256bits said:
Is he description from
@sophiecentaur called a tidal effects?
He is attributing the difference in pressure due to the revolution of the ship around the earth.
Nowhere did he mention the gravitational gradient of the earth that exists throughout the ship.
The two effects are of similar magnitude, but somewhat different in character.
If I have not muffed the analysis, the net of the centrifugal effect and the tidal effect is a pseudo-tidal effect which has 3 times the magnitude of the centrifugal effect alone in the radial direction and has zero magnitude in the tangential direction.
Let me show the work:
To be clear, ##r## is position relative to the center of mass of the space craft. ##R_0## is the distance of the craft's center of mass from the center of the Earth. When appropriate, ##R## will be the distance from the center of the Earth to a particular point in the craft. The craft is (we assume without evidence) consistently oriented with the same face toward the earth, rotating at the same rate as its orbital period.
The centrifugal force from rotation has magnitude ##\omega^2 r## and is repulsive in both the radial and tangential directions.
Tidal gravity in the radial direction results from the gradient of a ##\frac{1}{R^2}## force field which is equal to ##\omega^2 R_0## near the position of interest. So we can normalize the gravitational force to ##\frac{\omega^2{R_0}^3}{R^2}## and take the derivative with respect to ##R##. That will yield ##\frac{-2\omega^2{R_0}^3}{R^3}##. Evaluated at ##R = R_0##, this yields ##-2 \omega^2\ dR##.
This is to say that the gravitational tidal gradient in the radial direction is twice the centrifugal component. Both are repulsive. So they add to a net effect that is three times that of centrifugal alone.
Let us turn our attention to the gravitational tidal gradient in the tangential direction. Offhand, I do not see a convenient way to attack this. So I will be floundering a bit. However, I expect the result that is obtained will be proportional ##\omega^2 r## with a small integer as the constant of proportionality.
The tangential gradient will be given by ##\omega^2 R_0 \sin \theta## where ##\theta## is the angle subtended by an arc of length ##dR## on a circle of radius ##R_0##. The small angle approximation is appropriate, so we can rewrite the gradient as ##\frac{\omega^2 R_0\ dR}{R_0}##. Which simplifies to ##\omega^2\ dR##. [Which was surprisingly straightforward after the initial panic had worn off].
So the gravitational gradient in the tangential direction is equal to the centrifugal effect. But the two act in opposite directions, so the net effect is zero tidal gradient in the tangential direction (to first order).