# A Question on a South to North flight on a Rotating Earth

A Problem is posed for a flight from Melbourne to Cairns.

Using the formula for rotational speed at latitude -

1) The latitude of Melbourne is approx 38 degrees south with a rotation velocity of 820 miles/hr 2) The latitude of Cairns is approx 17 degrees south with a rotation velocity of 994 miles/hr

The above two locations have an Earth west to east bound rotation velocity difference of 174 miles/hr. The cruising velocity of a jetstar A321 is 536 miles/hr. How does the aeroplane land when the airport at Cairns is moving west to east at 174 miles/hr under the north travelling flight? The cruise speed is already 536 miles/hr when travelling north, therefore when the plane turns east, the flight must increase velocity by 174 miles/hr, to change the west to east velocity of 820 miles/hr to 994 miles/hr and land under a relatively stationary airport. Yet the additional 174 miles/hr cannot be obtained from a plane already travelling at maximum velocity.

The above problem may be translated into a more general question. How does the aeroplane take off and land on a spinning Earth with a rotation velocity which varies with latitude? Every north to south, or south to north flight must require an acceleration or deceleration of the aeroplane to equate the west to east rotational velocity of the plane's original take off latitude to the west to east velocity of the plane at the airport's latitude at the end of the flight.

JM

Both locations are spinning at the same angular velocity. Let's take a look at the rotation equation: $$\vec{M} = I\vec{\alpha}$$
As there is no momentum being acted on the plane (in the east direction), the angular acceleration would also be $$0$$, this means that the angular velocity $$\omega$$ would remain constant. We also know that if $$\omega$$ is constant we can use $$\frac{v}{r} = \omega$$
Hence, as $$\frac{v}{r}$$ would be constant, its linear velocity will increase if the plane goes higher.