If I travel for one day in a circle above ground at $0.99999999c$ and point a laser, will the ground see it for the whole time? First of all, I am not trying to prove or disprove anything. I know that relativity is true.
But I cannot get out of this mind experiment. It goes like this:
Let’s say I could handle the air friction and move in a circular orbit which has a 2 meter radius just above my friend. While I am orbiting, I point a very powerful and visible laser on the ground so my friend can see it. Let’s say the battery of the laser is just enough for 1 day of use. 
Now I calculated that if I move at $0.99999999c$ for one day, the time will pass to my friend as about 19 years. 
I wonder since my battery has just enough energy for 1 day, will my friend see the laser on the ground for 19 years? I mean, the energy for 19 years laser usage is known and I have as much as $1/7010$ of it. What will happen?

 A: According to your picture your friend will not see any light. It seems that your laser is directed straight down, at right angle to direction of your motion.
To give exact answer, whether your fried would see laser light, you must specify at which angle you turn your laser, because you must consider relativistic aberration of light.
https://en.wikipedia.org/wiki/Aberration_of_light
So as your friend would see laser light, you must turn your laser backward at angle $\sin\theta_s=v/c$.
At this velocity your laser should almost be "lying" on orbit.
Angles of emission and reception are tied with relativistic aberration formula:
$$ \cos {\theta_o} = \frac {\cos {\theta_s} + \frac v c} {1+\frac v c \cos \theta_s} $$
Received frequency will be:
$$ f_o = \frac {f_s} {\gamma (1+ \frac v c \cos \theta_0) } $$
Since light pulse approaches your friend at right angle $$\cos\theta_0=0$$ 
$$f_o= \frac {f_s}{\gamma}$$
So, light will be redshifted, because your clock dilates.
If your friend will point his laser at you, he must keep his laser at right angle to direction of your motion (tangential) and  the light pulse will approach you at oblique angle $\sin\theta_s=v/c$.
You will see radiation blueshifted then (since your clock dilates): 
$$ f_0 = \gamma (1 – \frac {v \cos \theta_s} {c}) f_s$$
since $$\cos\theta_s=0$$ we get
$$f_o= \gamma f_s$$
Transverse Doppler Effect: https://en.wikipedia.org/wiki/Relativistic_Doppler_effect
These effects have been confirmed by Mossbauer rotor experiments
https://en.wikipedia.org/wiki/Ives%E2%80%93Stilwell_experiment
In simplified model (we do not take into account gravitational time dilation) , if Michelson and Morley (and Bradley) were able to measure frequency of the source in the center of circumference (the Sun, or a star in zenith), they would see blueshift of radiation. The reason is actual dilation of their own clocks. Since their own clock dilates, they see frequency as increased.
This way we can see that due dilation of your clocks 19 years of your friend turn into just one day for you.
UPDATE:   If the laser pointer is directed at right angle to direction of motion of the source, the source will not be able to hit a target in the center of circumference. The source must emit laser light backward. In this case, an observer in the center would see the light redshifted.
What would be radius of circumference that this laser draws on a large white sheet of paper that lies 5 meters below the plane of rotation? 
We can consider an inertial – tangential source, which momentarily coincides with the rotating one at the moment of emission. Let they emit at the same (right) angle in their own frames, i.e. they directed their laser pointers “straight down”.  That pulse of light will always be “below” the inertial source, but in the reference frame of the paper it will have large x-velocity “away” from the observer. Apparently that means that this light pulse will travel very large distance before “landing” on the paper. Hence, radius of this circumference will be very large and it will approach infinitely large value as linear velocity of the source approaches that of light.
What would be color of the spot on the paper, if proper frequency – color of the laser was “green”?
Color of the spot will be “blue”.  We can explain “blue” color of the spot by presence of longitudinal component, as in the Relativistic Doppler Effect for moving source. However, this “blue” color will be less blue than it could be in classical case due to dilation of source’s clock, i.e. it also contains transverse component.
If “green” inertial source approaches an observer (not straight towards him, but it moves at parallel line, overhead), observer will see blueshift first.  The closer this source comes, the greater is the transverse component, the reddish will be light. At certain moment the observer will see proper “green" color. When light comes “straight from the top” the observer will see purely transverse redshift. This light was emitted by the source at points of closest approach with the observer.
References: 
Mathpages, Doppler Effect - Transverse Doppler Effect in particular.
R. C. Jennison. Ray path in a rotating system 1963 Nature No 4895 p. 739 
R  C  Jennison. Reflection from a transversely moving mirror 1974 Nature Vol 248 p. 661 
A: I appears to me that you are making a "simple" problem a lot more complicated by the ambiguous use of "day" and trying to inject relativity into the problem.  So, let me start by "over simplifying" the problem.
1- Lets start with a laser beam pointed to the ground and at a height of 5m.
The beam will shine in one spot until the battery runs out (one "normal" day).
2- Let the same beam rotate in a 2m radius, and parallel to the ground, at 30 rps.
The beam will create a visible (2m) circle on the ground until the battery runs out (one "normal" day).
3- Let the same beam rotate with a tangential velocity = .99999999c.
The beam will create the same circle as 2, but with equal or lesser luminosity than 2, until the battery runs out (one "normal" day).  The reason for this result is that the tangential velocity of the beam has no net effect on the beam's down direction.  
