Energy and angular momentum

Keep the reference, signs, and mass basis consistent

Gravitational potential energy compares positions using a chosen zero. Here the zero is infinitely far from the isolated central body. Bringing a spacecraft inward releases potential energy, so its value at a finite radius is negative. To move it outward slowly, an external agent must supply energy. A negative potential value does not mean a negative amount of matter or an instrument reading below a physical minimum.

Near the ground, the expression mass times surface gravity times height approximates a change in potential energy. It treats gravity as nearly constant over the height interval. The orbital expression instead allows gravity to weaken with radius. You cannot set the two expressions equal as absolute energies without accounting for their reference choices. They can agree approximately on small energy differences while assigning different zero levels.

Specific energy means energy divided by spacecraft mass. A thousand-kilogram vehicle and a hundred-kilogram vehicle on the same ideal orbit have the same specific energy, but their total orbital energies differ by a factor of ten. The energy needed to change their motion is therefore not identical in joules. Dividing by mass simplifies trajectory prediction; it does not make the physical mass irrelevant to propulsion or engineering.

Read each pair of energy bars as a signed sum. The kinetic bar is positive, while the gravitational bar represents a negative term. Its longer length means a larger negative magnitude, not a larger positive contribution. At periapsis the two large terms nearly cancel to the same negative total found at apoapsis. Neither the length of a single bar nor the speed alone is the conserved quantity.

At radius 7,000 km on the displayed ellipse, mu divided by radius is about 56.943 square kilometers per square second. The kinetic term is about 37.013, leaving negative 19.930. At radius 13,000 km, the potential term is about negative 30.662 and the kinetic term about 10.732. Their sum is again negative 19.930. Use the unrounded model values for the comparison, then round the final reported totals.

The inward-to-outward change in potential energy is about positive 26.281 MJ/kg. The kinetic term decreases by the same amount. For a 100 kg craft, multiplying by mass gives about 2,628 MJ exchanged between those terms during that coast. This is not energy supplied by an engine. It is a redistribution within the conserved mechanical energy of the gravity-only model. A burn would require a separate accounting step.

The semi-major-axis form provides another check. For a equal to 10,000 km, negative mu divided by 20,000 km gives the same negative 19.930 MJ/kg. Doubling a halves the magnitude of the negative energy, bringing it closer to zero. Thus a larger bound orbit has higher specific energy even though its circular speed, when the orbit is circular, is lower. Potential energy is the missing term in the apparent contradiction.

Derive vis-viva before substituting numbers

Write the same specific energy in two ways: one half speed squared minus mu over the current radius, and negative mu over twice the semi-major axis. Equate them. Add mu over radius to both sides so that the kinetic term stands alone. Multiply both sides by two. Speed squared is then twice mu over radius minus mu over semi-major axis. Factoring out mu gives the expression inside the vis-viva square root.

This rearrangement explains the different jobs of r and a. Radius tells you where the spacecraft currently is. Semi-major axis identifies the energy of the whole orbit. During an unpowered coast on one ellipse, a remains fixed while r changes. Replacing a with the current radius at every step would silently replace the ellipse with a sequence of different circular orbits. That is not the motion being calculated.

At periapsis in the example, twice the reciprocal radius is approximately 0.000285714 per kilometer. Subtracting the reciprocal semi-major axis, 0.000100000 per kilometer, leaves 0.000185714. Multiplying by mu gives about 74.026 square kilometers per square second. Its square root is 8.604 km/s. The subtraction belongs inside the parentheses before multiplication and before the square root.

At apoapsis, twice the reciprocal radius is about 0.000153846 per kilometer. Subtract the same 0.000100000, not a newly chosen semi-major-axis term. Multiplication by mu now gives about 21.463 square kilometers per square second, whose square root is 4.633 km/s. The smaller kinetic term corresponds to the less negative potential at the larger radius. This connects the calculation back to the energy bars.

You can predict the comparison with circular speed without evaluating either square root. Circular speed squared is mu over radius. Subtracting that from the ellipse's speed squared leaves mu times the difference between one over radius and one over a. When r is smaller than a, that difference is positive; the ellipse is faster. When r is larger than a, it is slower. At r equal to a, their speed magnitudes match.

Matching speed at r equal to a does not mean the ellipse becomes a circle there. An eccentric orbit generally has a radial velocity component at that location, unlike a circular orbit. The craft passes through the radius rather than remaining on it. This is a concrete counterexample to inferring an orbit from speed and altitude alone. Direction remains part of the state even after energy has determined the speed magnitude.

Vis-viva also has a domain of interpretation. For a positive a describing a bound ellipse, a proposed radius larger than twice a would make the quantity under the root negative. That signals inconsistent inputs rather than an imaginary physical speed. Even a nonnegative result is not enough to establish that the point belongs to a particular ellipse: its eccentricity, orientation, and angular momentum constrain the allowed positions. The scalar energy equation is only part of the state description.

Use components to distinguish orbits with equal energy

At a given point, choose one direction straight outward from the planet and another perpendicular to it in the orbital plane. Decompose the velocity into these radial and transverse parts. The radial part changes distance from the center. The transverse part changes direction around the center. The two components combine to produce the actual velocity, which need not point along either chosen direction.

For a concrete pair of states at radius 10,000 km, let craft A move entirely transversely at 5 km/s. Let craft B have outward radial speed 3 km/s and transverse speed 4 km/s. The right-triangle relation gives B the same total speed, 5 km/s. Both have kinetic energy 12.5 MJ/kg and potential energy about negative 39.860 MJ/kg, so their total specific energies match. Their paths nevertheless differ.

A's specific angular momentum is 10,000 times 5, or 50,000 square kilometers per second. B's is 10,000 times 4, or 40,000 in the same units. B's radial motion contributes nothing to this magnitude. Equal energy therefore does not imply equal angular momentum. The states have the same semi-major axis in the ideal bound model but different eccentricities. A complete orbit calculation must preserve both pieces of information.

Why is angular momentum conserved under gravity alone? Torque measures a force's tendency to turn momentum about a chosen center. A force directed exactly along the radius has no sideways lever arm about that center. Its torque there is zero. This preserves the angular momentum vector during the coast. Its direction also fixes the orbital plane. A thrust component outside that plane can change this vector, which is why plane changes require a maneuver.

The transverse speed itself is not generally constant. The product of radius and transverse speed is constant. As radius increases, transverse speed must fall to keep that product unchanged. The radial and transverse directions also rotate as the craft moves. Saying that gravity has no transverse component at each instant does not justify holding the numerical transverse speed fixed while changing the radius and the local coordinate directions.

At the apsides, radial speed vanishes, so total speed equals transverse speed. Multiplying 7,000 km by approximately 8.604 km/s and 13,000 km by approximately 4.633 km/s gives nearly the same angular momentum, with a small mismatch if you use rounded speeds. Taking the ratio before rounding gives the inverse radius ratio, thirteen sevenths. Between the apsides, multiplying radius by total speed would include radial motion and fail this check.

The swept-area argument now follows directly. Over a short interval, the radius is one side of a thin triangle and transverse displacement supplies its height. Its area is approximately half of radius times transverse speed times elapsed time. The first two factors form the conserved specific angular momentum. Equal elapsed times therefore produce equal areas. Energy explains how speed changes with radius; angular momentum explains how much of that motion is transverse.

When examining a simulation, check conserved quantities only during intervals where their assumptions hold. A successful burn should change the orbit's energy or angular momentum, so demanding the old values after the burn would be a faulty test. During a gravity-only coast, however, unexplained drift can indicate numerical or implementation error. Separate physical changes in the modeled system from errors introduced by the method used to calculate it.