Reconcile orbital energy with angular momentum

Choose the central-source approximation

A small satellite moves in the field of a dominant fixed spherical source, with GM=12 m³/s² and satellite mass 2 kg in this synthetic model. The potential is U=−GMm/r with zero at infinity. The central force has zero torque about the source, so L is conserved; the conservative gravitational force also preserves total E=K+U. Neither condition requires K and U to be individually constant.

Derive the circular special case

For a circular orbit, mv²/r=GMm/r² gives v_c=sqrt(GM/r). Then K=GMm/(2r), U=−GMm/r and E=−GMm/(2r). L=mr v_c and T=2pi sqrt(r³/GM). Radius, speed, K and U are all constant. Doubling circular radius reduces speed by sqrt(2), halves the magnitude of E and multiplies period by 2sqrt(2).

Use the initial turning-point packet

Initially radial speed is zero and tangential speed is f times v_c at radius r0. Thus E=(GMm/r0)(f²/2−1) and L=mf sqrt(GM r0). If f is below sqrt(2), E is negative and the point-source orbit is bound. At f=sqrt(2), E=0 is the escape threshold; larger f gives positive energy. For an eccentric bound orbit, speed and kinetic energy increase as radius decreases while potential becomes more negative.

Derive the escape threshold from energy

Setting E=mv²/2−GMm/r=0 gives escape speed v_escape=sqrt(2GM/r), or sqrt(2) times the circular speed at the same radius. On an outward escaping path with only the central gravitational interaction, this threshold energy approaches zero speed only as radius tends to infinity. It does not come to rest at a finite radius. Positive total energy instead leaves a nonzero asymptotic speed sqrt(2E/m).

For GM=12 and r=4, the threshold speed is sqrt(6) m/s. Predict the change if the starting radius becomes 16: the threshold becomes sqrt(1.5) m/s, one half as large. This is a central-field energy comparison, not a launch procedure; no drag, propulsion history or intervening physical surface is modeled.

Optional turning-radius enrichment

At any radial turning point, E=L²/(2mr²)−GMm/r. Substituting the initial E and L into this quadratic yields the second positive turning radius r_other=r0 f²/(2−f²) for bound motion. For f less than 1 the initial radius is the larger turning radius; for 1 less than f less than sqrt(2) it is the smaller. At f=1 the roots coincide in the circular orbit. This optional effective-potential calculation is not a timed trajectory or a realistic launch model; the assumed point-source field may be invalid inside an actual central body.

Worked original case

At r0=4 and f=1, v=sqrt(3), K=3 J, U=−6 J, E=−3 J and L=8sqrt(3) kg m²/s. At the same radius with f=.5, E=−5.25 J, L=4sqrt(3) and the other turning radius is 4/7 m in the ideal point-source model.

Investigation

Compare f=.5,1,1.25,1.5 at r0=4. Compute E and L independently, classify the orbit and test whether a finite second turning radius exists. Explain why changing K around a bound eccentric orbit does not imply energy loss. For the circular case, derive a period ratio for a changed radius without using a full numerical trajectory.

Public numerical transfer practice

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An ideal circular orbit around a fixed source has radius multiplied by 4. By what factor does its period change?

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Sources and evidence boundary

Original derivations and assigned numerical cases checked 30 October 2026 against the official AP Physics C Mechanics framework and OpenStax angular momentum. These are supplied or generated records, not physical measurements or performed laboratory time. Do not construct rotating equipment, release loads or conduct launch experiments. Public feedback and retained self-reports do not certify independent mastery.