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Circular Motion overview

Topic 5 of 5

Circular orbits and geostationary satellites

In a circular satellite orbit, gravity supplies the inward resultant. Use distance from the central body's centre, then calculate the speed and period required at that radius.

Newtonian gravitation gives GMm/r2 outside a spherical central body. Uniform circular motion needs inward resultant mv2/r. Model the satellite mass m as small compared with the central mass M, so the central body can be treated as fixed in an approximately inertial frame. Neglect drag, thrust and other gravitational sources.

On this page: calculate a low orbit; explain a geostationary orbit.

Gravity is the inward force

GMm/r2 = mv2/r
v2 = GM/r
v = √(GM/r)

The satellite mass cancels. At a fixed radius, changing that small mass changes the gravitational force but not the required orbital speed. Gravity is not balanced by an extra outward force in this description; it produces the inward acceleration.

Substitute v = 2πr/T, where T is time for one revolution:

GMm/r2 = mr(2π/T)2
T2 = 4π2r3/(GM)

For fixed central mass, a larger circular orbit has a lower speed but a longer period. The relation v = rω does not imply that larger freely orbiting radii have greater speed: their angular speeds are not held fixed.

Estimate an orbital speed

Use supplied rounded scales GM about 4 × 1014 m3/s2 and low-orbit centre distance r about 7 × 106 m. The square root of GM/r is about 8 × 103 m/s, or about 8 km/s. This checks the scale before exact substitution; it is not a universal speed for all satellites.

Worked circular orbit

400 km altitude is not a 400 km orbital radius

Use G = 6.67 × 10-11 N m2 kg-2, M = 5.97 × 1024 kg and Earth radius R = 6.37 × 106 m. Then GM = 3.98199 × 1014 m3/s2. The satellite is h = 4.00 × 105 m above the surface.

r = R + h
= 6.37 × 106 + 4.00 × 105
= 6.77 × 106 m

Measure the orbit from Earth's centre

The altitude is enlarged in this schematic drawing so its reference points remain clear. It is not a distance scale: the worked low orbit has r/R about 1.063.

Orbital radius is Earth radius plus altitudeAn explicitly schematic Earth and circular orbit share centre O. At the rightmost point, an 800-kilogram satellite has a blue upward tangent velocity arrow, corresponding to anticlockwise motion. Its single physical force arrow is purple and points left towards Earth's centre: gravity. No outward balancing force is drawn. Projected dimension lines distinguish Earth radius R from centre to surface, altitude h from surface to satellite, and total orbital radius r equals R plus h. Altitude is deliberately exaggerated; the drawing is not to distance scale. In the worked model R is 6.37 million metres and h is 400 thousand metres, so r is 6.77 million metres.EarthOvGravityAltitude exaggerated for clarityRhr = R + h

R = 6.37 × 106 m and h = 4.00 × 105 m give r = 6.77 × 106 m. In this ideal model, gravity alone supplies the inward resultant. The satellite remains in free fall with substantial gravitational acceleration.

The schematic separates Earth radius R, altitude h and centre distance r = R + h. At the rightmost position, velocity is tangent and gravity points towards the centre. Gravity is the one physical force in this ideal orbit model.
v = √[(3.98199 × 1014)/(6.77 × 106)]
= 7669.30 m/s ≈ 7.67 km/s
T = 2πr/v = 5546.42 s ≈ 92.4 min

Keep extra digits when calculating the period from the speed, then round the final answer.

a = v2/r = GM/r2
= 8.68806 m/s2
For m = 800 kg:
F = ma = 6950.45 N ≈ 6.95 × 103 N

The acceleration also agrees with 4π2r/T2. The local field is nearly 89% of the model surface value, so the absence of the usual support sensation in orbit cannot be explained by negligible gravity.

Optional check In the fixed-Earth circular-orbit model, a satellite 400 km above the surface has speed 7.67 km/s and period 92.4 min. What changes if its mass doubles while the orbit radius remains the same?
In the fixed-Earth circular-orbit model, a satellite 400 km above the surface has speed 7.67 km/s and period 92.4 min. What changes if its mass doubles while the orbit radius remains the same?

All the geostationary conditions matter

A geostationary satellite stays above the same equatorial ground location in the ideal model. Its orbit must be:

  • Circular.
  • In Earth's equatorial plane.
  • In the same rotational sense as Earth, from west to east.
  • Of the same period as Earth's rotation.

Matching the period alone does not establish a fixed apparent position. An inclined orbit changes the satellite's apparent location even if it has the correct period.

Keep the same angular position relative to Earth

The ideal satellite has a circular equatorial orbit, the same period as Earth's rotation and the same west-to-east sense. The two diagrams are schematic in distance.

View from above the north pole: both advance 60°

Ground marker and satellite remain on one radial line at both timesViewed from above Earth's north pole, the equatorial ground marker and satellite both move anticlockwise. Hollow markers G0 and S0 show the initial ground and satellite positions on the rightward radial line. Filled markers G1 and S1 show them one sixth of a rotation period later, each sixty degrees anticlockwise from its initial angle. They still share a radial line. A small curved Earth arrow and the highlighted satellite arc show the same rotation sense. The Earth and orbit are true circles in this view, but their relative radii are schematic rather than a physical distance scale. Equal angular speed does not imply equal tangential speed.After T/6: 60° anticlockwiseEarthG0G1S0S160°Hollow: initially. Filled: T/6 later.

G0 and G1 are the same ground location at two times; S0 and S1 are the same satellite. Both angular displacements are 60° in T/6, so the satellite remains above that equatorial location. It is still moving in an Earth-centred inertial description.

Matching period alone is not enough

A tilted orbital plane fails the geostationary conditionA side view shows Earth's rotation axis vertically through its centre. The equatorial orbital plane is horizontal and perpendicular to that axis. A contrasting tilted plane also passes through the centre but is inclined. The viewing direction is along the planes' common line of intersection, so the planes are seen edge-on as straight lines, not as flattened circular paths. One schematic satellite position is marked in each alternative plane. These are alternative orbit models, not two moments of one equatorial orbit. A tilted orbit with the same period still changes latitude and does not stay above one ground location. Distances and the illustrative tilt are not numerical model inputs.Rotation axisNorthTiltedEquatorialEarthSide view: planes are seen edge-on.

The orbit must be equatorial, circular and in Earth's rotation sense, as well as having the matching period. A tilted orbit with that period is not geostationary.

Viewed from above the north pole, Earth and a geostationary satellite advance by the same angle in the same time. The second panel distinguishes the equatorial plane from an inclined orbit; equal period alone does not make the inclined orbit geostationary.

The satellite has the same angular speed as the ground marker, not the same tangential speed. It is farther from the rotation axis. It appears stationary to the ground observer but still moves and accelerates in an Earth-centred inertial description.

A free circular orbit must have its inward force directed towards Earth's centre. A circle parallel to the equator but displaced above or below it would require an inward direction towards a different circle centre. Central gravity alone cannot maintain that proposed fixed-latitude circle.

Worked geostationary radius

Use the supplied rotation period

Earth's sidereal rotation period is about 23 h 56 min 4 s = 86164 s. The usual school description of 24 hours is rounded; use a more precise value when it is supplied.

With the same model G and M as above, rearrange the period equation:

r = [GMT2/(4π2)]1/3
= 4.214998 × 107 m
≈ 4.21 × 107 m

This is centre distance. Subtract Earth radius to obtain altitude:

h = r - R = 3.577998 × 107 m
≈ 3.58 × 107 m
v = 2πr/T = 3.07363 × 103 m/s
≈ 3.07 km/s

If a problem instead supplies T = 86400 s, the same model gives r = 4.222691 × 107 m. Keep one supplied period consistently throughout a calculation rather than mixing rounded and more precise values.

Connect the position to the use

Communication: a fixed apparent satellite position lets a ground antenna point in a fixed direction. Weather observation: a continuing view of the same broad region allows changing cloud and weather patterns to be followed over time.

A geostationary satellite is not directly overhead at an arbitrary latitude, and its equatorial position does not provide equally useful coverage of all polar regions. Choose the orbit for the intended view and communication geometry, not just for a long period.

Optional check A satellite has Earth's rotation period but follows an inclined circular orbit. Must it remain over one fixed ground location?
A satellite has Earth's rotation period but follows an inclined circular orbit. Must it remain over one fixed ground location?