Tidal forces acting on planet NGTS-10b

[Buzz Bloom]

TL;DR

NGTS-10b, illustrated generically, is the closest and fastest-orbiting giant planet yet discovered, circling its home star in only 18 hours. NGTS-10b is a little larger than Jupiter, but it orbits less than two times the diameter of its parent star away from the star’s surface. When a planet orbits this close, it is expected to spiral inward, pulled down by tidal forces to be eventually ripped apart by the star’s gravity.
https://apod.nasa.gov/apod/ap200226.html

I am hoping a PF participant can help me understand why the tidal forces cause the NGTS-10b planet to fall towards it's star. I more-or-less understand why our moon's tidal forces on the Earth cause the Earth's rotation period to decrease, and the conservation of angular momentum of the Earth-moon system causes the moon to move away from the Earth. The only ideas I can think of are probably wrong.
(1) If the orbital spin direction of NGTS-10b about it's star is in the opposite direction of the rotational spin direction of it's star, that might explain this effect.
(2) If the orbital spin direction of NGTS-10b about it's star is in the opposite direction of the rotational spin direction of NGTS-10b, that might explain this effect.

The quoted text says the phenomenon depends of the distance of the planet from it's star. I do not understand this at all.

 

[Janus]

You will also get the spiral in result if the period of the orbit is shorter than the period for the rotation of the star (with both orbiting in the same direction). A closer planet has a shorter orbital period. This same effect is causing Phobos to spiral in on Mars.

 

[Bandersnatch]

Expanding on the answer above, look at the picture:

Row one shows how the smaller body rises a tidal bulge on the larger one. The tides on the smaller one are ignored. Likewise ignored are the forces due to the far end of the tidal bulge, as they're symmetrical but lower, so the nett effect is as shown above.

Rows two and three show what happens if we take into account that the larger body is rotating, with angular velocity ω1, while the smaller one orbits with angular velocity ω2. Both are in the anti-clockwise direction.

It takes some finite time for the tidal bulge to be raised and lowered, so depending on which ω is larger, the tidal bulge gets displaced from the line connecting the centres of the two bodies.

The displaced bulge acts gravitationally on the orbiting body along a direction that is no longer purely radial, so the force will result in torque on the orbiting body either pro- or retro-grade. This causes the orbit to be raised/lowered respectively (adding/removing orbital energy).

notes:
1) at the same time, the force vector acting on the larger body, and its tangential component causes torque, retarding/accelerating its rotation
2) since the orbiting body follows Kepler's laws, being initially located at lower R means the orbital period is faster, so for any omega ω1 there is a R where ω2 becomes higher than ω1 and the orbiting body is doomed to crash into the larger one
3) a special case is where the orbiting body orbits in a direction opposite of the direction of rotation of the larger body, in which case the forces are like shown in row 2, but the direction of orbital motion V is opposite, so the orbital torque is always retrograde - the orbiter will eventually fall into the central body regardless of the magnitude of its orbital speed.
4) another special case is when ω1=ω2, i.e. it's the tidal lock scenario, in which case the bulge remains aligned along the radial direction and no torques are present.
5) analogous effect works on the other body, i.e. the larger raises a bulge on the smaller, with all the same relations involved.