For an airplane moving north from some latitude (say $30^{\circ}N$), why does rotation of the Earth cause an increase in apparent drag?

Question

I'm not able to understand the answer to this example:

Example 10 from the Curtis Orbital Mechanics text book :

An airplane of mass $70 000\ \mathrm{ kg}$ is traveling due north at latitude $30^\circ$ north, at an altitude of 10 km with a speed of $300\ \mathrm{ m/s}$. Calculate

(a) the components of the absolute velocity and acceleration along the axes of the topocentric-horizon reference frame

(b) the net force on the airplane.

The topocentric horizon reference frame is just East, North and Up. Assume Earth is perfect sphere. z axis intersects with Earth's center.

The answer to (b) is as below. These are the forces required to ensure we stay in the same path.

The force on the west direction is just whats required to balance the Coriolis force. The explanation for the other two forces are :

The forward and downward forces are in the directions of the airplane’s centripetal acceleration, caused by the earth’s rotation and, in the case of the downward force, by the earth’s curvature as well.

I understand how rotation and curvature decreases the apparent weight. But I don't understand how rotation of earth can increase the apparent drag (y direction)?

Answer 1

Draw a picture of a section through the earth's north and south poles.

The airplane is flying in a direction tangential to the earth's surface.

The centripetal force caused by the earth's rotation is acting perpendicular to the line joining the N and S poles, i.e. at an angle of 30 degrees to the airplane's direction of flight.

So in your "North/East/Up" coordinate system, the centripetal force has components in the North and Up directions.

Note in your picture the "North" component of the force acting on the airplane is reducing drag, not increasing it. It would be increasing the drag if the plane was flying south not north.