# How to calculate the inaccuracy in the "standard" turn radius equation, for supersonic airspeeds

I learned to calculate the radius of a level turn (r) using the function $$r = \frac{V^2} {g . tan(bank angle)}$$. V is true airspeed, g is the local gravity constant.

But now I have seen in the comments (from DeltaLima) to the answer (by Peter Kämpf), to a similar question How to calculate angular velocity and radius of a turn? that at very high speed, this equation underestimates the radius, because at such speeds, "your apparent weight is less". In the comments, the author states that for Mach 6,

the weight is about 10% less, so the radius is about 10% bigger if travelling eastwards at Mach 6 above the equator.

So at Mach 6 there is a 10% inaccuracy in the formula. How would I go about evaluating the inaccuracy at various Mach speeds, particularly between Mach 1 and Mach 6?

Ignoring the first term you can just add an additional Term to the Equilibrium of Forces (previous Answer) in the opposite direction of m*g ("becoming lighter") and do the same procedure. Additional Force: $$mV^2/R$$ with $$V$$ the Ground Speed and with $$R$$ the distance of the aircraft from the center of earth -> R = r+alt (radius of earth + altitude).
• The usual terminology does include the effect of rotation of Earth in the term gravity, though most calculations do just take the average value of $9.81 \frac{\mathrm{m}}{\mathrm{s}^2}$ instead of taking latitude into account and varying the value from $9.78$ at the equator to $9.83$ at the poles. Feb 7 at 8:48