Orbital Velocity Formula: vₒ = √(GM/r), Satellite Speed

The orbital velocity formula is $v_o = \sqrt{\frac{GM}{r}}$ , where M is the planet's mass and r is the radius of the orbit measured from the planet's centre. For a satellite at height h above a planet of radius R, $r = R + h$ , so $v_o = \sqrt{\frac{GM}{R + h}}$ . Close to Earth's surface it reduces to $\sqrt{gR} \approx 7.9$ km/s.

A satellite in a circular orbit is falling all the time, but it moves sideways so fast that the ground curves away beneath it just as quickly as it falls. Gravity supplies exactly the inward force needed for the circle. Satellites for weather, navigation and communication all move at speeds set by this one formula, which also appears as the orbital velocity formula Class 11 students derive in the gravitation chapter.

Quick facts

Formula
$v_o = \sqrt{\frac{GM}{r}}$
At height h
$v_o = \sqrt{\frac{GM}{R + h}} = R\sqrt{\frac{g}{R + h}}$
Near surface
$\sqrt{gR} \approx 7.9$ km/s for Earth
SI unit
m/s
Independent of
Mass of the satellite
Escape link
$v_e = \sqrt{2}\,v_o$

Deriving the orbital velocity formula

For a satellite of mass m in a circular orbit, gravity is the only force, and it must equal the needed centripetal force:

$\frac{GMm}{r^2} = \frac{mv_o^2}{r}$

Cancel m and one factor of r, then take the root: $v_o = \sqrt{\frac{GM}{r}}$ . Using $GM = gR^2$ , you can also write $v_o = R\sqrt{\frac{g}{R + h}}$ , handy when the question gives g instead of G and M.

Worked examples

Example 1: skimming the surface

Find the orbital speed for a satellite just above Earth's surface, ignoring air. Work from the surface value of g and Earth's radius.

$v_o = \sqrt{gR} = \sqrt{9.8 \times 6.37 \times 10^6} \approx 7.90 \times 10^3\ \text{m/s}$

About 7.9 km/s, or more than 28 000 km/h.

Example 2: a space station 400 km up

Find the orbital speed at a height of 400 km.

r = 6370 km + 400 km = 6770 km = 6.77 × 10⁶ m, and GM for Earth ≈ 3.98 × 10¹⁴ m³/s².

$v_o = \sqrt{\frac{3.98 \times 10^{14}}{6.77 \times 10^6}} \approx 7.67 \times 10^3\ \text{m/s}$

Example 3: a geostationary satellite

A geostationary satellite orbits at r = 4.22 × 10⁷ m. Find its speed.

$v_o = \sqrt{\frac{3.98 \times 10^{14}}{4.22 \times 10^7}} \approx 3.07 \times 10^3\ \text{m/s}$

Higher orbits are slower, which surprises many students.

Energy of a satellite in orbit

Energies for a satellite of mass m at radius r
Energy	Formula	Sign
Kinetic	$\frac{1}{2}mv_o^2 = \frac{GMm}{2r}$	Positive
Potential	$-\frac{GMm}{r}$	Negative
Total	$-\frac{GMm}{2r}$	Negative, so the satellite is bound
Binding energy	$+\frac{GMm}{2r}$	Energy needed to free it

The kinetic energy is exactly half the size of the potential energy. To move a satellite to a higher orbit you must add energy, yet it ends up slower. The potential energy page explains the minus sign.

Orbital speed, escape speed and period

Multiply $v_o$ by $\sqrt{2}$ and you get the escape velocity from the same radius. Divide the orbit's circumference by $v_o$ and you get the time period of a satellite, $T = \frac{2\pi r}{v_o}$ . For the station in Example 2 that is about 5.5 × 10³ s, roughly 92 minutes.

The formula holds only for circular orbits. For an ellipse the speed changes along the path, greatest at the closest point.

Symbols and what sets the speed

Inputs to vₒ = √(GM/r)
Letter	Stands for	Unit
$v_o$	Speed along a circular orbit	m/s
$G$	Gravitational constant	N m²/kg²
$M$	How massive the central body is	kg
$r$	Orbit radius, from the centre of that body	m

Only M and r appear. A satellite round the Moon at the same r as one round Earth goes much more slowly, because the Moon's mass is about 1/81 of Earth's, so its speed is roughly 1/9 as large. Quadruple r and the speed halves, since $v_o \propto \frac{1}{\sqrt{r}}$ .

Pitfalls

Using R where the formula needs R + h.
Forgetting to change height in km to metres.
Thinking a heavier satellite needs more speed. Mass cancels.
Using the Earth's mass when the satellite orbits the Moon or the Sun.

The source of the force is set out on the gravitational force formula hub.

Frequently asked questions

What is the formula of orbital velocity?

vₒ = √(GM / r), with r = R + h the distance from the planet's centre. Near the surface this is √(gR), about 7.9 km/s for Earth.

Will a heavier satellite need a faster orbit?

No. The satellite's mass cancels when gravity is set equal to the centripetal force, so all satellites at the same height move at the same speed.

Why do higher satellites move more slowly?

Gravity is weaker farther out, so less speed is needed to keep a circular path. vₒ falls as 1 / √r.

How is orbital velocity related to escape velocity?

At the same distance from the centre, escape velocity is √2 times the orbital velocity, about 1.41 times larger.