For a circular orbit, the period is T = 2π√(r³ ÷ GM). Everything must be in SI units: metres, kilograms and seconds, with r measured from the planet’s centre.
This lesson is part of gravitation. It builds on explaining centripetal force in orbital motion.
What are the conversions to check first?
Before any calculation, check three things.
- The radius is measured from the centre, so add the planet’s radius to any altitude.
- Distances are in metres, so 600 km becomes 6.0 × 10⁵ m.
- The answer is in seconds until you convert it.
Checking these three catches most wrong-power errors.
Worked example: a satellite 600 km up
A satellite orbits 600 km above Earth’s surface. Earth’s radius is 6.37 × 10⁶ m and GM = 3.98 × 10¹⁴ N m² kg⁻¹. Find its period.
- Orbital radius: r = 6.37 × 10⁶ + 6.0 × 10⁵ = 6.97 × 10⁶ m.
- r³ = (6.97 × 10⁶)³ = 3.39 × 10²⁰ m³.
- r³ ÷ GM = 3.39 × 10²⁰ ÷ 3.98 × 10¹⁴ = 8.50 × 10⁵.
- T = 2π × √(8.50 × 10⁵) = 2π × 922 = 5.79 × 10³ s, which is 96.6 minutes.
If the altitude alone (6.0 × 10⁵ m) had been used as the radius, the answer would have been far too small.
The ratio method
Suppose satellite B orbits at four times the radius of satellite A around the same planet. Since T² ∝ r³, T_B² ÷ T_A² = 4³ = 64, so T_B ÷ T_A = 8.
If A has a period of 2.0 hours, B has a period of 16 hours. No constants were needed.
The mistake that costs marks
The slips come from units and from the wrong radius.
| Slip | Effect | Fix |
|---|---|---|
| Using altitude as r | Period far too short | Add the planet’s radius |
| Leaving 600 km as 600 | Answer off by 10⁹ in r³ | Convert to 6.0 × 10⁵ m |
| Reporting T in seconds as “hours” | Wrong unit | Divide by 3600 |
| Squaring the ratio, not cubing | Wrong factor | T² ∝ r³ |
Sense-check the result. A low Earth orbit should take about an hour and a half, and a geostationary orbit should take a day.
Check yourself
A geostationary satellite has an orbital radius of 4.2 × 10⁷ m. Use GM = 3.98 × 10¹⁴ N m² kg⁻¹ and find its period in hours.
Answer
r³ = (4.2 × 10⁷)³ = 7.41 × 10²² m³. Then r³ ÷ GM = 7.41 × 10²² ÷ 3.98 × 10¹⁴ = 1.86 × 10⁸.
T = 2π × √(1.86 × 10⁸) = 2π × 1.364 × 10⁴ = 8.57 × 10⁴ s.
In hours: 8.57 × 10⁴ ÷ 3600 = 23.8 hours, which is close to one day, as expected.
What to study next
Go back to distinguishing mass, weight and gravitational field strength if g itself is still unclear. Then try the gravitation practice set.
To have a teacher go through your conversions, see online one-to-one Physics tuition. The units and significant figure checker helps you test powers of ten.