Does an orbital trajectory reduce my required lift for high mach numbers?

Question

when an aircraft is flying from let us say Paris to New York on a fixed altitude it is not flying on a straight line but on an orbital / circular trajectory around the earth. Assuming, the aircraft is flying Mach 5 or even faster, does this significantly reduce the amount of required lift, since the centrifugal force is acting on the aircraft?

Answer 1

We can calculate this by comparing the centrifugal force to the force of gravity:

$$ F_\text{centrifugal} = m \omega^2 r $$ $$ F_\text{gravity} = G \frac{m M}{r^2} $$

The ratio is then given by

$$ \frac{F_\text{centrifugal}}{F_\text{gravity}} = \frac{m \omega^2 r}{G m M / r^2} = \frac{\omega^2 r^3}{G M} $$

Let us assume we fly at FL600 (where Concorde flew), so $ h \approx 18 \, 288 \, \text{m} $ and then $ r = R + h$ (with Earth radius $R$ and mass $M$). The speed of sound at FL600 is $ c \approx 295.1 \, \text{m/s} $, which gives a speed at Mach 5 of $ v = 5c \approx 1476 \, \text{m/s} $. On a circular "orbit", we have $\omega = v/r $. Plugging in the numbers, I get

$$ \frac{F_\text{centrifugal}}{F_\text{gravity}} \approx 3.48 \, \% $$

So flying at Mach 5 would reduce the required lift by about 3.5%.

However, as Ralph pointed out in the comments, you need to take the Earth's own rotation into account. If you are on the equator, the above effect is already ~0.3% without moving at all (w.r.t. the Earth's surface). You need to add or subtract the Earth's motion from the velocity above. On the equator, that gives a difference of about 463 m/s (add when moving East, subtract when moving West). Repeating the calculation above then gives:

$$ \text{East:} \; \frac{F_\text{centrifugal}}{F_\text{gravity}} \approx 6.02 \, \% \qquad \text{West:} \; \frac{F_\text{centrifugal}}{F_\text{gravity}} \approx 1.64 \, \% $$