Orbits: Period, Speed and Kepler's Third Law

Choose the model before choosing an orbital formula. Here the satellite is much less massive than the spherical body it circles. We treat that body as fixed and neglect atmospheric drag and other bodies. Measure the radius r from the central body's centre. A height above the surface is a different distance.

Gravity bends the path inward. For a circle, the required inward force is mv²/r, and gravity supplies GMm/r². Equating them gives v² = GM/r after m cancels. Satellite mass therefore does not set the circular speed in this model.

The distance travelled in one orbit is 2πr, so its period is T = 2πr/v. Substitution gives T² = 4π²r³/(GM). Comparing satellites around the same body cancels the common GM. If their central masses differ, that cancellation is not available. These equations compare circular paths; a transfer between them need not itself be circular.

Tycho Brahe recorded planetary positions over many years. Kepler studied those observations and found mathematical patterns. His laws described the motion before Newton explained the force behind it.

The first law gives an elliptical orbit with the Sun at one focus. The semimajor axis a is half the ellipse's longest diameter. In the special case of a circle, a equals the radius.

The second law says that the line joining a planet to the Sun sweeps out equal areas in equal time intervals. Along an elliptical path, this means faster motion near the Sun and slower motion farther away.

The third law relates period to orbital size. For small bodies orbiting the same central mass, T² is proportional to a³. A circular orbit uses r in place of a. Newton's inverse-square gravitational law explains these patterns in an ideal two-body system. Other bodies can disturb real orbits, so the ideal description has limits.

A low circular Earth orbit has a high speed and a short period. Thin air can still exert drag, removing mechanical energy and changing the orbit. Moving beyond most of the atmosphere does not remove Earth's gravity.

A geostationary satellite stays above one point on the equator. Its circular orbit lies in Earth's equatorial plane, and it moves eastward with Earth's rotation. Its period is about 86,164 seconds, matching one rotation relative to distant stars. Its radius is about 42,200 km from Earth's centre, and its height is about 35,800 km above the surface.

A geosynchronous orbit matches the rotation period but need not meet all the other conditions. An inclined or elliptical geosynchronous orbit does not keep a fixed ground position.

Comparing two circular Earth orbits, the higher one has a lower speed but a higher total mechanical energy. An engine burn changes motion during a transfer. Its speed changes cannot be inferred from the final circular speeds alone.

Consider a small satellite of mass m orbiting a much larger mass M. Ignore drag, thrust and other bodies. Set gravitational potential energy to zero at infinite separation. At distance r, potential energy is U = -GMm/r.

A circular orbit has v² = GM/r. Its kinetic energy is therefore K = mv²/2 = GMm/(2r). Adding the two energies gives E = K + U = -GMm/(2r). The negative total means the satellite is bound by gravity in this model.

At a larger circular radius, kinetic energy is lower. Potential energy rises by a greater amount, so the total is higher, meaning less negative. Reaching that higher circular orbit requires added mechanical energy despite the lower final speed.

Along an unpowered orbit without drag, total mechanical energy stays constant while kinetic and potential energy may change. In the ideal inverse-square field, an outward escape path with nonzero angular momentum is parabolic at zero total energy and hyperbolic at positive total energy.