With fixed radius-cubed, only the matching period-square preserves the normalized ratio. Exact arithmetic here means exact results for the stated model inputs; measured inputs still carry uncertainty and significant-figure limits.

highlighted = computed this step

Right period-square row 1

The right radius-cubed is fixed at 64. Right period-square 32 gives the checked right ratio.

TB2rB3=32/64=12,a=1\frac{T_B^{2}}{r_B^{3}}=32/64=\frac{1}{2},\quad a=1
Kepler row 1The right period-square is checked against fixed radius-cubed.keplerConstant=1/2 s^2/m^3leftRadiusCubed=8 m^3leftPeriodSquared=4 s^2leftRatio=1/2 s^2/m^3rightRadiusCubed=64 m^3rightPeriodSquared=32 s^2rightRatio=1/2 s^2/m^3acceptedBit=1 bit

Right period-square row 2

The right radius-cubed is fixed at 64. Right period-square 36 gives the checked right ratio.

TB2rB3=36/64=916,a=0\frac{T_B^{2}}{r_B^{3}}=36/64=\frac{9}{16},\quad a=0
Kepler row 2The right period-square is checked against fixed radius-cubed.keplerConstant=1/2 s^2/m^3leftRadiusCubed=8 m^3leftPeriodSquared=4 s^2leftRatio=1/2 s^2/m^3rightRadiusCubed=64 m^3rightPeriodSquared=36 s^2rightRatio=9/16 s^2/m^3acceptedBit=0 bit

Right period-square row 3

The right radius-cubed is fixed at 64. Right period-square 40 gives the checked right ratio.

TB2rB3=40/64=58,a=0\frac{T_B^{2}}{r_B^{3}}=40/64=\frac{5}{8},\quad a=0
Kepler row 3The right period-square is checked against fixed radius-cubed.keplerConstant=1/2 s^2/m^3leftRadiusCubed=8 m^3leftPeriodSquared=4 s^2leftRatio=1/2 s^2/m^3rightRadiusCubed=64 m^3rightPeriodSquared=40 s^2rightRatio=5/8 s^2/m^3acceptedBit=0 bit

Only one right-period row matches the normalized constant

The helper keeps the two radius-cubed sources fixed. The scan changes only the right period-square, so one-half accepts and nearby ratios reject.

TBTBrBrBrBTBTB/rBrBrBKa3264121213664916120406458120\begin{array}{c|c|c|c|c}T_B\cdot T_B&r_B\cdot r_B\cdot r_B&T_B\cdot T_B/r_B\cdot r_B\cdot r_B&K&a\\32&64&\frac{1}{2}&\frac{1}{2}&1\\36&64&\frac{9}{16}&\frac{1}{2}&0\\40&64&\frac{5}{8}&\frac{1}{2}&0\\\end{array}
Kepler period-square boundary scanThe displayed row is a rejected right-period candidate.36/64 rejects