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Orbital velocity of a satellite Formula, SPM Physics

Formula: v = √(GM / r). Not given in the exam, recall it.

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Orbital velocity of a satellite
v = √(GM / r)

Not given in the exam, recall it.

What it is for

Orbital velocity of a satellite is used in Gravitation (SPM Physics Form 4). Keep the SI units in every line of working. Gravitation

Symbols and SI units

SymbolMeaningSI unit
vorbital velocitym s⁻¹
Ggravitational constantN m² kg⁻²
Mmass of planetkg
rorbital radiusm

Rearrangements

  • v² = GM / r
  • r = GM / v²
  • M = v²r / G

Worked example

A satellite orbits Earth (M = 6.0 × 10²⁴ kg) at r = 7.0 × 10⁶ m (G = 6.67 × 10⁻¹¹). v = √(GM / r) = √((6.67 × 10⁻¹¹ × 6.0 × 10²⁴) / 7.0 × 10⁶) = 7.56 × 10³ m s⁻¹.

Common trap

r is measured from the planet's centre, not the surface. The orbital speed does not depend on the satellite's own mass.

Understanding orbital velocity of a satellite

The formula v = √(GM / r) gives the speed a satellite must keep to stay in a circular orbit. Here v is the orbital velocity in metres per second (m s⁻¹), G is the gravitational constant, 6.67 × 10⁻¹¹ N m² kg⁻², M is the mass of the planet in kilograms (kg), and r is the orbital radius in metres (m) measured from the planet's centre.

The relation comes from setting gravity equal to the centripetal force, which is why the satellite's own mass cancels out. To find other quantities, square both sides to get v² = GM / r, then rearrange to r = GM / v² or M = v²r / G.

A larger radius always means a slower orbit.

A worked example, step by step

A satellite orbits Earth (M = 6.0 × 10²⁴ kg) at a radius r = 4.2 × 10⁷ m, near geostationary height. Take G = 6.67 × 10⁻¹¹ N m² kg⁻².

Write the formula: v = √(GM / r). Substitute with units: v = √((6.67 × 10⁻¹¹ N m² kg⁻² × 6.0 × 10²⁴ kg) / 4.2 × 10⁷ m).

Work the top first: GM = 4.00 × 10¹⁴. Divide by r: 4.00 × 10¹⁴ / 4.2 × 10⁷ = 9.53 × 10⁶.

Take the square root: v = √(9.53 × 10⁶) = 3.09 × 10³ m s⁻¹. So the satellite travels at about 3090 m s⁻¹, quoted to three significant figures.

Common mistakes and how this is tested

The main trap is measuring r from the planet's surface. It must be from the centre, so add the planet's radius to the altitude.

The orbital speed also does not depend on the satellite's own mass, a fact examiners like to test with a distractor value.

Other errors: forgetting to take the square root at the end and leaving the answer as v²; dividing before multiplying G and M; and mixing up r in kilometres with r in metres. Always convert to metres first.

Under the command word calculate you show the substitution and units. Under explain you may be asked why a higher orbit means a slower satellite, or why mass cancels.

Source: DSKP KSSM Physics Form 4 and 5 (Versi English), Sijil Pelajaran Malaysia: Format Pentaksiran mulai 2021, Fizik (4531) (Bahagian Pembangunan Kurikulum (BPK), KPM)

Written by the spmphysics.com.my editorial team.· Updated 5 Sept 2026

Frequently asked questions

Is this formula given in the exam?
No. SPM Physics papers provide no formula sheet, so this must be recalled.
Why does the satellite's mass not appear in the formula?
Gravity provides the centripetal force, so mv²/r = GMm/r². The satellite mass m cancels on both sides, leaving v² = GM/r. Two satellites at the same radius orbit at the same speed regardless of their mass.
Is r the height above the ground?
No. r is the distance from the planet's centre to the satellite. Add the planet's radius to the orbital altitude before substituting, and keep everything in metres.
How do I find the orbital period from v?
The satellite travels one circumference each period, so T = 2πr / v. Find v first with v = √(GM / r), then divide the circumference 2πr by that speed.

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