ORBIT
v²r = GM, E = −GMm/2r, and 2KE + PE = 0 to 1e-12 — the virial ledger, audited.
Use the simulation above to change the variables and play through the guided stages. The explanation below describes the default starting values; the simulation updates its explanation as you experiment.
Setup
A satellite on a perfect circle around a central mass. Three questions follow it around every loop: how fast must it move, how long does one orbit take, and what does its energy account actually look like? All three answers are single lines — and the third is quietly profound.
Centripetal
On a circle there is exactly one job for the one force: gravity must supply the entire centripetal demand, no more and no less. Write that balance and the satellite’s own mass cancels off both sides — which is why a bowling ball and a bus at the same altitude orbit at precisely the same speed.
Solve
Speed is the square root of GM over r. The period follows from distance over speed and reproduces Kepler’s third law exactly. Then the energy: kinetic is positive, potential is twice as large and negative, and their sum — the total — lands exactly halfway, negative, the signature of a bound orbit.
The ledger
Watch the bars as the satellite circles. Kinetic energy stands up; potential hangs down twice as deep; total sits at the midpoint. Here 2KE + PE = 0 at every instant because the orbit is circular; for a general bound orbit the virial theorem states the time-averaged relation.
Audit
Speed-squared times radius returns GM, and the period-squared law returns four-pi-squared r-cubed over GM, both across the whole radius slider to a part in a trillion. The virial identity is bitwise, and total energy is minus the kinetic exactly. The ledger even predicts the famous paradox: drop to a lower orbit and the satellite speeds up, because losing energy buys more kinetic.