4PH1

Motion in the Universe

Astrophysics

Exam Frequency Analysis

Past paper frequency (2018 to 2024)

This topic accounts for approximately 5% of your exam marks.

stable
Rare
Stable5%

Orbital motion, gravitational fields and circular motion in the solar system tested as shorter questions.

Definition

Orbital period (T) = the time taken for an object to complete one full orbit

  • Units: seconds (s) in SI, but in practice often expressed in minutes, hours, days or years depending on the orbit
  • Examples:
    • International Space Station: T ≈ 90 minutes
    • The Moon around Earth: T ≈ 27.3 days
    • Earth around the Sun: T = 1 year
    • Halley's Comet around the Sun: T ≈ 76 years

Orbital speed equation

  • In one complete orbit, the body sweeps out a path whose length is the circumference of the orbit (treating it as a circle):

distance per orbit = 2πr

  • where r is the average orbital radius (measured from the centre of the body being orbited to the orbiting body, not from the surface)
  • Combine this with speed = distance / time to get the orbital speed equation:

v = 2πr / T

  • Where:
    • v = orbital speed (m/s)
    • r = orbital radius (m)
    • T = orbital period (s)
The orbital radius r is measured from the centre of the Sun to the planet, and the orbital speed v acts tangentially to the circular orbit
Source: Orbital speed formula by Save My Exams

Watch out for the radius

  • r is from centre to centre, not from the planet's surface
  • For a satellite in low Earth orbit at a height h above the surface, the orbital radius is:

r = R_Earth + h

  • where R_Earth ≈ 6400 km. Forgetting to add R_Earth is the single most common mistake in this calculation

Why faster-orbiting bodies are closer in

  • Looking at v = 2πr / T together with the fact that gravity falls off with distance, you can see the pattern:
    • Close to the central body, gravity is strong, needs high speed to balance, period is short
    • Far from the central body, gravity is weak, only slow speed needed to balance, period is long
  • That is why Mercury has a year of 88 days and Neptune has a year of 165 years; and why low satellites zip around the Earth in 90 minutes while geostationary satellites take 24 hours
  • Quantitatively, Kepler's third law says T² ∝ r³, but the qualitative rule "closer = faster" is enough here
Worked example

Calculating orbital speed from radius and period

A moon orbits its planet with an orbital radius of 600 000 km and a period of 4 days. Calculate the orbital speed of the moon in m/s.

Solution:

  • Convert the period to seconds: 4 days × 24 × 60 × 60 = 345 600 s
  • Convert the orbital radius to metres: 600 000 km × 1000 = 6.0 × 10⁸ m
  • Apply v = 2πr / T: v = (2 × π × 6.0 × 10⁸) / 345 600
  • v = 10 900 m/s (to 3 s.f.)