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4.3 · Analyze uniform circular orbits

Learn to analyze uniform circular orbits through clear examples and targeted practice.

University of Alberta EN PH 131: Engineering Mechanics: Dynamics

Gravitation and Orbital Motion

How gravity supplies the inward acceleration

A uniform circular orbit is motion along a circular path at constant speed. The word uniform describes the speed, not the velocity: as the particle travels, its velocity direction changes, so it accelerates. In the basic model used here, a spherical central body attracts an orbiting particle, and gravity supplies the inward force required for circular motion. We treat the orbiting body as a particle and use a frame centred on the central body that does not rotate. These assumptions let us analyse the orbit using Newton's law of gravitation and Newton's second law.

What you will learn

  • Describe a circular orbit using a defined system, observer, reference frame, radius, and positive direction.
  • Explain why constant speed on a circular path still involves acceleration.
  • Use Newton's law of gravitation and Newton's second law to find orbital speed, period, radius, or central-body mass.
  • Check orbital results for consistent units, directions, assumptions, and limiting behaviour.

1. Describe the circular motion

Let the central body's mass be MM and the orbiting particle's mass be mm. The system is the orbiting particle. An observer uses a non-rotating reference frame whose origin is at the central body's centre; for this introductory analysis, treat the frame as inertial. The orbital radius rr is the constant distance from that origin to the particle. If a problem gives the particle's height hh above a spherical body's surface of radius RR, then the orbital radius is r=R+hr=R+h.
Choose radial outward as the positive radial direction. At any point on the path, velocity is tangent to the circle. Acceleration points toward the centre, so its radial component is negative with this sign convention. Its inward magnitude is v2/rv^2/r, where vv is the particle's constant speed. Thus, in signed radial form, ar=−v2/ra_r=-v^2/r.
Position is the particle's location relative to the chosen origin. Displacement is the vector from its initial position to its final position. Path length is the distance travelled along the orbit: one full revolution covers 2πr2\pi r, even though the displacement after that revolution is zero. Velocity is the rate of change of position, while acceleration is the rate of change of velocity. A changing velocity direction explains why acceleration is nonzero even when speed is constant.
ar=−v2ra_r=-\frac{v^2}{r}
  • Measure orbital radius from the central body's centre, not from its surface.
  • Velocity is tangent to the circular path; acceleration points inward.
  • A full revolution has path length 2πr2\pi r and zero net displacement.

2. Match gravity to the required acceleration

For a spherical central body, Newton's law of gravitation gives the attractive force magnitude GMm/r2GMm/r^2, where GG is the universal gravitational constant. The force is inward. With radial outward positive, its radial component is −GMm/r2-GMm/r^2.
Newton's second law in the radial direction states that the net radial force equals mass times radial acceleration. In this model, gravity is the only force on the orbiting particle. Equating the inward force magnitude to the force needed for circular acceleration gives the circular-orbit condition.
The particle's mass cancels from this condition. Therefore, in this ideal model, the speed needed at a particular radius depends on the central mass and radius, not on the orbiting particle's mass. “Centripetal force” is a name for the net inward force required by circular motion; it is not an extra force to add to gravity. The model applies only when the orbit is circular and the speed is constant.
GMmr2=mv2r\frac{GMm}{r^2}=\frac{mv^2}{r}
  • Gravity is the net inward force in the model.
  • Use consistent signs: inward radial force and acceleration are negative when outward is positive.
  • Centripetal force describes the required net inward force; it is not an additional interaction.

3. Find orbital quantities and check them

Rearranging the circular-orbit condition gives the speed. One revolution covers a circumference at constant speed, so period—the time for one revolution—is circumference divided by speed. These relationships let you solve for speed or period when the central mass and radius are known. Rearranging them also gives the central mass from a measured period and radius, or the radius from a known speed and central mass.
Check dimensions before trusting a result. The quantity GM/rGM/r has units of speed squared, and the period has units of time. At fixed central mass, a larger radius gives a lower circular speed and a longer period. As radius grows without bound, the required speed tends toward zero and the period grows without bound. These trends agree with weaker gravitational attraction at greater distance.
Use SI units throughout: kilograms for mass, metres for distance, seconds for time, and GG in N m2/kg2\mathrm{N\,m^2/kg^2}. Keep magnitudes and directions distinct. Speed is a positive scalar; velocity is tangent to the path, and gravitational force and acceleration point inward. Before using the model, confirm that the stated motion is a circular orbit at constant speed.
v=GMr,T=2πr3GMv=\sqrt{\frac{GM}{r}},\qquad T=2\pi\sqrt{\frac{r^3}{GM}}
  • Speed and period depend on the central mass and the centre-to-particle radius.
  • A larger circular orbit around the same central body has lower speed and a longer period.
  • Check dimensions, inward direction, and whether the assumptions of the model fit the stated motion.

Worked example

Find a satellite's speed and period

A satellite is modelled as a particle in a uniform circular orbit 7.00×106 m7.00\times10^6\,\mathrm{m} from a planet's centre. The planet's mass is 5.97×1024 kg5.97\times10^{24}\,\mathrm{kg}. Find the satellite's speed and orbital period. Use G=6.67×10−11 N m2/kg2G=6.67\times10^{-11}\,\mathrm{N\,m^2/kg^2}.
Satellite circular path
Satellite circular pathxyrPlanetv tangenta inwardvelocityacceleration

At the rightmost point of the stated circular path, the counterclockwise velocity is upward and the acceleration is inward.

  1. Define the system and frame
    The system is the satellite, modelled as a particle. Use a non-rotating frame centred on the planet, with radial outward positive. The initial and final positions for one period are the same point after one complete revolution. The given radius is already measured from the planet's centre.
  2. Choose the orbit relationships
    The stated path is circular and the speed is constant. Gravity supplies the inward net force, so the circular-orbit relation gives the speed. The circumference divided by that speed gives the period.
    v=GMr,T=2πrvv=\sqrt{\frac{GM}{r}},\qquad T=\frac{2\pi r}{v}
  3. Substitute the values
    Substitute the given SI values. The speed is a magnitude; at the illustrated point the velocity is tangent to the path, while acceleration points inward.
    v=(6.67×10−11)(5.97×1024)7.00×106=7.54×103 m/s,T=2π(7.00×106)7.54×103=5.83×103 sv=\sqrt{\frac{(6.67\times10^{-11})(5.97\times10^{24})}{7.00\times10^6}}=7.54\times10^3\,\mathrm{m/s},\qquad T=\frac{2\pi(7.00\times10^6)}{7.54\times10^3}=5.83\times10^3\,\mathrm{s}
  4. Check units and direction
    The period is about 97.297.2 minutes. The inward acceleration calculated from circular motion agrees with the gravitational acceleration at this radius.
    v2r=GMr2≈8.12 m/s2\frac{v^2}{r}=\frac{GM}{r^2}\approx8.12\,\mathrm{m/s^2}
Answer: The orbital speed is approximately 7.54×103 m/s7.54\times10^3\,\mathrm{m/s} and the period is approximately 5.83×103 s5.83\times10^3\,\mathrm{s}, or 97.297.2 minutes.
Check: Both force and acceleration are inward. The acceleration check agrees with gravity, and the period is positive and has units of seconds.

Worked example

Estimate a planet's mass from a probe's orbit

A small probe moves in a uniform circular orbit of radius 1.20×107 m1.20\times10^7\,\mathrm{m} with a measured period of 1.80×104 s1.80\times10^4\,\mathrm{s}. Estimate the planet's mass using G=6.67×10−11 N m2/kg2G=6.67\times10^{-11}\,\mathrm{N\,m^2/kg^2}.
Probe circular path
Probe circular pathxyrPlanetv tangenta inwardvelocityacceleration

The probe follows a stated circular path; at the rightmost point its velocity is tangent and its acceleration is inward.

  1. Define the system and knowns
    The system is the probe. Use a non-rotating frame centred on the planet, with radial outward positive. The initial and final positions are one revolution apart, so the elapsed time is the period. The radius is the centre-to-probe distance.
    r=1.20×107 m,T=1.80×104 sr=1.20\times10^7\,\mathrm{m},\qquad T=1.80\times10^4\,\mathrm{s}
  2. Relate period to central mass
    The probe travels one circumference in one period at constant speed. Combining this with the gravity-based circular-orbit condition eliminates the speed and gives the planet's mass.
    M=4π2r3GT2M=\frac{4\pi^2r^3}{GT^2}
  3. Calculate the mass
    Substitute the radius, period, and gravitational constant in SI units. Keep the radius cubed and the period squared as required by the rearranged relationship.
    M=4π2(1.20×107)3(6.67×10−11)(1.80×104)2=3.16×1024 kgM=\frac{4\pi^2(1.20\times10^7)^3}{(6.67\times10^{-11})(1.80\times10^4)^2}=3.16\times10^{24}\,\mathrm{kg}
  4. Verify the result
    The mass equation reduces to kilograms. The corresponding speed and inward acceleration give the same acceleration as gravity at the stated radius.
    v=2πrT≈4.19×103 m/s,GMr2=v2r≈1.46 m/s2v=\frac{2\pi r}{T}\approx4.19\times10^3\,\mathrm{m/s},\qquad \frac{GM}{r^2}=\frac{v^2}{r}\approx1.46\,\mathrm{m/s^2}
Answer: The estimated planet mass is approximately 3.16×1024 kg3.16\times10^{24}\,\mathrm{kg}.
Check: The mass is positive, the dimensions reduce to kilograms, and the gravitational and circular-motion accelerations match and point inward.

Worked example

Find the radius from orbital speed

A particle orbits a spherical body of mass 4.00×1024 kg4.00\times10^{24}\,\mathrm{kg} at constant speed 5.00×103 m/s5.00\times10^3\,\mathrm{m/s}. Find the radius of the uniform circular orbit and its period. Use G=6.67×10−11 N m2/kg2G=6.67\times10^{-11}\,\mathrm{N\,m^2/kg^2}.
Inward gravity on particle
Inward gravity on particleradial outwardtangentialOrbiting particlegravity inward

The gravitational force acts inward; the orbit is circular as stated in the problem.

  1. Set the system and coordinates
    The system is the orbiting particle, observed in a non-rotating frame centred on the spherical body. Choose radial outward as positive. The initial and final states for the period are the same position after one complete revolution. The speed is constant.
    M=4.00×1024 kg,v=5.00×103 m/sM=4.00\times10^{24}\,\mathrm{kg},\qquad v=5.00\times10^3\,\mathrm{m/s}
  2. Solve for the radius
    Gravity supplies the inward acceleration for the circular orbit. Rearranging the circular-orbit condition gives the centre-to-particle radius from the known speed and central mass.
    r=GMv2r=\frac{GM}{v^2}
  3. Calculate radius and period
    First find the radius. Then divide the circular path length by the constant speed to find the time for one revolution.
    r=(6.67×10−11)(4.00×1024)(5.00×103)2=1.07×107 m,T=2πrv=1.34×104 sr=\frac{(6.67\times10^{-11})(4.00\times10^{24})}{(5.00\times10^3)^2}=1.07\times10^7\,\mathrm{m},\qquad T=\frac{2\pi r}{v}=1.34\times10^4\,\mathrm{s}
  4. Check the acceleration
    The radius has units of metres and the period has units of seconds. The calculated inward acceleration from the circular motion agrees with the gravitational acceleration.
    v2r=GMr2≈2.34 m/s2\frac{v^2}{r}=\frac{GM}{r^2}\approx2.34\,\mathrm{m/s^2}
Answer: The required orbital radius is approximately 1.07×107 m1.07\times10^7\,\mathrm{m} from the body's centre, and the period is approximately 1.34×104 s1.34\times10^4\,\mathrm{s}.
Check: The radius and period are positive. The gravitational acceleration and the acceleration required for circular motion agree and point inward.

Common mistakes and how to avoid them

Treating constant speed as zero acceleration.
Correction: Velocity changes direction continuously around the orbit, so acceleration is nonzero and points inward.
Adding a separate centripetal force to gravity.
Correction: Centripetal force describes the net inward force required for circular motion. In this model, gravity supplies that force.
Using height above the surface as the orbital radius.
Correction: Use the distance from the central body's centre. If height is given, add the body's radius.
Giving inward force or acceleration a positive radial sign when outward is defined as positive.
Correction: Their radial components are negative under that convention, although their magnitudes are positive.

Lesson summary

  • Uniform circular motion has constant speed but changing velocity, so it has inward acceleration.
  • For a circular orbit under gravity alone, gravitational attraction supplies the required inward force.
  • Use the radius measured from the central body's centre and maintain consistent SI units.
  • Check dimensions, directions, and agreement between gravitational and circular-motion acceleration.

Check your understanding

Question 1

A particle moves at constant speed in a circular orbit. Which statement about its acceleration is correct?
  1. It is zero because the speed is constant.
  2. It points tangent to the path in the direction of travel.
  3. It points toward the centre of the orbit.
  4. It points away from the centre because the particle is moving outward.
Show answer and explanation
It points toward the centre of the orbit.
The velocity direction changes continuously, so the particle accelerates inward even though its speed is constant.

Question 2

For a circular orbit under gravity alone, what supplies the required inward force?
  1. A second force in addition to gravity.
  2. The inward gravitational force.
  3. The particle's velocity.
  4. An outward force equal to gravity.
Show answer and explanation
The inward gravitational force.
Gravity is the net inward force in this model and supplies the acceleration required for circular motion.

Question 3

For the same central mass, what happens to circular-orbit speed as orbit radius increases?
  1. It increases.
  2. It stays constant.
  3. It decreases.
  4. It becomes directed radially outward.
Show answer and explanation
It decreases.
The circular-orbit speed varies as the inverse square root of radius, so a larger orbit requires a lower speed.

Key terms

Uniform circular orbit
Motion along a circular path at constant speed.
Orbital radius
The distance from the central body's centre to the orbiting particle.
Period
The time required to complete one full revolution.
Centripetal acceleration
The inward acceleration required for circular motion, with magnitude equal to speed squared divided by radius.

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