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I'm at the equator. Suppose I am driving at x kph, in the direction of the Earth's rotation (i.e. due East), I presume that the force my vehicle exerts on the ground is less than when I am stationary. If on the other hand I drive West (i.e. against the direction of the Earth's rotation), then I presume I will exert a greater force on the road surface in comparison with an equal speed in the opposite direction.

Assumptions

My mass is 80 kg. My speed is a variable x kph The acceleration due to gravity = 9.8 m/s² The surface of the earth at the equator moves East at a speed of 460 meters per second.

Question

How can I calculate the force I exert on the ground? Will it be at its greatest when my vehicle is travelling at 460 meters per second westwards?

Bonus question

How can this be generalised (a) to an arbitrary direction at the equator (b) an arbitrary point on the Earth's surface.

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  • $\begingroup$ I presume that the force my vehicle exerts on the ground is less than when I am stationary. Why? $\endgroup$ – user214814 Jan 6 at 12:47
  • $\begingroup$ @ StudyStudy - Because the rotation of the Earth at the equator is throwing me away from Earth. This explains the somewhat flattened shape of the globe. As I understand it, satellites are better launched in the direction of a planet's rotation for this reason. If I go faster then I will be thrown outwards more. $\endgroup$ – chasly from UK Jan 6 at 12:52
  • $\begingroup$ Okay:) As you say, for objects lauched vertically, "satellites are better launched in the direction of a planet's rotation for this reason." A related post is this one: astronomy.stackexchange.com/questions/2313/…. I just wasn't sure how much details you were getting into. +1 Best of luck with it. $\endgroup$ – user214814 Jan 6 at 13:18
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How can I calculate the force I exert on the ground? Will it be at its greatest when my vehicle is travelling at 460 meters per second westwards?

Closer to 465 meters per second, but yes.

One way to look at a vehicle at the equator moving either eastward or westward is from the perspective of a rotating frame such that the vehicle appears to be stationary. The rotation rate of this frame with respect to inertial is $\omega = \Omega + v_e/R$ where $\Omega$ is the Earth's sidereal rotation rate, one revolution per sidereal day, $v_e$ is the vehicle's eastward velocity (negative if the vehicle is moving westward), and $R$ is the Earth's equatorial radius.

There's no Coriolis effect in this frame since the vehicle is stationary from the perspective of this frame. There is however an outward centrifugal acceleration of $a_c = \omega^2 R = \Omega^2 R + 2 \Omega\,v_e + {v_e}^2/R$. The first term on the right, $\Omega^2 R$, is the centrifugal acceleration in a frame rotating with the Earth; this is by convention incorporated into the local value of $g$. The latter two terms, $2 \Omega v_e + v_e^2/R$, change the apparent weight of an object.

For positive values $v_e$, this upward acceleration increases monotonically with increasing eastward velocity. For negative values of $v_e$ (i.e., motion is to the west), $v_e$ is negative but ${v_e}^2$ remains positive. Differentiating with respect to $v_e$ and setting the result to zero to find the extremum yields $2\Omega + 2 v_e/R = 0$, or $v_e = -R\,\Omega$. Going westward at a speed that exactly cancels the effects of the Earth's rotation maximizes apparent weight.

How can this be generalised (a) to an arbitrary direction at the equator (b) an arbitrary point on the Earth's surface.

The apparent upward acceleration at an arbitrary latitude $\phi$ given an arbitrary horizontal direction $\vec v = v_n \hat n + v_e \hat e$ where $v_n \hat n$ is the local northward component of the horizontal velocity and $v_e \hat e$ is the local eastward component is

$$a_u = 2 \cos\phi\,\Omega\,v_e + \frac{{v_e}^2+{v_n}^2}R$$

This is the Eötvös effect.

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