Irodov Problem 1.220

Problem Statement

A particle of mass m moves in a conservative force field. Prove that in a circular orbit, the kinetic energy equals minus one-half the potential energy (virial theorem for inverse-square law).

Given

Circular orbit, F = α/r² (inverse-square). Virial theorem.

Concepts & Formulas

For circular orbit: mv²/r = α/r² → KE = mv²/2 = α/(2r). PE = −α/r. Therefore KE = −PE/2.

Step-by-Step Solution

Step 1: Centripetal: mv²/r = α/r².
Step 2: KE = mv²/2 = α/(2r).
Step 3: U = −α/r (gravitational/Coulomb).
Step 4: KE = −U/2 → 2KE + U = 0 → E = KE + U = −KE.

Worked Calculation

KE = α/(2r), U = −α/r → KE = −U/2. E = −KE = U/2.

Boxed Answer

KE = −U/2; Total energy E = −KE

Physical Interpretation

This is the virial theorem for 1/r potentials. A bound particle’s kinetic energy is half the magnitude of its potential energy — fundamental to atomic physics and astrophysics.