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

Published by Patrick Mutisya · 14 days ago

Cambridge IGCSE Physics 0625 – Topic 6.1.1 The Earth

6.1.1 The Earth

Learning Objective

Define the average orbital speed of a planet using the equation

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

and be able to recall and apply this equation in calculations.

Key Concepts

  • Orbital speed (v) – the average speed at which a planet travels around the Sun.

  • Average orbital radius (r) – the mean distance from the planet to the Sun; for a nearly circular orbit it is taken as the radius of the orbit.

  • Orbital period (T) – the time taken for one complete revolution around the Sun.

Derivation of the Equation

The path of a planet in a circular orbit is a circle. The distance travelled in one complete orbit is the circumference of that circle:

\$\text{distance} = 2\pi r\$

If this distance is covered in a time \$T\$, the average speed \$v\$ is distance divided by time:

\$v = \frac{\text{distance}}{T} = \frac{2\pi r}{T}\$

Using the Equation

To find any one of the three variables, rearrange the formula as required:

  • For speed: \$v = \dfrac{2\pi r}{T}\$

  • For radius: \$r = \dfrac{vT}{2\pi}\$

  • For period: \$T = \dfrac{2\pi r}{v}\$

Example: Earth’s Average Orbital Speed

Given data for Earth:

QuantitySymbolValueUnit
Average orbital radius\$r\$1.496 × 1011m
Orbital period\$T\$3.156 × 107s

Calculate the average orbital speed:

\$v = \frac{2\pi (1.496 \times 10^{11}\,\text{m})}{3.156 \times 10^{7}\,\text{s}}\$

Carrying out the calculation:

\$v \approx \frac{9.40 \times 10^{11}\,\text{m}}{3.156 \times 10^{7}\,\text{s}} \approx 2.98 \times 10^{4}\,\text{m s}^{-1}\$

Thus, Earth’s average orbital speed is about \$29.8\ \text{km s}^{-1}\$.

Suggested diagram: A circle representing Earth’s orbit with radius \$r\$, showing one full revolution and labeling the distance \$2\pi r\$ and period \$T\$.

Practice Questions

  1. Calculate the average orbital speed of Mars, given \$r = 2.279 \times 10^{11}\ \text{m}\$ and \$T = 5.94 \times 10^{7}\ \text{s}\$.

  2. If a hypothetical planet has an orbital speed of \$15\ \text{km s}^{-1}\$ and an orbital period of \$2.0 \times 10^{7}\ \text{s}\$, find its average orbital radius.

  3. Explain why the orbital speed of a planet closer to the Sun is greater than that of a planet farther away, using the equation \$v = 2\pi r/T\$.