Define average orbital speed from the equation v = 2 π r / T where r is the average radius of the orbit and T is the orbital period; recall and use this equation

6.1.1 The Earth – Average Orbital Speed (S6.1.1)

Learning objective

Define the average orbital speed of a planet and recall, rearrange and use the equation

v = \(\displaystyle\frac{2\pi\,\color{#1976d2}{r}}{\color{#388e3c}{T}}\)

in calculations (AO1 + AO2) as required by the Cambridge IGCSE Physics syllabus.

Why the equation matters

The formula is a direct consequence of Kepler’s 3^rd law – the square of the orbital period is proportional to the cube of the semi‑major axis (\(T^{2}\propto r^{3}\)).

For a near‑circular orbit the distance travelled in one revolution is the circumference \(2\pi r\); dividing by the period \(T\) therefore gives the average orbital speed.

Key definitions

• Orbital speed (v) – average speed of a planet (or satellite) as it moves round the Sun.
• Average orbital radius (r) – mean distance from the planet to the Sun. For a nearly circular orbit we treat the path as a circle of radius \(r\).
• Orbital period (T) – time taken for one complete revolution around the Sun.

Derivation (circular orbit)

1. Distance travelled in one full revolution = circumference = \(2\pi r\).
2. Time taken for that distance = orbital period \(T\).
3. Average speed = distance ÷ time → \(v = \dfrac{2\pi r}{T}\).

Formula box – rearrangements

Unit‑verification tip

Check that the units reduce to metres per second (m s⁻¹):

• \(r\) in metres (m)
• \(T\) in seconds (s)
• \(\dfrac{2\pi r}{T}\) → \(\dfrac{\text{m}}{\text{s}} = \text{m s}^{-1}\)

Worked example – Earth’s average orbital speed

Calculation

\[

\begin{aligned}

v &= \frac{2\pi (1.496\times10^{11}\,\text{m})}{3.156\times10^{7}\,\text{s}}\\[4pt]

&\approx \frac{9.40\times10^{11}\,\text{m}}{3.156\times10^{7}\,\text{s}}\\[4pt]

&\approx 2.98\times10^{4}\,\text{m s}^{-1}

\end{aligned}

\]

Sanity‑check: 30 km s⁻¹ is the commonly quoted value for Earth’s orbital speed – our result (29.8 km s⁻¹) is consistent.

Unfamiliar context – exoplanet estimate

A newly discovered exoplanet orbits its star with a period of 200 days and a semi‑major axis of 1.2 AU**. Estimate its average orbital speed.

1. Convert the data:

   • 1 AU = 1.496 × 10¹¹ m → \(r = 1.2 \times 1.496\times10^{11} = 1.795\times10^{11}\) m.
   • 200 days = 200 × 24 × 3600 = 1.728 × 10⁷ s → \(T = 1.73\times10^{7}\) s.

2. Insert into the formula:

   \[

   v = \frac{2\pi (1.795\times10^{11}\,\text{m})}{1.73\times10^{7}\,\text{s}}

   \approx 6.5\times10^{4}\,\text{m s}^{-1}

   = 65\ \text{km s}^{-1}

   \]

This type of “unfamiliar” data (AU and days) mirrors the style of Cambridge exam questions and tests the student’s ability to convert units before applying the equation.

Comparison of inner and outer planets – link to Kepler’s 3^rd law

Because \(T^{2}\propto r^{3}\) (Kepler’s 3^rd law), a smaller radius implies a much shorter period. Substituting the smaller \(r\) and the correspondingly smaller \(T\) into \(v = 2\pi r/T\) therefore yields a larger speed for inner planets.

Real‑world application – satellite launch windows

Concept‑check question

“A communications satellite orbits the Earth at an average altitude of 20 000 km. If the satellite must complete one revolution every 12 hours, what is its average orbital speed?”

• Identify the symbols: \(r\) = Earth radius + altitude, \(T\) = 12 h, \(v\) = ?
• Use the rearranged form \(v = 2\pi r/T\) to find the unknown.

Practice questions (AO1 + AO2)

1. Calculate the average orbital speed of Mars given \(r = 2.279 × 10^{11}\) m and \(T = 5.94 × 10^{7}\) s.
2. A hypothetical planet has an orbital speed of 15 km s⁻¹ and an orbital period of 2.0 × 10^{7}\) s**. Find its average orbital radius.
3. Explain, using \(v = 2\pi r/T\) and Kepler’s 3^rd law, why a planet closer to the Sun moves faster than one that is farther away.
4. For a low‑Earth‑orbit satellite, \(r = 6.78 × 10^{6}\) m and \(v = 7.8 km s^{-1}\). Determine its orbital period \(T\).
5. Exoplanet challenge (see example above): An exoplanet orbits its star with a period of 300 days and a semi‑major axis of 0.9 AU. Estimate its average orbital speed.

Suggested diagram