Please, explain "Tangent to the star". I was puzzled by these words in Forward's explanation.
They give me a picture of a plane that touches the star's surface at one point, like a ball on a table. An orbiting body will pass through such a plane, briefly and at an angle relative to the direction of the star's centre. Is the acceleration applied at an angle?

If this tangent plane always touches the star at a point immediately below the satellite, it will never pass through it. In which case, I can't see how acceleration can be applied in a plane that doesn't pass through the satellite.

If the plane is parallel to a tangent to the star, then this is the same as though the centre of the satellite, at right angles to the direction of the stars centre. In other words sideways. No, I really can't see that happening!

So what is meant by "tangent to the star"?

Thanks for pointing me to the other thread - I'll go and be educated there too.
John

PS Well, I went and I was educated and I have enormous respect for RichardMB as a teacher. His three-tethered-balls-at-the-distance-of -the-Moon thought experiment, with the simple calculations appended, is JUST what I need to get a better picture in my mind of what is happening.
AND it reveals that even if my half baked view of 'different orbits' is wrong it was half right! The tension between the three balls nearly doubles if they are in orbit, rather than merely falling straight in! Does this mean that the faster the orbit, the greater the tide? I'll have to try to follow Richard's calculations better.
But I still don't understand 'tangent to the star"!

Thanks again to all.
John

Yes, that's it, and it really does happen. We know this from actual measurements.

That effect is pretty much balanced, on the equator, by a centrifictional force produced by a rotation of the satellite that coincides with the time of orbit.

cool :P