Why does the E-field of radiation vary with inverse distance?

[Fyreth]

Feynman_Lectures_on_Physics_Volume_1_Chapter_28

In 28-2 Radiation Feynman starts talking about the third term of Eq. 28.3 and why that it varies with the inverse of the distance. On page 28-4 he says that the unit vector e_'r is pointed to the apparent position of the charge. I understand that. The unit vector is just the normalized distance vector between the apparent position of the charge and the point of measurement P.
Then he says that the end of the unit vector goes on a slight curve. I assume he means that if the tail of the unit vector is on point P and the head that vector is on a sphere with radius = 1, the head is moving in along a curve.
But what does he mean with that the acceleration has a transverse component and a radial component? Is the transverse component the change in the direction of the unit vector? And the radial vector the change in the length? But isn't the length constant since it's a unit vector? And why does the former vary with the inverse of the distance and the latter with the inverse of the square?

I've been searching on Wikipedia and Hyperphysics but I can't find anything about it. I haven't seen anything that even looks remotely like Eq. 28.3.
I just started my physics study. I'm still in my first year of university. But these lectures are so much more interesting than my other books for some reason.

 

[Future Bruno]

The transverse component and radial component of the acceleration that Feynman is referring to are related to the different parts of Eq. 28.3. The transverse component is the part of the acceleration that changes the direction of the unit vector e'r. This part is proportional to 1/r, where r is the distance between the charge and the point at which the field is being measured. The radial component is the part of the acceleration that changes the magnitude of the unit vector, which is kept constant at 1 in this equation. This part is proportional to 1/r2, where r is the same distance as before. The 1/r and 1/r2 terms come from the fact that the strength of the electric field decreases as the distance increases. As the charge moves away from the point P, the electric field gets weaker and the acceleration of the unit vector decreases. Since the unit vector is constantly pointing towards the charge, its direction changes in response to the changing electric field, and has a transverse component. Since the magnitude of the unit vector is kept constant, it has a radial component as well.