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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?

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Some assumptions were made for this formula, not including the rotation speed of the earth and the rotational speed of the aircraft around the earth. The first term would be influenced by the position and heading you are having on earth. The second Term would be influenced by the Mach number adding an additional centrifugal force (the aircraft, becoming a satellite around the earth [As far as I remember this influence is only particularly interesting starting from Ma=3]).

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).

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    $\begingroup$ If you’re travelling westwards then the aircraft’s weight will initially increase with increasing speed until the groundspeed equals that of the Earth’s rotation. During a turn, the effective weight will therefore vary continuously as the aircraft turns. $\endgroup$
    – Frog
    Feb 5 at 5:02
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    $\begingroup$ 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. $\endgroup$
    – Jan Hudec
    Feb 7 at 8:48

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