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

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?

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:

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

• Is this a case where the rotational velocity of the Earth would actually affect the numbers, i.e. adding it to an eastbound jet, subtracting it for a westbound, and (roughly) no change for a north or southbound jet?
– Ralph J
Jan 16 at 12:13
• @RalphJ Good point! In space, there is no difference between retrograde and prograde orbits (Space.SE). But in the atmosphere, it can make quite a difference. I added a paragraph with some more details. Jan 16 at 13:45
• So the tl;dr might be something like "Yes, but not much, and mostly if you're flying towards the East. The accountants will be pleased at a bit of fuel saved." Jan 16 at 20:34
• @WayneConrad: The extra lift scales with speed squared, so Mach 10 would have 4x the effect, about 14% +- east / west difference. But yeah I guess that's still "not much", and much higher speeds Mach 20 is more like re-entry from actual orbit. Even Mach 10 is much less realistic than Mach 5. Jan 16 at 23:34
• @WayneConrad I think the accountants will be pretty mad at you for flying at Mach 5 because that really increases fuel consumption :D Jan 17 at 7:15