3 ms·
There is a Coriolis force here too; it's just much smaller than the centrifugal force. Consider a plane flying at ground level along the equator at groundspeed
by bzbarsky 7y ago
There is a Coriolis force here too; it's just much smaller than the centrifugal force.
Consider a plane flying at ground level along the equator at groundspeed velocity v in the direction of the Earth's rotation. In the Earth's (rotating) reference frame there is a centrifugal acceleration based on the Earth's angular velocity and a Coriolis acceleration given by -2 W x v, where W is the Earth's angular velocity and v is the plane's velocity. This ends up pointing straight up in this configuration.
So in the plane's reference frame there is a straight-up centrifugal acceleration, given by (W + v/R)^2 * R = W^2 * R + 2W x v + v^2/R which is balanced by the aerodynamic and gravitational forces on the plane (which in this frame is non-accelerating), while in the Earth's reference frame there is a straight-up centrifugal acceleration given by W^2 * R and a straight-up Coriolis acceleration of 2W x v, and the plane is accelerating radially down at v^2/R due to gravity and lift combined.
Now to put numbers to this, the speed record for an X15 is about 2000 m/s, R is 6.4 * 10^6 m, and W is 1 revolution per 24 hours, or 1/86400 s^1. So the three force components look like this, if I am doing my numbers right:
W^2 * R is .00086 m/s^2
2W x v is 0.04 m/s^2
v^2/R is 0.63 m/s^2.
So the v^2/R "centrifugal force due to the plane's motion" term is by far the biggest contributor here.
- pdonis 7y ago> There is a Coriolis force here too; it's just much smaller than the centrifugal force. Yes, agreed.