Topic 5 of 5
Circular orbits and geostationary satellites
A circular orbit needs the correct tangential speed at its radius. Gravity supplies the inward acceleration; an energy account then distinguishes reaching that radius from reaching the complete orbital state.
Model a small satellite m outside a spherical central mass M that is effectively fixed. Neglect drag, thrust and other gravitational sources. The radius r is measured from the source centre, not its surface.
On this page: orbital energy; radius-period data; geostationary conditions.
Use gravity as the inward resultant
Uniform circular motion requires inward acceleration v2/r = rω2. Gravity supplies the actual force:
v = √(GM/r)
The small satellite mass cancels. At a fixed radius around the same source, its required speed is independent of its mass, although the gravitational force is proportional to that mass.
As in the radial-force method, centripetal describes the resultant's role, not an extra interaction to add to gravity. There is no outward balancing force in this inertial-frame account. Constant speed still involves changing velocity and inward acceleration.
The period T is time for one complete revolution. Substitute v = 2πr/T into the radial equation:
T2 = 4π2r3/(GM)
T = 2π√[r3/(GM)]
Larger circular orbits around the same source have lower speed and longer period. The relation v = rω does not imply greater speed at larger r here, because the different orbits do not have the same angular speed.
Estimate the orbit scale
For supplied rough scales GM about 4 × 1014 m3/s2 and centre distance r about 8 × 106 m, the speed is about 7 × 103 m/s. The circumference divided by that speed gives a period of about 7 × 103 s, of order 2 hours. These checks precede the more precise calculation.
Worked orbit and energy account
A 400 kg body at centre radius 8.0 × 106 m
Use GM = 4.0 × 1014 m3/s2 and source radius R = 6.4 × 106 m. The orbit altitude is h = r - R = 1.6 × 106 m.
= 7071.07 m/s ≈ 7.07 km/s
T = 2πr/v = 7108.61 s ≈ 1.97 h
The field magnitude there is 6.25 N/kg, so the 400 kg body's gravitational force is 2500 N inward. Its velocity is tangent to the orbit.
A circular orbit needs both the correct inward force and kinetic energy
Use centre radius r = R + h
R = 6.4 × 106 m and h = 1.6 × 106 m give r = 8.0 × 106 m. The circular speed is about 7.07 km/s; this is different from the 10.0 km/s minimum escape speed at that same radius.
The full surface-rest to circular-orbit energy account
The model surface is nonrotating. All bars share one energy scale, in 1010 J. Negative values extend left of zero; positive values extend right. E = K + UG is total mechanical energy.
The mechanical-energy increase between these stated states is 1.50 × 1010 J. This is the energy that must be supplied before allowing for losses. The 5.00 × 109 J potential increase alone omits the final orbital kinetic energy.
Include the final kinetic energy
Using the circular result v2 = GM/r, the final kinetic energy is:
= 1.00 × 1010 J
The potential energy uses zero at infinity:
Emechanical,final = Ek,final + UG,final
= -1.00 × 1010 J
These are consequences of the radial force equation and the potential-energy expression. The total is negative relative to the infinity reference, consistent with a bound orbit.
Starting at rest on the nonrotating model surface, the initial kinetic energy is zero and initial potential energy is -2.50 × 1010 J. Therefore:
= -1.00 × 1010 - (-2.50 × 1010)
= +1.50 × 1010 J
The increase consists of 5.00 × 109 J in potential energy and 1.00 × 1010 J in kinetic energy. A calculation that gives only the potential-energy change misses the required final motion. This is an ideal mechanical-energy change before losses, not a complete propulsion-energy calculation.
At this same radius, circular speed is 7.07 km/s, while minimum escape speed is 10.0 km/s. A maintained circular orbit and an escape to indefinitely large separation have different final conditions.
Optional check A 400 kg body starts at rest on a nonrotating model surface with U_G = -2.50 x 10^10 J. It ends in a circular orbit with U_G = -2.00 x 10^10 J and kinetic energy 1.00 x 10^10 J. What is its mechanical-energy increase before losses?
Test the radius-period relationship
For a common central mass, T2 is proportional to r3. The table gives periods generated from the supplied GM model and rounded to the nearest second. They are model values, not satellite observations.
| Centre radius r / m | Period T / s |
|---|---|
| 8000000 | 7109 |
| 10000000 | 9935 |
| 12000000 | 13059 |
| 16000000 | 20106 |
Plot T2 vertically against r3 horizontally. Before rounding, the exact model is a straight line through the origin with gradient:
= 9.869604401 × 10-14 s2/m3
The gradient represents the inverse of GM multiplied by 4π2. It is not an orbital speed. Rearranging gives GM = 4π2/gradient.
For genuine observations, differences from the circular prediction could reflect uncertainty in radius or period, a noncircular orbit, or other neglected influences. Check those conditions before attributing every mismatch to a measurement zero error.
A geostationary orbit needs more than a matching period
In the ideal model, a geostationary satellite remains above the same equatorial longitude. It has a circular orbit in Earth's equatorial plane, moves west to east in Earth's rotational sense, and has the same period as Earth's rotation.
These conditions give equal angular rates and a fixed apparent ground position. A matching period alone is insufficient: an inclined orbit changes the apparent position, while an eccentric orbit does not maintain the required constant angular rate. Reversing the orbital sense also prevents the fixed alignment.
A fixed apparent position needs matching angular motion and the correct plane
The ideal orbit is circular, equatorial and in Earth's west-to-east rotation sense, with the same period as Earth. These geometry panels are schematic in distance.
North-pole view: Earth and satellite advance 60°
G0 and G1 are the same equatorial ground location at two times. Its radial alignment with the satellite is unchanged because both angular speeds match.
A tilted orbit does not keep the same apparent position
Matching the rotation period alone is insufficient. The orbit must also be circular, equatorial and in Earth's rotation sense.
The satellite is still moving and accelerating in an Earth-centred inertial description. Gravity provides its inward resultant. Its angular speed matches Earth's rotation; its tangential speed is not the same as that of a ground marker at a smaller radius.
Supplied-period calculation
Distinguish geostationary radius from altitude
Use the supplied sidereal period T = 86164 s, about 23 h 56 min 4 s. The often-used 24 hours is a rounded approximation; use 86400 s consistently if that is what a problem supplies.
With this chapter's GM = 4.0 × 1014 m3/s2:
≈ 4.2213 × 107 m
h = r - R ≈ 3.5813 × 107 m
These extra digits describe the supplied rounded model, not precise measured Earth-orbit constants. The calculated r is centre distance; h uses the same model surface radius R = 6.4 × 106 m.
Communications: the fixed apparent direction allows a ground antenna to keep pointing the same way. Weather observation: a continuing view lets changes in the same broad visible region be followed over time.
The satellite is not directly overhead every observer: its fixed ground location is equatorial. The viewing geometry is also less useful for far polar regions. The application depends on the region that must be seen, not just on the period.