Physics · Orbital mechanics

Orbital velocity in a circular orbit

Q: How can I check a calculation involving Circular-orbit speed?

Model values: $GM$=4.00×10¹⁴ m³/s², central radius R=6.40×10⁶ m and altitude h=1.60×10⁶ m. Satellite mass is negligible; the orbit lies outside the atmosphere. Add radius and altitude before evaluating the root. Divide the full circumference by speed. v≈7.07 km/s and T≈118.5 min. Using h alone would describe an incorrect orbit inside the central radius. $\frac{v^2}r$=6.25 m/s² agrees with $\frac{GM}{r^2}$=6.25 m/s². Independently, $2\pi\sqrt{\frac{r^3}{GM}}$ gives the same period. For a light satellite, $\mu\approx GM$ may be used. All quantities must belong to the same model and separation.

Q: How do I check units and scale for Circular-orbit speed?

For a light satellite, $\mu\approx GM$ may be used. All quantities must belong to the same model and separation. A tabulated $GM$ is convenient because G and central mass need not be multiplied separately. Use its stated units: km³/s² converts to m³/s² by multiplying by 10⁹. The following parameters are explicitly rounded model values, not claimed precision measurements. At fixed μ, $v\propto r^{-1/2}$ but $T\propto r^{3/2}$. A circular-orbit body farther out moves more slowly and also needs more time for its longer path. Here μ is the gravitational parameter, not reduced mass. r is constant body separation; rm and rM are their distances from the barycentre.

For a light satellite around a dominant central mass, gravity supplies circular acceleration; $\sqrt{\frac{GM}{r}}$ is the required speed.

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Formula

Calculate

v_{orb} = \frac{GM}{r}

v in $\mathrm{m\,s^{-1}}$ ; G in $\mathrm{N\,m^2\,kg^{-2}}$ ; M in kg; r in m.

Variables & units

Variables & units
Symbol	Meaning	Unit
vorb	Circular speed of the light satellite relative to the dominant central body	m/s
$G$	Gravitational constant	N·m²/kg²
$M$	Dominant central-body mass, assuming m≪M	kg
$r$	Centre-to-centre separation; for altitude h above radius R, r=R+h	m

LaTeX

v_{\mathrm{orb}}=\sqrt{\frac{GM}{r}}

QUANTA STUDYquanta-study.de

Two bodies on concentric circular orbits about common center O: M left at radius rM, m right at three times the radius. Tangential velocities are opposite, with magnitude ratio 1:3; separation r is the sum of radii. — Two-body circular orbit around common center of mass O, here M = 3m. Radii and speed magnitudes have ratio 1:3; vrel = √(G(M+m)/r) is relative speed, not each body’s speed.

Circular orbital velocity calculator

Find the orbital speed, orbit radius or period of a satellite on a circular orbit, also for two similar masses circling their centre of mass.

Calculation model

Find

Gravitational parameter μ = GM

μ = G · M of the central body, 398,600 km³/s² for the Earth, 1.327 · 10¹¹ km³/s² for the Sun.

Orbit radius r (from the centre)

Distance from the centre, not the height: for the ISS 6371 km + 400 km = 6771 km.

Filled with example values. Type your own and the result updates straight away.

Working · rearrange for v, substitute, evaluate

1. Rearrange for v
$v=\sqrt{\frac\mu r}$
2. Substitute values (in SI units)
- μ = 398 600 km³/s² = 3.986 · 10¹⁴ m³/s²
- r = 6771 km = 6 771 000 m
$\sqrt{\frac{398600000000000}{6771000}}=7672.59$
3. Result
v = 7673 m/s

Result: Orbital speed v

v = 7673 m/s

With μ = 398 600 km³/s² and r = 6771 km, v = 7673 m/s.

Quick tips

Four times the orbit radius, half the speed: v is proportional to 1/√r. The period grows with r to the power 3/2, so four times the radius gives eight times the period.
The radius is measured from the centre of the Earth: 6371 km plus height. The ISS at 400 km has r ≈ 6771 km and travels at about 7.7 km/s.

Assumptions and limits

Light test body orbiting a stationary dominant central mass, ideal Newtonian circular orbit in its spherical exterior field with r>0 and μ=GM>0. Velocity is purely tangential, gravity the only force; no atmosphere, propulsion or third bodies. Result is a magnitude, not orbital direction or a launch-manoeuvre model.

Derivation & background

An ideal Newtonian circle requires tangential velocity and gravity as the only force. A light satellite has $v=\sqrt{\frac{GM}r}$ , where $r=R+h$ is centre separation. Two moving partners instead have $v_{\mathrm{rel}}=\sqrt{\frac{G(M+m)}r}$ ; their barycentric speeds differ. Ellipses additionally require semimajor axis. At the same separation, escape speed in the loss-free model exceeds circular speed by factor $\sqrt2$ .

Exam blueprint

Validity range

Ideal Newtonian circular orbit at constant separation with tangential velocity and no drag, thrust or third bodies. $\sqrt{\frac{GM}{r}}$ assumes m≪M; generally relative speed is √(G(M+m)/r). r is centre-to-centre separation, not altitude.

Derivation steps

In the test-mass model, gravity alone provides the radial acceleration required for a circular orbit.

1Consider a light satellite with $m\ll M$ outside a spherical central mass.
2Measure centre separation; altitude h above radius R gives $r=R+h$ .
3The purely tangential circular orbit needs radial acceleration $\frac{v^2}r$ .
4Gravity is the only force: $\frac{GMm}{r^2}=\frac{mv^2}r$ .
5Cancel satellite mass and rearrange to $v^2=\frac{GM}r$ .
6The positive root gives speed. For two moving partners use $v_{\mathrm{rel}}=\sqrt{\frac{G(M+m)}r}$ instead.

Rearrangements

Orbital radius

r=\frac{GM}{v_{\mathrm{orb}}^2}

Applies to a circular orbit.

Task variant

A light satellite follows an ideal circle at centre separation $r=7.0\cdot10^6\,\mathrm m$ . Use $G=6.674\cdot10^{-11}\,\mathrm{m^3\,kg^{-1}\,s^{-2}}$ and $M=5.97\cdot10^{24}\,\mathrm{kg}$ . Find tangential speed.

Approach: $v=\sqrt{\frac{GM}r}$ .
Substitution gives approximately $7544.51\,\mathrm{m\,s^{-1}}$ , rounded to 7.54 $\mathrm{km\,s^{-1}}$ .

Check: $\frac{v^2}r=\frac{GM}{r^2}$ . The calculation alone does not specify direction; a circle requires tangential motion.

Common mistakes

Using altitude instead of centre separation, or adding gravity and centripetal force as separate forces.

Use $r=R+h$ . Gravity provides the entire radial force here; the calculated speed also requires tangential velocity.

Exam context

Estimate circular orbital speeds of satellites and planets.

Formula cluster

Orbital mechanics

Related relations: kepler-drittes-gesetz, gravitationsgesetz, zentripetalkraft-kreisbewegung.

T²/a³ = 4π²/[G(M+m)]

Kepler Third Law

F = G·m₁m₂/r²

Newton's Law of Universal Gravitation

Fz = m·v²/r

Centripetal Force (Circular Motion)

Worked example

For a light satellite at $r=7.0\cdot10^{6}\,\mathrm{m}$ with $G=6.674\cdot10^{-11}\,\mathrm{m^3\,kg^{-1}\,s^{-2}}$ and $M=5.97\cdot10^{24}\,\mathrm{kg}$ , $v\approx7544.51\,\mathrm{m\,s^{-1}}$ ≈7.54 $\mathrm{km\,s^{-1}}$ . r is centre-to-centre distance.

Applications

Estimate circular orbital speeds of satellites and planets.

How do you calculate with Circular-orbit speed?

Here is how to work through a typical Circular-orbit speed (vorb = √(GM/r)) task step by step:

1
Task
A light satellite follows an ideal circle at centre separation $r=7.0\cdot10^6\,\mathrm m$ . Use $G=6.674\cdot10^{-11}\,\mathrm{m^3\,kg^{-1}\,s^{-2}}$ and $M=5.97\cdot10^{24}\,\mathrm{kg}$ . Find tangential speed.
Solution path
1. Approach: $v=\sqrt{\frac{GM}r}$ .
2. Substitution gives approximately $7544.51\,\mathrm{m\,s^{-1}}$ , rounded to 7.54 $\mathrm{km\,s^{-1}}$ .
Check: $\frac{v^2}r=\frac{GM}{r^2}$ . The calculation alone does not specify direction; a circle requires tangential motion.

Understand and apply the equation

A circular orbit requires a specific tangential speed: gravity bends the trajectory while separation stays constant. The familiar $\sqrt{\frac{GM}{r}}$ describes a light satellite around a dominant mass. When masses are comparable, both bodies move, so their relative motion must be calculated first.

Why a satellite does not keep moving straight

A satellite needs the correct tangential speed for a circular orbit. Gravity continuously turns its velocity without changing the speed. Distance from the centre is decisive, rather than altitude above the surface. Begin with a light satellite, then distinguish both moving partners in a two-body orbit.[1]

Prior knowledge: circular motion and gravitational force. “Centripetal” means directed toward the centre; here it names gravity’s role rather than a second force.

How do you calculate tangential circular-orbit speed and the correct radius?

v_{\mathrm{orb}}\simeq\sqrt{\frac{GM}{r}}\qquad(m\ll M)

Here v is the light satellite’s speed relative to the dominant central body in the Newtonian circular-orbit approximation.[1]

r measures separation between centres of mass. For a spherical central body of radius R and a satellite at altitude h, $r=R+h$ . The formula applies outside the object where a central field describes its gravity. Velocity is tangential and acceleration points radially inward. Velocity therefore changes direction even though its magnitude stays constant.[1]

View diagram: Larger circular orbits require lower speed

Larger circular orbits require lower speed

QUANTA STUDYquanta-study.de

GM

=4.00·10¹⁴

\mathrm{m^3\,s^{-2}}

from example 1. At r=8 million metres,

v\approx7.071\,\mathrm{km\,s^{-1}}

; at four times that radius,

v\approx3.536\,\mathrm{km\,s^{-1}}

. Each point is a different circle in the same idealised exterior field, not a spontaneous radial flight.[1]

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Gravity supplies circular acceleration

A light satellite of mass m needs centripetal acceleration $\frac{v^2}{r}$ . The only force in the ideal model is gravity $\frac{GMm}{r^2}$ . Equate their magnitudes and cancel satellite mass. Centripetal force names the radial resultant here; it is not an extra force alongside gravity.[1]

\begin{aligned}\frac{mv^2}{r}&=\frac{GMm}{r^2}\\v^2&=\frac{GM}{r}\\v&=\sqrt{\frac{GM}{r}}\end{aligned}

The positive root gives a magnitude. Either orbital direction is possible; the velocity vector must additionally be tangential.[1]

Cancellation of m means only that satellites at equal r need equal speed in the test-mass model. It does not mean a second massive object has no influence in the general two-body problem. In that case the first mass also accelerates appreciably.[2]

Explore further: Two moving bodies and their shared centre of mass

\begin{aligned}&\mu=G(M+m)\\&v_{\mathrm{rel}}=\sqrt{\frac{\mu}{r}}\\&r_m=\frac{M}{M+m}r\\&r_M=\frac{m}{M+m}r\end{aligned}

Here $\mu$ is the gravitational parameter, not reduced mass. r is constant body separation; $r_m$ and $r_M$ are their distances from the barycentre.[2]

\begin{aligned}&v_m=\frac{M}{M+m}v_{\mathrm{rel}}\\&v_M=\frac{m}{M+m}v_{\mathrm{rel}}\end{aligned}

These magnitudes are measured in the stationary barycentric frame; the bodies’ velocity vectors point oppositely.[2]

Relative acceleration is the sum of the two acceleration magnitudes towards each other: $\frac{G(M+m)}{r^2}$ . Relative circular motion has radius r. Each individual body instead circles the barycentre at the same $\omega$ but at its own smaller radius. Do not mix $v_{\mathrm{rel}}$ with $r_m$ in a centripetal equation.[2]

How are gravitational parameter, orbital radius and period related?

\begin{aligned}[\mu]&=\mathrm{m^3/s^2}\\{} [\sqrt{\frac{\mu}{r}}]&=\mathrm{m/s}\\{} r&=\frac{\mu}{v_{\mathrm{rel}}^2}\\{} T&=\frac{2\pi r}{v_{\mathrm{rel}}}=2\pi\sqrt{\frac{r^3}{\mu}}\end{aligned}

For a light satellite, $\mu$ ≈ $GM$ may be used. All quantities must belong to the same model and separation.[1][2]

A tabulated $GM$ is convenient because G and central mass need not be multiplied separately. Use its stated units: $\mathrm{km^3\,s^{-2}}$ converts to $\mathrm{m^3\,s^{-2}}$ by multiplying by 10⁹. The following parameters are explicitly rounded model values, not claimed precision measurements. At fixed $\mu$ , $v\propto r^{-1/2}$ but $T\propto r^{3/2}$ . A circular-orbit body farther out moves more slowly and also needs more time for its longer path.[1]

Use altitude correctly and track both bodies

Light satellite in a rounded central field

Given: Model values: $GM$ =4.00×10¹⁴ $\mathrm{m^3\,s^{-2}}$ , central radius $R=6.40\cdot10^{6}\,\mathrm{m}$ and altitude $h=1.60\cdot10^{6}\,\mathrm{m}$ . Satellite mass is negligible; the orbit lies outside the atmosphere.
Find: Find orbital speed and period.

Approach: For a light satellite in circular orbit use $v=\sqrt{\frac{GM}{r}}$ , $r=R+h$ , and $T=\frac{2\pi r}{v}$ . Add radius and altitude before evaluating the root.
$\begin{aligned}r&=R+h=8.00\times10^6\,\mathrm m\\{} v&=\sqrt{\frac{4.00\times10^{14}}{8.00\times10^6}}\approx7071.07\,\mathrm{m/s}\end{aligned}$
Divide the full circumference by speed.
$T=\frac{2\pi(8.00\times10^6)}{7071.07}\approx7108.61\,\mathrm s$

Result: $v\approx7.07\,\mathrm{km\,s^{-1}}$ and T≈118.5 min. Using h alone would describe an incorrect orbit inside the central radius.

Check: $\frac{v^2}{r}=6.25\,\mathrm{m\,s^{-2}}$ agrees with $\frac{GM}{r^2}=6.25\,\mathrm{m\,s^{-2}}$ . Independently, $T=2\pi\sqrt{\frac{r^3}{GM}}$ gives the same period.

Two bodies of comparable mass

Given: $M=6.00\cdot10^{24}\,\mathrm{kg}$ , $m=2.00\cdot10^{24}\,\mathrm{kg}$ and separation $r=2.00\cdot10^{8}\,\mathrm{m}$ . Use model value $G=6.6743\cdot10^{-11}\,\mathrm{m^3\,kg^{-1}\,s^{-2}}$ ; the bodies are compact relative to r.
Find: Find relative speed and both barycentric speeds.

Approach: For comparable masses use $v_{\mathrm{rel}}=\sqrt{\frac{G(M+m)}r}$ , then distribute speed using barycentric mass factors. The relative orbit uses the sum of masses.
$\begin{aligned}\mu&=6.6743\times10^{-11}(8.00\times10^{24})=5.33944\times10^{14}\,\mathrm{m^3/s^2}\\{} v_{\mathrm{rel}}&=\sqrt{\frac{\mu}{r}}\approx1633.93\,\mathrm{m/s}\end{aligned}$
Split separation and speed using the barycentric factors.
$\begin{aligned}r_m&=1.50\times10^8\,\mathrm m\\{} r_M&=0.50\times10^8\,\mathrm m\\{} v_m&\approx1225.45\,\mathrm{m/s}\\{} v_M&\approx408.48\,\mathrm{m/s}\end{aligned}$

Result: Relative speed is about 1.634 $\mathrm{km\,s^{-1}}$ ; in the barycentric frame the lighter body travels at about 1.225 $\mathrm{km\,s^{-1}}$ and the heavier at 0.4085 $\mathrm{km\,s^{-1}}$ .

Check: Speed magnitudes add to $v_{\mathrm{rel}}$ ; $Mv_M=mv_m$ . For the lighter body, $\frac{v_m^2}{r_m}\approx0.01001\,\mathrm{m\,s^{-2}}$ , also $\frac{GM}{r^2}$ . This checks the correct individual radius.

A circular orbit is not an escape orbit

\begin{aligned}&\varepsilon=\frac{v_{\mathrm{rel}}^2}{2}-\frac{\mu}{r}=-\frac{\mu}{2r}\\&v_{\mathrm{esc,rel}}=\sqrt{\frac{2\mu}{r}}=\sqrt2\,v_{\mathrm{circ,rel}}\end{aligned}

ε is relative orbital energy per reduced mass in the isolated Newtonian two-body system. For a test mass, it is energy per satellite mass.[3][2]

A circular-orbit body is bound: its energy is negative. Escape speed at the same separation is greater and corresponds to zero energy in the loss-free limit. This relation does not give the full rocket launch requirement: central-body rotation, atmosphere, propulsion history and existing velocity belong to a separate mission analysis. Gravitational potential energy must use the same zero at infinity as total energy.[3]

v_{\mathrm{rel}}^2=\mu\left(\frac2r-\frac1a\right)

An ellipse additionally requires semimajor axis a. Only for a circle with $a=r$ does the right-hand side simplify to $\frac\mu r$ .[1][2]

Five exercises with solutions

Rearrange · Foundation

In a central field with $\mu=3.20\cdot10^{14}\,\mathrm{m^3\,s^{-2}}$ , a light satellite has circular speed $v=4.00\,\mathrm{km\,s^{-1}}$ . Find r.

Show hint

First convert v to $\mathrm{m\,s^{-1}}$ , then square.

Learning goal: Reconstruct circular radius from speed and gravitational parameter.

Show solution and calculation

Approach: Rearrange the circular-orbit condition to $r=\frac\mu{v^2}$ , converting kilometres per second to metres per second before squaring.
$r=\frac{\mu}{v^2}=\frac{3.20\cdot10^{14}}{4000^2}\,\mathrm m=2.00\cdot10^7\,\mathrm m$ .
This is centre-to-centre distance, not altitude; altitude would require the central radius.

r=2.00\times10^7\,\mathrm m

Interpret · Application

An ideal circular orbit is changed to four times its radius at fixed $\mu$ . How do circular speed and period change?

Show hint

Compare two circular states; transfer motion is not circular.

Learning goal: Distinguish speed and period scaling for two circular orbits.

Show solution and calculation

Approach: For fixed $\mu$ , $v\propto r^{-1/2}$ , whereas $T\propto r^{3/2}$ .
$\frac{v_2}{v_1}=\frac1{\sqrt4}=\frac12$ .
$\frac{T_2}{T_1}=4^{3/2}=8$ .
Enlargement requires an appropriate orbital manoeuvre; the relation does not imply a satellite spontaneously drifts outward.

\begin{aligned}\frac{v_2}{v_1}&=\frac12\\{} \frac{T_2}{T_1}&=8\end{aligned}

Transfer · Challenge

Two equal-mass bodies have relative circular speed 2.0 $\mathrm{km\,s^{-1}}$ . How fast does each move relative to the barycentre, and what is relative escape speed at the same separation?

Show hint

Equal masses split $v_{\mathrm{rel}}$ into two equal magnitudes.

Learning goal: Separate relative orbital speed, barycentric speed and escape speed.

Show solution and calculation

Approach: Equal masses split relative speed equally: $v_M=v_m=\frac12v_{\mathrm{rel}}$ ; relative escape threshold is $\sqrt2\,v_{\mathrm{circ,rel}}$ .
Each travels at 1.0 $\mathrm{km\,s^{-1}}$ in opposite directions.
Relative escape speed is √2·2.0≈2.83 $\mathrm{km\,s^{-1}}$ ; for symmetric escape in the barycentric frame each would travel at about 1.41 $\mathrm{km\,s^{-1}}$ .

\begin{aligned}v_m&=v_M=1.0\,\mathrm{km/s}\\{} v_{\mathrm{esc,rel}}&\approx2.83\,\mathrm{km/s}\end{aligned}

Interpret · Application

A test body orbits a point mass on a Newtonian circle with $\mu=4.0\cdot10^{14}\,\mathrm{m^3\,s^{-2}}$ and $r=1.0\cdot10^{7}\,\mathrm{m}$ . A free-body diagram shows gravity inward and an equal “centripetal force” outward. Correct the diagram and find acceleration magnitude. Does this acceleration increase the body’s speed along the circular orbit?

Show hint

Centripetal force means the resultant inward force. In an inertial frame it is not an additional force canceling gravity.

Learning goal: Understand gravity as the required centripetal force rather than a balanced opposing force.

Show solution and calculation

Approach: calculate gravitational acceleration at the orbital radius.
It is 4.0 $\mathrm{m\,s^{-2}}$ inward; the correct free-body diagram contains gravity alone.
Gravity is itself the centripetal force.
Velocity changes direction, not magnitude.

Check: $v\approx6325\,\mathrm{m\,s^{-1}}$ gives the same acceleration through $\frac{v^2}r$ ; perpendicular force and velocity imply no instantaneous power transfer.

\begin{aligned}a_r&=\frac\mu{r^2}=4.0\,\mathrm{m\,s^{-2}}\\{}v_{\mathrm{circ}}&=\sqrt{\frac\mu r}\approx6325\,\mathrm{m\,s^{-1}}\\{}\frac{v_{\mathrm{circ}}^2}{r}&=a_r\\{}\vec F\cdot\vec v&=0\end{aligned}

[1]

Error analysis · Challenge

A test body follows an ellipse with semimajor axis $a=1.0\cdot10^{7}\,\mathrm{m}$ and farthest distance $r_a=1.5\cdot10^{7}\,\mathrm{m}$ , with $\mu=3.0\cdot10^{14}\,\mathrm{m^3\,s^{-2}}$ . A solution calls $\sqrt{\frac\mu{r_a}}$ its speed there. Check using the vis-viva equation and compare actual speed with local circular speed.

Show hint

For the ellipse, use $v^2=\mu\left(\frac2r-\frac1a\right)$ . Only a circle has $a=r$ everywhere.

Learning goal: Avoid applying the circular-speed formula to an arbitrary point on an ellipse.

Show solution and calculation

Approach: insert the farthest distance into the elliptic energy relation.
Actual speed is about 3162 $\mathrm{m\,s^{-1}}$ ; local circular speed is about 4472 $\mathrm{m\,s^{-1}}$ .
Their ratio is $\frac1{\sqrt2}$ .

Check: at apocentre the ellipse has less speed than a circle at the same radius. Instantaneous radial acceleration remains gravitational, but local curvature is not that of a circle of radius r_a.

\begin{aligned}v_a^2&=\mu\left(\frac2{r_a}-\frac1a\right)=1.0\cdot10^7\,\mathrm{m^2/s^2}\\v_a&\approx3162\,\mathrm{m/s}\\v_{\mathrm{circ}}&=\sqrt{\frac\mu{r_a}}\approx4472\,\mathrm{m/s}\\\frac{v_a}{v_{\mathrm{circ}}}&=\sqrt{2-\frac{r_a}{a}}=\frac1{\sqrt2}\end{aligned}

[3][1]

Validity, orbital direction and real departures

A circular orbit requires constant separation and precisely matched purely tangential relative velocity. A radial component or different speed generally produces another Kepler orbit. Merely stating “speed is $\sqrt{\frac{GM}{r}}$ ” is insufficient without direction. Higher speed does not automatically mean escape: between circular and escape speed a bound ellipse can result.[1][3]

The model neglects drag, thrust, third bodies and departures from spherical symmetry. Inside an extended mass, $\frac{GM}{r^2}$ using its entire mass is generally wrong; Newton’s formula is also insufficient in strongly relativistic fields. Actual orbit planning therefore needs the appropriate gravitational parameter, coordinate frame and, where relevant, perturbation or relativistic models.[1]

What to remember

For a light satellite, equate gravity with the required radial circular-orbit force. The root gives speed, and velocity must also be tangential. Distinguish altitude from centre separation. With comparable masses, relative speed uses total mass while each partner uses its own barycentric distance.[1]

Sources for this explanation

Quanta · Updated October 2, 2026

Editorial responsibility: Quanta Study editorial team

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This explanation, 2 worked examples and 5 exercises are available in German and English. Structure, source references and mathematical notation undergo automated checks. Calculations, model assumptions and both language versions receive a separate AI review.

An internal human subject review is documented for an earlier release of this page. That confirmation does not apply automatically to the revised explanation and new exercises.

An independent, named human subject review has not yet been documented. Our exercise numbers are worked examples, rather than measurements published by the cited sources.

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Change history

October 2, 2026Highlighted tangential orbital direction, centre separation and period in question headings; added paragraph connections to centripetal force and central-field energy.
October 1, 2026Added two exercises; labelled all five by task type, difficulty and learning goal.
October 1, 2026Added an individual introduction and conclusion, checked existing calculations, clarified notation and model references, and added an original learning figure.

Cite this reference page

Quanta. Circular-orbit speed. Updated October 2, 2026. https://quanta-study.de/en/formula-sheet/kreisbahn-geschwindigkeit

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Frequently asked questions about Circular-orbit speed

What does Circular-orbit speed describe physically?+

Here v is the light satellite’s speed relative to the dominant central body in the Newtonian circular-orbit approximation. r measures separation between centres of mass. For a spherical central body of radius R and a satellite at altitude h, $r=R+h$ . The formula applies outside the object where a central field describes its gravity. Velocity is tangential and acceleration points radially inward. Velocity therefore changes direction even though its magnitude stays constant. Cancellation of m means only that satellites at equal r need equal speed in the test-mass model. It does not mean a second massive object has no influence in the general two-body problem. In that case the first mass also accelerates appreciably.

How can I check a calculation involving Circular-orbit speed?+

Model values: $GM$ =4.00×10¹⁴ m³/s², central radius R=6.40×10⁶ m and altitude h=1.60×10⁶ m. Satellite mass is negligible; the orbit lies outside the atmosphere. Add radius and altitude before evaluating the root. Divide the full circumference by speed. v≈7.07 km/s and T≈118.5 min. Using h alone would describe an incorrect orbit inside the central radius. $\frac{v^2}r$ =6.25 m/s² agrees with $\frac{GM}{r^2}$ =6.25 m/s². Independently, $2\pi\sqrt{\frac{r^3}{GM}}$ gives the same period. For a light satellite, $\mu\approx GM$ may be used. All quantities must belong to the same model and separation.

What assumptions and limits apply to Circular-orbit speed?+

A circular orbit requires constant separation and precisely matched purely tangential relative velocity. A radial component or different speed generally produces another Kepler orbit. Merely stating “speed is $\sqrt{\frac{GM}{r}}$ ” is insufficient without direction. Higher speed does not automatically mean escape: between circular and escape speed a bound ellipse can result. The model neglects drag, thrust, third bodies and departures from spherical symmetry. Inside an extended mass, $\frac{GM}{r^2}$ using its entire mass is generally wrong; Newton’s formula is also insufficient in strongly relativistic fields. Actual orbit planning therefore needs the appropriate gravitational parameter, coordinate frame and, where relevant, perturbation or relativistic models.

How do I check units and scale for Circular-orbit speed?+

For a light satellite, $\mu\approx GM$ may be used. All quantities must belong to the same model and separation. A tabulated $GM$ is convenient because G and central mass need not be multiplied separately. Use its stated units: km³/s² converts to m³/s² by multiplying by 10⁹. The following parameters are explicitly rounded model values, not claimed precision measurements. At fixed μ, $v\propto r^{-1/2}$ but $T\propto r^{3/2}$ . A circular-orbit body farther out moves more slowly and also needs more time for its longer path. Here μ is the gravitational parameter, not reduced mass. r is constant body separation; rm and rM are their distances from the barycentre.

When should I combine Circular-orbit speed with another relation?+

ε is relative orbital energy per reduced mass in the isolated Newtonian two-body system. For the circular orbit, $\varepsilon=\frac{v_{\mathrm{rel}}^2}2-\frac\mu r=-\frac\mu{2r}$ , where $\mu=G(M+m)$ . For a test mass, it is energy per satellite mass. A circular-orbit body is bound: its energy is negative. Escape speed at the same separation is greater and corresponds to zero energy in the loss-free limit. This relation does not give the full rocket launch requirement: central-body rotation, atmosphere, propulsion history and existing velocity belong to a separate mission analysis. An elliptical Kepler orbit obeys $v_{\mathrm{rel}}^2=\mu\left(\frac2r-\frac1a\right)$ , with semimajor axis a; setting $a=r$ gives $v_{\mathrm{rel}}^2=\frac\mu r$ .

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Related formulas

v^2=\mu\left(\frac{2}{r}-\frac{1}{a}\right)

Vis-Viva Equation for Elliptical Orbits

More formulas: Mechanics

\sum \vec F_{\mathrm{ext}} = \vec F_{\mathrm{res}} = m \vec a

Newton's Second Law

E_{kin} = \frac{1}{2} m v^2

Kinetic Energy

E_{pot} = m \cdot g \cdot h

Potential Energy (Gravitational)

F = k \cdot x

Hooke's Law (Spring Force)

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Create a curated FSRS exam set for vorb = √(GM/r): formula recall, variables, derivation, rearrangement, worked example, common mistakes and exam context.

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Orbital velocity in a circular orbit

Variables & units

Circular orbital velocity calculator

Quick tips

Derivation & background

Exam blueprint

Validity range

Derivation steps

Rearrangements

Task variant

Common mistakes

Exam context

Orbital mechanics

Worked example

Applications

How do you calculate with Circular-orbit speed?

Why a satellite does not keep moving straight

How do you calculate tangential circular-orbit speed and the correct radius?

Larger circular orbits require lower speed

Gravity supplies circular acceleration

Explore further: Two moving bodies and their shared centre of mass

How are gravitational parameter, orbital radius and period related?

Use altitude correctly and track both bodies

Light satellite in a rounded central field

Two bodies of comparable mass

A circular orbit is not an escape orbit

Five exercises with solutions

In a central field with μ=3.20⋅1014 m3 s−2\mu=3.20\cdot10^{14}\,\mathrm{m^3\,s^{-2}}μ=3.20⋅1014m3s−2, a light satellite has circular speed v=4.00 km s−1v=4.00\,\mathrm{km\,s^{-1}}v=4.00kms−1. Find r.

An ideal circular orbit is changed to four times its radius at fixed μ\muμ. How do circular speed and period change?

Two equal-mass bodies have relative circular speed 2.0 km s−1\mathrm{km\,s^{-1}}kms−1. How fast does each move relative to the barycentre, and what is relative escape speed at the same separation?

Validity, orbital direction and real departures

What to remember

Sources for this explanation

Frequently asked questions about Circular-orbit speed

Subject references and further reading

Related formulas

More formulas: Mechanics

Retain Circular-orbit speed for exams

In a central field with $\mu=3.20\cdot10^{14}\,\mathrm{m^3\,s^{-2}}$ , a light satellite has circular speed $v=4.00\,\mathrm{km\,s^{-1}}$ . Find r.

An ideal circular orbit is changed to four times its radius at fixed $\mu$ . How do circular speed and period change?

Two equal-mass bodies have relative circular speed 2.0 $\mathrm{km\,s^{-1}}$ . How fast does each move relative to the barycentre, and what is relative escape speed at the same separation?