Velocity of a car relative to the center of the Earth

Question

If we approximate the Earth's circumference to be $40,075\;km$ we could say that the the amount of time it takes for a full rotation is about $24\; hours$. We could then approximate the rotation of the Earth at the equator to be $\sim464 \;m/s$.

Using frames of reference, how could one apply this to the velocity of a vehicle moving on a highway in the same direction of the rotation of the Earth in North America? (let's say New York). Say, the car was traveling at $\sim28\;m/s$, would it be justified to add their velocities together? Or does it make a difference that New York is so far North of the Equator?

Answer 1

Yes, it does matter that the car is in North America. The amount of time it takes for rotation about the Earth's axis of rotation is the same in North America (still 24 hours), but the distance traveled during this rotation is less so the angular velocity is smaller.

Answer 2

The latitude of New York is 40°N, which makes a substantial difference in its speed around the earth's axis. The velocity $v_\text{earth}$ of the earth along the a circle of latitude $\theta$ is $$v_\text{earth} = \frac{40\,000\,\mathrm{km}}{24\,\mathrm{hours}}\cdot\cos\theta$$ (assuming a spherical earth). At the equator ($\theta$=0°) that's just what you calculated, but at 40° latitude this evaluates to a velocity of only 355 m/s. To this value you can add the speed of your car if it is driving eastward, i.e. the same direction as earth's rotation. But take care, because eastward of New York there comes a lot of water ;)