Orbital Velocity

Orbital velocity is the velocity required for an object to remain in orbit around another body under the influence of gravity.

For a satellite in a circular orbit, gravity continuously pulls the satellite toward the centre of the planet. Instead of falling straight down, the satellite has enough sideways velocity that it continually falls around the planet.

For a circular orbit, the orbital speed is:

v_orbit = √(GM/r)

where:

• v_orbit = orbital speed in m/s
• G = gravitational constant = 6.67 × 10⁻¹¹ N·m²/kg²
• M = mass of the central body in kg
• r = orbital radius measured from the centre of the central body in m

The satellite's own mass does not appear in the final equation.

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What Is an Orbit?

An orbit is the curved path followed by an object moving under the gravitational influence of another body.

Examples include:

• the Moon orbiting Earth
• Earth orbiting the Sun
• artificial satellites orbiting Earth
• moons orbiting other planets
• binary stars orbiting one another

An orbiting object is constantly accelerating because its velocity is continuously changing direction.

Even in a perfectly circular orbit where its speed remains constant, its velocity does not.

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Speed and Velocity in an Orbit

Speed is a scalar quantity.

Velocity is a vector quantity.

For a circular orbit:

• speed can remain constant
• direction continuously changes
• velocity therefore continuously changes
• the satellite has an acceleration

This acceleration points toward the centre of the orbit and is called centripetal acceleration.

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Satellites Are Falling

A satellite in orbit is in continuous free fall.

Gravity pulls it toward Earth, but the satellite also has a large sideways velocity.

As the satellite falls toward Earth, Earth's curved surface falls away beneath it.

The result is an orbit.

This idea is sometimes illustrated using Newton's cannon.

A projectile launched slowly falls back to Earth.

A faster projectile travels farther before hitting Earth.

At a sufficiently high horizontal speed, the projectile falls around Earth rather than striking the surface.

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Gravity Provides the Centripetal Force

Circular motion requires a force directed toward the centre of the circle.

This is called the centripetal force.

For an orbiting satellite, gravity provides this force.

Therefore:

gravitational force = required centripetal force

This relationship allows us to derive the orbital-speed equation.

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Gravitational Force

For a satellite of mass m orbiting a planet of mass M:

F_gravity = GMm/r²

where r is the distance between the centres of the two objects.

The force points toward the centre of the planet.

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Centripetal Force

For circular motion:

F_c = mv²/r

where:

• m = satellite mass
• v = orbital speed
• r = orbital radius

The required centripetal force also points toward the centre of the circular path.

For a gravitational orbit, these two forces are not separate forces.

Gravity is the centripetal force.

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Deriving the Orbital-Speed Equation

For a circular orbit:

F_gravity = F_c

Therefore:

GMm/r² = mv²/r

Cancel the satellite mass m:

GM/r² = v²/r

Multiply both sides by r:

GM/r = v²

Therefore:

v² = GM/r

Taking the square root:

v_orbit = √(GM/r)

This is the equation for circular orbital speed.

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Why Satellite Mass Cancels

The satellite mass m appears in both:

F_gravity = GMm/r²

and:

F_c = mv²/r

A more massive satellite experiences a greater gravitational force.

However, a more massive satellite also requires proportionally more force to produce the same centripetal acceleration.

These effects cancel.

Therefore:

orbital speed does not depend on satellite mass

for the same circular orbital radius around the same central body.

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What Determines Orbital Speed?

From:

v_orbit = √(GM/r)

orbital speed depends on:

• the mass M of the central body
• the orbital radius r

It does not depend on the satellite's mass.

Therefore:

larger M → greater orbital speed

and:

larger r → lower orbital speed

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Effect of Central Mass

If orbital radius remains constant:

v ∝ √M

Suppose Planet B has four times the mass of Planet A.

At the same orbital radius:

vB/vA = √4

= 2

A satellite would need twice the orbital speed around Planet B.

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Effect of Orbital Radius

For the same central mass:

v ∝ 1/√r

If orbital radius increases by a factor of 4:

v_new/v_old = 1/√4

= 1/2

Therefore, a satellite four times farther from the centre travels at half the circular orbital speed.

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Lower Orbits Are Faster

For satellites around the same planet:

smaller r → greater orbital speed

larger r → smaller orbital speed

This may initially seem surprising.

A satellite farther from Earth has a larger path to travel around its orbit, but it moves more slowly because Earth's gravitational field is weaker there.

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Orbital Radius Is Not Altitude

One of the most important details in orbital calculations is the meaning of r.

The equation:

v = √(GM/r)

uses distance from the centre of the planet.

Therefore:

r = R_planet + h

where:

• R_planet = planetary radius
• h = satellite altitude above the surface

Never substitute altitude alone unless the problem specifically defines r that way.

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Worked Example: Finding Orbital Radius

Suppose a satellite orbits:

400 km

above Earth's surface.

Earth's radius is approximately:

6370 km

Therefore:

r = 6370 + 400

= 6770 km

Convert to metres:

r = 6.77 × 10⁶ m

This is the value used in the orbital-speed equation.

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Worked Example: Low Earth Orbit

Calculate the orbital speed of a satellite 400 km above Earth.

Use:

M = 5.97 × 10²⁴ kg

G = 6.67 × 10⁻¹¹ N·m²/kg²

r = 6.77 × 10⁶ m

Then:

v = √(GM/r)

v = √[(6.67 × 10⁻¹¹)(5.97 × 10²⁴)/(6.77 × 10⁶)]

v ≈ 7.67 × 10³ m/s

Therefore:

v ≈ 7.7 km/s

A low-Earth-orbit satellite therefore travels at several kilometres every second.

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Gravity Is Still Strong in Orbit

Satellites are not orbiting because they are beyond Earth's gravity.

At an altitude of only a few hundred kilometres, Earth's gravitational field remains quite strong.

The satellite is weightless in the sense of apparent weight because the satellite and everything inside it are falling together.

Gravity is actually what keeps the satellite in orbit.

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Orbital Velocity Is Tangential

For a circular orbit, the satellite's instantaneous velocity points tangent to the orbital path.

The gravitational force points toward the centre.

Therefore:

velocity ⟂ gravitational force

at every instant.

Gravity continuously changes the direction of the velocity.

In a perfectly circular orbit, it does not change the speed.

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Why Gravity Does No Work in a Circular Orbit

Work is:

W = Fd cos θ

For an infinitesimal displacement in a circular orbit:

• gravitational force points radially inward
• displacement is tangential
• θ = 90°

Since:

cos 90° = 0

gravity does no work.

Therefore:

• kinetic energy remains constant
• gravitational potential energy remains constant
• total mechanical energy remains constant
• orbital speed remains constant

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Circular Orbits and Equipotentials

A circular orbit around a spherical mass remains at constant radius.

Since:

V = −GM/r

constant r means constant gravitational potential.

Therefore, the circular orbit follows an equipotential surface.

The satellite moves along an equipotential while gravity acts perpendicular to its motion.

This connects orbital motion with the ideas of gravitational potential and equipotential surfaces.

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Orbital Speed and Field Strength

We know:

g = GM/r²

and:

v² = GM/r

Therefore:

GM = gr²

Substituting:

v² = gr²/r

so:

v = √(gr)

Here g is the gravitational field strength at the orbital radius.

This provides another useful way to calculate orbital speed.

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Worked Example Using Field Strength

At a satellite's orbital radius:

g = 4.0 N/kg

and:

r = 1.0 × 10⁷ m

Then:

v = √(gr)

= √[(4.0)(1.0 × 10⁷)]

= √(4.0 × 10⁷)

≈ 6.32 × 10³ m/s

Therefore:

v ≈ 6.3 km/s

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Orbital Period

The orbital period, T, is the time required to complete one orbit.

For a circular orbit, the distance travelled in one complete orbit is:

2πr

Since:

speed = distance/time

we have:

v = 2πr/T

Therefore:

T = 2πr/v

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Connecting Period and Orbital Radius

Using:

v = √(GM/r)

in:

T = 2πr/v

gives:

T = 2πr / √(GM/r)

which simplifies to:

T = 2π√(r³/GM)

Squaring:

T² = 4π²r³/(GM)

Therefore:

T² ∝ r³

for objects orbiting the same central mass.

This is the mathematical form of Kepler's Third Law for circular orbits.

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What Kepler's Third Law Means

Satellites farther from a planet:

• travel more slowly
• have larger orbital paths
• take much longer to complete an orbit

Therefore, orbital period increases strongly with orbital radius.

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Worked Example: Orbital Period

Suppose a satellite has:

r = 7.0 × 10⁶ m

v = 7.5 × 10³ m/s

Then:

T = 2πr/v

= 2π(7.0 × 10⁶)/(7.5 × 10³)

≈ 5860 s

Convert to minutes:

5860/60 ≈ 97.7 min

The satellite completes approximately one orbit every:

98 minutes

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Geostationary Orbit

A geostationary satellite remains above approximately the same point on Earth's equator.

For this to happen, it must:

• orbit above the equator
• orbit in the same direction as Earth's rotation
• have an orbital period equal to Earth's rotation period
• follow an approximately circular orbit

Its altitude is approximately:

35,786 km above Earth's surface

and its orbital speed is about:

3.1 km/s

The much lower speed compared with low Earth orbit is a consequence of its much larger orbital radius.

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Orbital Velocity Around the Moon

The same equation applies around any approximately spherical celestial body:

v = √(GM/r)

For the Moon:

M ≈ 7.35 × 10²² kg

Suppose a spacecraft orbits at:

r = 2.0 × 10⁶ m

Then:

v = √[(6.67 × 10⁻¹¹)(7.35 × 10²²)/(2.0 × 10⁶)]

≈ 1.57 × 10³ m/s

Therefore:

v ≈ 1.6 km/s

This is much lower than typical low-Earth orbital speed.

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Orbital Velocity Around the Sun

Planets are also orbiting bodies.

For a roughly circular planetary orbit:

v = √(GM_sun/r)

Earth's orbital speed around the Sun is approximately:

30 km/s

The same gravitational physics that governs artificial satellites also governs planetary motion.

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Orbital Velocity and Escape Speed

Orbital speed and escape speed are closely related but describe different motions.

For a circular orbit:

v_orbit = √(GM/r)

For escape from the same radius:

v_escape = √(2GM/r)

Therefore:

v_escape = √2 v_orbit

Since:

√2 ≈ 1.414

the escape speed at a particular radius is about:

1.414 times the circular orbital speed at that same radius.

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Worked Comparison

Suppose a satellite has a circular orbital speed of:

7.5 km/s

The escape speed at the same radius is:

v_escape = √2 × 7.5

≈ 1.414 × 7.5

≈ 10.6 km/s

The satellite would therefore need a higher speed to escape completely than it needs to remain in circular orbit.

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Why Escape Speed Is Greater

A circularly orbiting satellite is still gravitationally bound.

Its total mechanical energy is negative.

To escape, additional energy must be supplied until:

E_total ≥ 0

Therefore, escape requires a greater speed than circular orbit at the same radius.

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Energy in a Circular Orbit

For a circular orbit:

Eₖ = GMm/(2r)

Gravitational potential energy is:

Eₚ = −GMm/r

Therefore:

Eₚ = −2Eₖ

Total mechanical energy is:

E_total = −GMm/(2r)

Since total energy is negative, the satellite is gravitationally bound.

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Energy at Escape Speed

At escape speed:

E_total = 0

Therefore:

1/2 mv_escape² − GMm/r = 0

This is why:

v_escape = √(2GM/r)

Circular orbit has negative total energy.

Minimum escape has zero total energy.

This energy difference explains the factor of √2 between the two speeds.

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Circular Orbit Versus Escape

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What Happens at Different Launch Speeds?

Imagine an object launched horizontally from high above a planet where atmospheric effects can be ignored.

Low Speed

The object follows a curved path and eventually strikes the planet.

Correct Circular-Orbit Speed

The object's curvature matches the curvature needed for a circular orbit.

It remains at constant r.

Faster Than Circular Speed but Below Escape Speed

The object can enter an elliptical orbit.

Escape Speed

The object is at the threshold of escaping indefinitely.

Greater Than Escape Speed

The object escapes with kinetic energy remaining at infinity.

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Elliptical Orbits

The equation:

v = √(GM/r)

applies specifically to circular orbital speed.

In an elliptical orbit, the object's speed changes.

Near the central body:

• gravitational potential energy is lower
• kinetic energy is higher
• orbital speed is greater

Farther away:

• gravitational potential energy is higher
• kinetic energy is lower
• orbital speed is smaller

Therefore, an elliptical orbit cannot be described using one constant orbital speed.

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The Vis-Viva Equation

For a general elliptical orbit, orbital speed can be calculated using the more general vis-viva equation:

v² = GM(2/r − 1/a)

where:

• r = current distance from the central body
• a = semi-major axis of the orbit

For a circular orbit:

a = r

so:

v² = GM(2/r − 1/r)

which gives:

v² = GM/r

Therefore, the familiar circular-orbit equation is a special case of the more general orbital relationship.

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Speed in an Elliptical Orbit

At periapsis, the satellite is closest to the central body.

Its speed is greatest.

At apoapsis, the satellite is farthest away.

Its speed is smallest.

This change in speed follows from conservation of both energy and angular momentum.

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Changing Orbits

To change a satellite's orbit, spacecraft engines change its velocity.

A short engine firing is often called a burn.

Increasing velocity does not simply make the spacecraft stay at the same position while moving faster.

Instead, changing velocity changes the shape and size of its orbit.

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Speeding Up to Move Higher

An interesting feature of orbital mechanics is that a spacecraft may initially speed up in order to reach a higher orbit.

The engine adds energy.

The spacecraft then climbs outward:

• GPE increases
• some kinetic energy is converted into GPE
• its speed decreases as it reaches the higher part of the transfer orbit

A second engine burn can then establish a new circular orbit.

This is why orbital mechanics can seem counterintuitive.

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Higher Orbit, Lower Final Speed

Suppose two satellites are in circular orbits.

Satellite A:

r

Satellite B:

4r

Then:

vB/vA = √(r/4r)

= 1/2

So the satellite at 4r moves at:

half the orbital speed

of the satellite at r.

However, the higher satellite has greater total mechanical energy because its total energy is less negative.

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Orbital Speed and Gravitational Field Strength

As altitude increases:

• gravitational field strength decreases
• required centripetal acceleration decreases
• orbital speed decreases

But the satellite must also travel around a larger orbit.

Both effects contribute to longer orbital periods at greater distances.

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What Happens if a Satellite Slows Down?

Suppose a satellite in a circular orbit suddenly loses some speed.

It no longer has enough speed to remain in that circular orbit.

Gravity curves its path more strongly than required for the original circle.

The satellite enters a new orbit with a lower point.

If enough speed is lost, the trajectory may intersect the atmosphere or the planet.

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What Happens if a Satellite Speeds Up?

If a satellite receives a forward velocity increase:

• kinetic energy increases
• total mechanical energy increases
• its original circular orbit changes
• the opposite side of the new orbit generally rises

If enough energy is added:

E_total reaches zero

and the spacecraft can escape.

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Orbital Velocity Is a Vector

The term orbital speed describes the magnitude of the velocity.

Orbital velocity also includes direction.

For a circular orbit:

• magnitude is constant
• direction changes continuously

Therefore, the velocity vector continuously changes even though the speed remains constant.

This changing velocity means the satellite is accelerating.

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Centripetal Acceleration

For circular motion:

a_c = v²/r

For an orbit:

v² = GM/r

Therefore:

a_c = GM/r²

But:

g = GM/r²

So:

a_c = g

The gravitational acceleration is exactly the centripetal acceleration required to maintain the circular orbit.

This is the fundamental connection between gravity and orbital motion.

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Worked Example: Centripetal Acceleration

A satellite orbits with:

v = 7000 m/s

r = 8.0 × 10⁶ m

Then:

a_c = v²/r

= 7000²/(8.0 × 10⁶)

= 49 × 10⁶/(8.0 × 10⁶)

≈ 6.1 m/s²

The gravitational field strength at that orbital radius must therefore be approximately:

6.1 N/kg

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Why Astronauts Feel Weightless

Astronauts in orbit are not beyond gravity.

Instead:

• spacecraft accelerates toward Earth
• astronauts accelerate toward Earth
• equipment accelerates toward Earth

Everything is falling together.

Because there is little normal contact force supporting the astronauts, they experience apparent weightlessness or microgravity.

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Real Satellite Orbits

Real orbital motion can be affected by:

• atmospheric drag
• Earth's non-perfectly spherical shape
• the Moon
• the Sun
• other planets
• radiation pressure
• engine burns

The circular-orbit equation is therefore an idealized model.

However, it provides an excellent starting point for understanding orbital mechanics.

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Orbital Decay

Low-orbiting satellites may experience small amounts of atmospheric drag.

Drag removes mechanical energy.

As total orbital energy decreases:

• the orbit becomes lower
• the satellite eventually moves into denser atmosphere
• drag increases further

Without correction, the satellite may eventually re-enter the atmosphere.

Interestingly, a satellite moving into a lower circular orbit ultimately has a higher orbital speed, even though drag initially removes energy.

This happens because gravitational potential energy decreases as the orbit lowers.

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A Useful Calculation Strategy

For circular-orbit problems:

Identify the central mass M.

Then determine the orbital radius:

r = planetary radius + altitude

Use:

v = √(GM/r)

If the problem gives gravitational field strength at the orbital position, use:

v = √(gr)

For orbital period:

T = 2πr/v

For escape speed at the same location:

v_escape = √2 v_orbit

Always check whether the problem asks for orbital radius or altitude.

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Common Mistakes

Thinking gravity and centripetal force are two separate forces

Gravity provides the centripetal force.

Using altitude as r

Orbital radius is measured from the centre of the planet:

r = R + h

Thinking a satellite has no acceleration because its speed is constant

Velocity changes direction, so the satellite accelerates.

Thinking astronauts are weightless because gravity is absent

They experience apparent weightlessness because they are in continuous free fall.

Thinking higher satellites move faster

For circular orbits around the same body:

greater r → lower orbital speed

Thinking a larger satellite needs a higher orbital speed

Satellite mass cancels from the orbital-speed equation.

Using the circular-orbit equation for every orbit

Elliptical orbits have changing speeds.

Confusing orbital speed with escape speed

At the same radius:

v_escape = √2 v_orbit

Thinking a satellite needs continuous thrust to remain in an ideal orbit

An ideal satellite can remain in orbit under gravity alone.

Thinking gravity does work in a circular orbit

Gravity is perpendicular to instantaneous displacement, so it does no work in the ideal circular case.

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Key Terms

Orbital velocity — velocity of an object moving along an orbit around another body.

Orbital speed — magnitude of orbital velocity.

Orbit — curved gravitational trajectory of one object around another.

Circular orbit — orbit with constant distance from the central body.

Orbital radius (r) — distance from the centre of the central body to the orbiting object.

Centripetal force — net inward force required for circular motion.

Centripetal acceleration — inward acceleration of an object undergoing circular motion.

Tangential velocity — velocity directed tangent to a circular path.

Orbital period (T) — time required to complete one orbit.

Geostationary orbit — equatorial circular orbit in which a satellite remains above approximately the same point on Earth's surface.

Elliptical orbit — orbit in which distance from the central body changes.

Periapsis — closest point of an orbit to the central body.

Apoapsis — farthest point of an orbit from the central body.

Vis-viva equation — equation relating orbital speed, current radius, and semi-major axis for an orbit.

Escape speed — minimum idealized initial speed required to escape indefinitely without further propulsion.

Microgravity — condition of very small apparent weight experienced by freely falling objects in orbit.

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Key Takeaways

• Orbital velocity is the velocity required for an object to follow an orbit under gravity.
• A satellite in orbit is continuously falling toward the central body.
• In a circular orbit, gravity provides the required centripetal force.
• Setting gravitational force equal to centripetal force gives v_orbit = √(GM/r).
• Orbital speed does not depend on satellite mass.
• Orbital radius is measured from the centre of the central body.
• Higher circular orbits have lower orbital speeds.
• Lower circular orbits have higher orbital speeds.
• Circular orbital velocity is tangential while gravitational force is radial.
• Gravity does no work in an ideal circular orbit because force and displacement are perpendicular.
• Gravitational acceleration provides the required centripetal acceleration: g = v²/r.
• Orbital period increases strongly with orbital radius.
• For circular orbits, T² ∝ r³.
• Elliptical orbits have changing orbital speeds.
• Objects move fastest near periapsis and slowest near apoapsis.
• Circular orbital speed is v_orbit = √(GM/r).
• Escape speed at the same radius is v_escape = √(2GM/r).
• Therefore, v_escape = √2 v_orbit.
• A circular orbit has negative total mechanical energy and is gravitationally bound.
• Increasing or decreasing a satellite's velocity changes its orbit.
• Orbital mechanics connects gravitational force, circular motion, energy conservation, and gravitational potential.

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Check Your Understanding

1. Define orbital velocity and explain how it differs from orbital speed.
2. Explain why a satellite in a circular orbit is described as being in continuous free fall.
3. Explain why gravity provides the centripetal force for an orbiting satellite.
4. Starting with GMm/r² = mv²/r, derive v = √(GM/r).
5. Explain why the mass of the satellite does not affect its circular orbital speed.
6. A satellite orbits 500 km above Earth. Earth's radius is 6370 km. Calculate the satellite's orbital radius in metres.
7. Using the radius from Question 6, calculate its orbital speed. Use M_Earth = 5.97 × 10²⁴ kg and G = 6.67 × 10⁻¹¹ N·m²/kg².
8. A satellite orbits at radius r with speed v. Another satellite orbits the same planet at radius 4r. Determine its circular orbital speed in terms of v.
9. Explain why satellites in higher circular orbits move more slowly than satellites in lower circular orbits.
10. A satellite travels at 6.0 km/s in a circular orbit. Calculate the escape speed at the same orbital radius.
11. Explain why gravity does no work on a satellite in an ideal circular orbit even though gravity continuously acts on it.
12. A satellite has orbital radius 8.0 × 10⁶ m and orbital speed 7.0 × 10³ m/s. Calculate its centripetal acceleration and therefore the gravitational field strength at that location.
13. Explain how kinetic energy, gravitational potential energy, and total mechanical energy compare for a satellite in a circular orbit.
14. Describe what happens to the orbit of a satellite if its forward speed is suddenly increased but remains below the local escape speed.
15. Compare circular orbital speed and escape speed at the same radius, and explain the difference using total mechanical energy.