon is between the Sun and Earth we see a new Moon; when it is opposite the Sun we see a full Moon.

Orbital‑speed formula

For a body in (approximately) circular orbit

\[

v=\frac{2\pi r}{T}

\]

where

• \(r\) = orbital radius (average distance) in metres (m)
• \(T\) = orbital period in seconds (s)
• \(v\) = orbital speed in metres per second (m s⁻¹)

Example – Mars

\[

r_{\text{Mars}} = 1.52\;\text{AU}=1.52\times1.496\times10^{11}\text{ m}=2.28\times10^{11}\text{ m}

\]

\[

T_{\text{Mars}} = 1.88\;\text{yr}=1.88\times3.156\times10^{7}\text{ s}=5.94\times10^{7}\text{ s}

\]

\[

v_{\text{Mars}}=\frac{2\pi(2.28\times10^{11})}{5.94\times10^{7}}

\approx 2.4\times10^{4}\text{ m s}^{-1}=24\text{ km s}^{-1}

\]

6.2 The Solar System (extended)

Learning Objective

Students should be able to:

• Define a solar system.
• List the eight planets in order from the Sun and identify dwarf planets and the asteroid belt.
• Compare the sizes and average distances of the major solar‑system bodies.
• Explain, using Newton’s law of gravitation, why planets remain in orbit and why inner planets move faster.
• State Kepler’s 3rd law (\(T^{2}\propto r^{3}\)) and use it qualitatively to relate period and distance.

Definition

A solar system consists of a star (the Sun) together with all objects that are gravitationally bound to it – planets, dwarf planets, moons, asteroids, comets, the Kuiper‑belt objects and the interplanetary medium.

Order of the Planets

1. Mercury
2. Venus
3. Earth
4. Mars
5. Jupiter
6. Saturn
7. Uranus
8. Neptune

Minor bodies

• Dwarf planets: Pluto, Eris, Haumea, Makemake, Ceres (the latter also lies in the asteroid belt).
• Asteroid belt: Region between Mars and Jupiter (≈ 2–3.5 AU) containing millions of rocky bodies.
• Kuiper‑belt & Oort cloud: Reservoir of icy bodies beyond Neptune; the source of most comets.

Comparative Size & Distance Table

Why do planets stay in orbit?

• Newton’s law of universal gravitation:

  \[

  F = G\frac{M_{\odot}m}{r^{2}}

  \]

  where \(M_{\odot}\) is the Sun’s mass, \(m\) the planet’s mass, \(r\) the centre‑to‑centre distance and \(G = 6.674\times10^{-11}\,\text{N m}^{2}\text{kg}^{-2}\).

• This gravitational force provides the required centripetal force for circular motion:

  \[

  \frac{mv^{2}}{r}=G\frac{M_{\odot}m}{r^{2}}

  \;\;\Longrightarrow\;\;

  v=\sqrt{\frac{GM_{\odot}}{r}}

  \]

  The same speed is obtained from the orbital‑speed formula \(v=2\pi r/T\).

• Inverse‑square law: As \(r\) increases, the gravitational pull falls off as \(1/r^{2}\). Consequently inner planets (small \(r\)) experience a stronger pull and must travel faster to balance it – exactly what the syllabus expects students to state.
• Kepler’s 3rd law (qualitative):

  \[

  T^{2}\propto r^{3}

  \]

  This expresses the same relationship in a form often used in exam questions (e.g., comparing periods of two planets).

Worked Example – Orbital speed of a planet at 2 AU with a period of 2 yr

\[

r = 2\;\text{AU}=2\times1.496\times10^{11}\text{ m}=2.992\times10^{11}\text{ m}

\]

\[

T = 2\;\text{yr}=2\times3.156\times10^{7}\text{ s}=6.312\times10^{7}\text{ s}

\]

\[

v=\frac{2\pi r}{T}

=\frac{2\pi(2.992\times10^{11})}{6.312\times10^{7}}

\approx 2.98\times10^{4}\text{ m s}^{-1}=29.8\text{ km s}^{-1}

\]

6.3 The Universe

Learning Objective

Students should be able to:

• State that the Milky Way is one of billions of galaxies in the observable Universe.
• Identify the Milky Way as a barred spiral galaxy and recognise the other main galaxy types (elliptical, irregular).
• Give the approximate diameter of the Milky Way (≈ 100 000 light‑years) and convert this distance into kilometres.
• Recall the approximate number of galaxies in the observable Universe and the age of the Universe (~13.8 billion years).
• Interpret scale diagrams that compare Earth, the Solar System, the Milky Way and the observable Universe.

Key Facts

• Observable Universe contains an estimated 2–3 trillion (≈ 2 × 10¹²) galaxies.
• Our home galaxy, the Milky Way, is a barred spiral galaxy (type SBc).
• Other common galaxy types:

  • Elliptical – smooth, ellipsoidal shape, little gas or dust.
  • Spiral – flat disc with winding arms (e.g., Andromeda, M81).
  • Irregular – no regular shape (e.g., Large Magellanic Cloud).

• Diameter of the Milky Way ≈ 100 000 light‑years (ly).
• 1 light‑year = distance light travels in one Julian year:

  \[

  1\text{ ly}=c\times(1\text{ yr})\approx 9.461\times10^{12}\text{ km}=9.461\times10^{15}\text{ m}

  \]

• Milky Way diameter in kilometres:

  \[

  100\,000\text{ ly}\times9.461\times10^{12}\frac{\text{km}}{\text{ly}}

  \approx 9.5\times10^{17}\text{ km}

  \]

• Age of the Universe ≈ 13.8 billion years (derived from cosmic microwave background and expansion measurements).

Comparative Size Table (Cosmic Scale)

Understanding Scale

1. If the Earth’s diameter were reduced to a 1 cm line, the Milky Way’s diameter would be about 7 km** on the same scale.
2. Even at that size, the Milky Way is only one of billions of galaxies, each containing billions of stars.
3. The faint milky band we see across the night sky is the combined light of these stars spread over the 100 000‑ly width of our galaxy.

Quick Revision Questions

• Approximately how many galaxies are thought to exist in the observable Universe?
• What type of galaxy is the Milky Way?
• State the approximate diameter of the Milky Way in light‑years.
• Convert the Milky Way’s diameter to kilometres using \(1\text{ ly}=9.461\times10^{12}\text{ km}\).
• Why does the Milky Way appear as a faint band rather than a bright disc in the night sky?
• Using \(v=2\pi r/T\), calculate the orbital speed of a planet that orbits at 2 AU with a period of 2 years.
• State Kepler’s 3rd law and explain what it tells you about the relationship between a planet’s orbital period and its distance from the Sun.
• Give the approximate age of the Universe.