Run Run Away

Question:

The original question was: Would a flight completely around the earth?s equator flying at 35,000 feet, in ABSENCE of any atmospheric conditions such as wind, humidity etc.. take slightly less time flying against the rotation of the earth and take slightly more time flying with the rotation of the earth, given the aircraft and flight parameters were identical each way?

There’s one teeny, tiny effect no one has mentioned…coriolis acceleration. It’s effect will depend on what assumptions are made about the "flight". But it can be seen if you imagine an earth with no atmosphere (i.e., vacuum). If a flying machine only lifts straight up against gravity to 35,000 as postulated, it will actually be travelling at 2.5 ft/sec (approx. 1.5 knots) in a westward direction relative to the ground (if it begins at the equator). This is because the flying machines initial velocity is based on the radius of the earth. As it ascends, this linear velocity is not enough for it to keep up with the surface, so it looses ground. Depending on the assumptions about the nature of your flying vehicle, there will be some slight effect on which direction around the globe you wish to travel…vanishingly small, but it’s there. Bill Levenson PP-ASEL-IA

Response:

See!  I knew I was right.  :)

Go East Young Man (To Extend Range) A airplane cruising in the equatorial plane at 35,000 feet elevation going East would get by using less fuel than its twin heading West. If you consider the accelerated frame of reference to be the rotating earth, it’s NOT THE CENTRIFUGAL FORCE which provides the impetus to modify the actual external forces on the two airplanes, IT’S THE CORIOLIS FORCE. [Note: I recently submitted a paper on this subject to The Physics Teacher. They returned it, saying it was correct and interesting but a little too advanced for their audience. They suggested I submit it to American Journal of Physics and I intend to do that.] Physics Background Pseudo-forces are those which appear because one insists on looking at particle motion in a non-inertial frame of reference. For example, relative to a rotating (and smooth) merry-go-round, a block of ice would slide away outwards unless restrained. That pseudo-force is the so-called "centrifugal" (away from the center) force. For the block of ice to remain "unaccelerated" with respect to (w.r.t.) the (non-inertial) merry-go-round, a counteracting centripetal (towards the center) force must be applied to the block of ice, say a barrier to its sliding. In an inertial frame, the unconstrained released block simply continues it’s initial motion in a straight line while the merry-go-round turns under it and "away off to the side." This is sort of, but with interesting wrinkles, the situation with the equatorial Eastbound and Westbound airplanes. The usual way of describing the influence of frames of reference rotating with angular speed OMEGA — the origin of that frame unaccelerated w.r.t. the inertial frame, and in our case coincident with it — in physics texts (Goldstein, Corben & Stehle, Symon, Kittel, Sommerfeld) or in flight dynamics texts (Thompson, Etkin) is to argue towards and derive:  a = F/m = OMEGA x OMEGA x R + [a] + 2*OMEGA x V + OMEGA_DOT x R Here vector a is the acceleration in FI, the inertial frame, and vector [a] is acceleration of the particle viewed in the rotating frame (for the rotating earth case, call it FE). R is the radius vector from the origin of FI to the particle, V is the speed of the particle as seen in the rotating frame, and OMEGA_DOT is the angular acceleration of the rotating frame w.r.t. the inertial one. Although the earth’s rotation rate is slowing (due to tidal friction), ignore that last very small term. Symbols ‘x’ denote cross or vector products. The pseudo forces, "seen" in the accelerated frame, are expressed best by solving for [a] (and, here, neglecting the angular acceleration): [a] = F/m – OMEGA x OMEGA x R – 2*OMEGA x V = (inertial) acceleration from external actually-applied forces + centrifugal pseudo-force + Coriolis pseudo-force. As further simplification we also assume the center of the earth is unaccelerated, so that without the rotation about its axis (also ignore precession and nutation), it would be a suitable inertial frame. That is, we ignore the orbital acceleration and any acceleration of the sun as it revolves in the Milky Way and anything else. In the first big equation above, the first term on the right is the centripetal acceleration, the negative of the centrifugal force, and the third term on the right is the (negative of the) Coriolis acceleration, commonly thought of as manifesting itself (only) by forcing moving objects in the Northern hemisphere to the right and those in the Southern hemisphere to the left. (Not quite correct; there will turn out to be more to that story.) Specialization to Motion in the Equatorial Plane The earth rotates "from West to East." About its axis, seen from above the North pole, that is considered a positive rotation with angular velocity OMEGA = 2*Pi/(24*60*60) = 7.27E-5 rad/sec. The radius of the earth is about 3960 statute miles or R = 2.091E7 ft. Add h = 35,000 to it and you get (R+h) = 2.094E7 ft. OMEGA x (R+h) points East and is of length 1522.3. OMEGA x that previous result, the centripetal term, points inwards, towards the center of the earth, and is of length 0.111 ft/sec^2 = 0.0034 g, where g = 32.174 ft/sec^2. Notice there is NO DEPENDENCE OF THE CENTRIPETAL TERM ON THE VELOCITIES of the eastbound or westbound airplanes. The centripetal term is there, true, but it’s precisely the same for the two airplanes. That’s an objection I made earlier, and it was correct, but I neglected something (else) important. The Coriolis acceleratiion, -2*OMEGA x V! Evaluated, using FE, for the eastbound airplane, this is an OUTWARD (from the earth’s center to the airplane) force of size 2*(W/g)*OMEGA*V. Not right, not left, but outward. On the westbound airplane, an inward force the same size. So the difference between the two airplanes is that the eastbound one, if it is to maintain its flight path, enjoys an extra lift of size 4*)W/g)*OMEGA*V. Numerical Example Let’s take for example airplanes with air speed 500 ft/sec (very close to 300 KTAS). For simplicity, no wind. Then the effect difference is 4*OMEGA*V = 4*7.27E-5*500 = 0.1454 ft/sec^2 = 0.0045 g. Of course it’s negligible, practically speaking, but that’s not the point. And just because it’s a pseudo-force doesn’t mean it doesn’t have real effects. Rivers in the Northern hemisphere wear away their right banks quicker and are a little, but measurably, higher on the right side than on the left. Another Wrinkle We do always need to consider FI, if we are to get Newton’s Laws to work, but nothing says we are restricted to considering FE, the rotating earth frame. Or any ONE frame. Instead, for the eastbound airplane, we can consider frame Fe, rotating (with respect to FI), with the eastbound airplane, at angular speed (OMEGA + omega), where the angular speed of the airplane (relative to the rotating earth) is omega = V/(R+h) = 500/(2.094E7) = 2.388E-5 rad/sec. Then there is NO CORIOLIS force, since no V in that frame Fe, and the centrifugal acceleration is -(OMEGA + omega) x (OMEGA + omega) x (R+h), outwards of size (7.27E-5 + 2.388E-5)^2*2.094E7 = 0.1953 ft/sec^2. For the westbound airplane, consider frame of reference Fw rotating (with respect to FI) at (OMEGA – omega), along with the westbound airplane, and you’ll find the centrifugal acceleration is -(OMEGA – omega) x (OMEGA – omega) x (R+h), inwards of size (7.27E-5 – 2.388E-5)^2*2.094E7 = 0.0449 ft/sec^2. The difference is that the eastbound airplane has an excess centrifugal force (compared with the westbound one) of 0.1454 ft/sec^2 = 0.0045 g. Precisely as before, using FI and FE. (If you go through the algebra, squaring the binomials and then taking differences, you’ll find that the difference between the eastbound and westbound airplanes is always 4*V*OMEGA exactly.) John L. John T. Lowry, PhD Flight Physics; Box 20919; Billings MT 59104

Response:

The original question was: Would a flight completely around the earth