Know that the force that keeps an object in orbit around the Sun is the gravitational attraction of the Sun

6.1.2 The Solar System

Learning Objective

Explain why the force that keeps any object (planet, moon, asteroid, comet) in orbit around the Sun is the Sun’s gravitational attraction.

Syllabus Overview – Solar‑System Bodies

• The Sun – contains ≈ 99.86 % of the total mass of the Solar System (M⊙ = 1.989 × 10³⁰ kg).
• Eight planets (in order from the Sun):

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

• Dwarf planets – e.g. Ceres (asteroid belt), Pluto, Eris, Haumea, Makemake.
• Minor planets / asteroids – mainly in the asteroid belt between Mars and Jupiter.
• Moons (natural satellites) – e.g. Earth’s Moon (orbital period ≈ 27.3 days), Jupiter’s Galilean moons, Saturn’s Titan.
• Comets – icy bodies from the Kuiper Belt and Oort Cloud that develop tails when near the Sun.

Key Physical Concepts

• Gravitational attraction between two masses.
• Newton’s law of universal gravitation.
• Centripetal force required for circular (or elliptical) motion.
• Equating the two forces to obtain orbital relationships.
• Kepler’s third law in its classic form : \(T^{2}=k\,r^{3}\) where \(k=\dfrac{4\pi^{2}}{GM_{\odot}}\).

Newton’s Law of Universal Gravitation

The attractive force between two masses \(M\) and \(m\) separated by a distance \(r\) is

\[

F_{\text{gravity}} = G\frac{M\,m}{r^{2}}

\]

where

• \(G = 6.67\times10^{-11}\ \text{N m}^{2}\,\text{kg}^{-2}\) (gravitational constant),
• \(M\) – mass of the Sun (or any other body),
• \(m\) – mass of the orbiting object,
• \(r\) – distance between the centres of the two bodies.

Centripetal Force for Circular Motion

An object of mass \(m\) moving in a circular orbit of radius \(r\) with speed \(v\) needs a centripetal force

\[

F_{\text{centripetal}} = \frac{m v^{2}}{r}.

\]

Orbital Condition – Derivation of the Orbital‑Speed Formula

For a stable orbit the Sun’s gravity supplies exactly the required centripetal force:

\[

G\frac{M_{\odot} m}{r^{2}} \;=\; \frac{m v^{2}}{r}.

\]

Step 1 – Cancel the planet’s mass \(m\) (a common point of confusion for AO2 candidates):

\[

G\frac{M_{\odot}}{r^{2}} = \frac{v^{2}}{r}.

\]

Step 2 – Solve for the orbital speed \(v\):

\[

v = \sqrt{\frac{G M_{\odot}}{r}}.

\]

Using the relation \(v = \dfrac{2\pi r}{T}\) (where \(T\) is the orbital period) gives Kepler’s third law:

\[

T^{2} = \frac{4\pi^{2}}{G M_{\odot}}\,r^{3}

\;=\; k\,r^{3},

\qquad

k = \frac{4\pi^{2}}{G M_{\odot}}.

\]

Typical Planetary Data (average orbital radius, period and mass)

Note: the Sun’s mass \(M_{\odot}=1.989\times10^{30}\ \text{kg}\) is required for all calculations.

Why Gravity Keeps Planets (and Other Bodies) in Orbit

1. The Sun’s enormous mass creates a strong gravitational pull toward its centre.
2. Each planet (or other body) possesses forward (tangential) momentum. In the absence of a force it would travel in a straight line (Newton’s first law).
3. The Sun’s gravity constantly pulls the body toward the Sun, continuously changing the direction of its motion.
4. This continual change of direction provides the exact centripetal acceleration needed for a circular or elliptical orbit, so the body does not fall straight into the Sun.

Limitations of the Circular‑Orbit Approximation

• Real planetary orbits are elliptical (Kepler’s first law). The circular‑orbit derivation is a useful approximation because most eccentricities are small.
• When higher precision is required (e.g., for spacecraft trajectories) the full elliptical‑orbit equations must be used.

Related Topics Required by the Syllabus

• Earth’s axial tilt (≈ 23.5°) and seasons – the tilt, together with Earth’s rotation, explains the variation in solar elevation and daylight length over the year.
• The Earth–Moon system – the Moon orbits Earth in ≈ 27.3 days; the Earth–Moon barycentre lies inside Earth, so the Moon’s motion is governed by Earth’s gravity, while the Earth–Moon pair together orbit the Sun.

Common Misconceptions

• “Gravity only pulls objects down to the ground.” In space “down” means toward the centre of mass of the dominant body – for planets this is the Sun.
• “An object needs a continuous push to stay in orbit.” No push is required; the Sun’s gravitational attraction continuously provides the necessary centripetal force.
• “If gravity pulls a planet inward, it must eventually crash into the Sun.” The planet’s tangential velocity balances the inward pull, producing a stable orbit.

Worked Example – Orbital Speed of the Earth

Using \(v = \sqrt{\dfrac{G M_{\odot}}{r}}\):

1. Insert the known values:
   

   \[

   v = \sqrt{\frac{(6.67\times10^{-11}\ \text{N m}^{2}\,\text{kg}^{-2})(1.989\times10^{30}\ \text{kg})}{1.496\times10^{11}\ \text{m}}}

   \]

2. Calculate the numerator (the Sun’s standard gravitational parameter):
   

   \[

   G M_{\odot} = 1.327\times10^{20}\ \text{m}^{3}\,\text{s}^{-2}

   \]

3. Divide by \(r\):
   

   \[

   \frac{1.327\times10^{20}}{1.496\times10^{11}} = 8.87\times10^{8}\ \text{m}^{2}\,\text{s}^{-2}

   \]

4. Take the square root (remember to keep track of units):
   

   \[

   v = \sqrt{8.87\times10^{8}} \approx 2.98\times10^{4}\ \text{m s}^{-1}

   \]

5. Convert to kilometres per second:
   

   \[

   v \approx 29.8\ \text{km s}^{-1}

   \]

Suggested Diagram

Check Your Understanding

1. Write the formula for the gravitational force between the Sun and a planet, defining every symbol.
2. In one sentence, explain why a planet does not spiral into the Sun despite the continual gravitational attraction.
3. Calculate the orbital speed of the Earth using the data given above. Show each step and give the final answer in km s⁻¹.
4. State Kepler’s third law in its classic form and give the numerical value of the constant \(k\) (use \(G\) and \(M_{\odot}\) provided).
5. Explain how Earth’s axial tilt leads to the seasons.
6. State the orbital period of the Moon and describe how the Earth–Moon system orbits the Sun.

Summary

The Sun’s gravitational attraction supplies the centripetal force required for planets, moons, asteroids and comets to remain in orbit. By equating the gravitational force \(G M{\odot} m/r^{2}\) with the centripetal force \(m v^{2}/r\) we obtain the orbital‑speed relation \(v=\sqrt{GM{\odot}/r}\). Combining this with \(v=2\pi r/T\) yields Kepler’s third law \(T^{2}=k\,r^{3}\) (with \(k=4\pi^{2}/GM_{\odot}\)). These relationships, together with an understanding of Earth’s tilt and the Earth‑Moon dynamics, form a core part of the Cambridge IGCSE/A‑Level physics syllabus.