If the Sun moves across the ground at 0.25 nautical miles per second, why do shadows move very much slower? [closed]

This title has been discussed previously on this site here.

Question: Shadows from the sun moves fairly slow across the ground. Maybe a centimeter per sec. however, I calculated that the sun is moving .25 Nautical miles per sec. over the circumference of the earth 21600 NM. Why does the shadow move so slow?

• The speed of the shadow will depend on the height of the object causing the shadow, and the time of day - with the shadow moving faster the nearer the time to dawn or dusk, and slowest at local noon. Where does 1 cm per second come from? Nov 2 '18 at 19:52
• The Sun does not move over the circumference of the Earth. What you are talking about is a particular projection, of the Sun onto the surface of the Earth, and when you talk about the moving shadow, you are talking about a different projection. Nov 2 '18 at 21:22
• I don't understand (at all) the relevance of the question's title and first sentence to what's actually being asked. You should either remove those parts or do some serious restructuring to make their relevance clear. Nov 14 '18 at 15:35
• Even if the computation was based on correct assumptions/estimates and methods, I don't see how "Why does the shadow move so slow?" is a reasonable question: the math says so, and intuition isn't a great tool for such orders of magnitude.
– user191954
Nov 14 '18 at 15:48

The Sun moves across the sky at a constant angular velocity, and that means that the shadows it casts also move at a constant angular velocity. This means, in turn, that the linear velocity of a given shadow on the ground will be directly proportional to how far it is from the object that's casting the shadow. Thus, if in a given situation a $$1\:\rm m$$ vertical pole casts a shadow that moves at $$1\:\rm cm/s$$ on flat ground, the shadow of a $$2\:\rm m$$ pole at the same location will move at $$2\:\rm cm/s$$.
The linear speed that you want, $$0.25\:\rm M/s$$, corresponds to a pole height equal to the radius of the Earth (give or take).