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1. *Corresponding Author: Paul T E Cusack, Email: St-michael@hotmail.com
REVIEW ARTICLE
Available Online at
Galaxies, Asteroids, and Solar Flares
www.ajms.in
Asian Journal of Mathematical Sciences 2017; 1(3):170-173
*
Paul T E Cusack1
*1
Figure 1: The shape of galaxies (a) & (b)
SHAPE OF GALAXIES
Consider the shape of a galaxy. Refer to Figure 1 (a). The “arms” spin out of the center because of the
radial acceleration. Refer to Figure 1(b). We show below the mechanical calculations behind this
phenomenon.
P=Mv=F=Ma
Ma=omegaR
Ma=d theta/dt *R
BScE, DULE, 1641 Sandy Point Rd., Saint John, NB E2K 5E, Canada
Received on: 29/03/2017, Revised on: 21/04/2017, Accepted on: 25/05/2017
ABSTRACT
I provide here a few calculations on the shape of galaxies; on possible asteroids collisions; and on the
cause of solar flares. Galaxies have the shape they do because the universe rotates at d theta/ dt=0.86/04.
The asteroid should be between Neptune and Jupiter that could cross Earth’s path every 65 million years.
Solar Flares can be modeled by Fluid Mechanics, namely Bernoulli’s equation.
Keywords: Galaxies, Asteroids, Solar Flares
INTRODUCTION
In this paper, we examine three problems proposed in the list of unsolved problems in Astronomy and
Astrophysics, namely why galaxies have the shape they do; the presence of an asteroid that may collide
with Earth; and the cause of Solar flares. Post this paper; there are researchers at Caltech who suggest that
there is a planet between Neptune and Uranus that is ten times the mass of the earth. We begin with why
galaxies are the shape that they are.

2. Paul T E Cusack et al. Galaxies, Asteroids, and Solar Flares
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M 0.86=0.86/0.4 *R
R/M=0.4
For a galaxy that is 1 cm scale =1.5 arc sec
1.5sec /60 sec =0.025=R
R/M=0.4
0.025/M=0.4
M= 16
omega=dtheta/dt=0.86/0.4=2.15
Omega R=2.15R
R=0.4 M =0.4(16) = 6.4
Omega=2.5 (0.4M) =0.86 M=0.86 R=2.15
For a certain galaxy: 1 cm=1.5 arc sec
1.5 sec /60 sec =0.025
Ma=omega R
a=0.86=0.86 (R/M) = R/M=1
M=2.15=R
2.15 *2Pi=13.50
1-13.50=0.8650 cf 0.866omega R=2.15R
r/M =0.4
R=0.4M
2.5(0.4M)=0.86
R=2.15
omega R=Ma
a=0.86*2.15/M
a=0.86*2.15 /2.15
a=0.86
v=a=omega=0.86
THIS THEN IS WHY THE GALAXYS HAVE THE SHAPE THEY DO. ITS BECAUSE THE
UNIVERSE ROTATES AT dtheta/dt=0.86/0.4
POSSIBLE ASTROID COLLISION
MEAN ANOMALY
M=360 Y/Tr=360(65,000,000)=1.047
1/1.047=0.9551
0.9551*360=343.8 degrees
343.8-180=163.8-29.4 (Congenital drift) =134.4 degrees
OR
M= 5000 *2Pi/65=0.4833215
0.4833215 *360=173.99
180-173.99=6.0042
6.0042-29.4=35.4
100degrees longitude +35.4=135.4 degrees (a difference of 1 degree)
I calculate that the asteroid is 1.618 billion km away on Easter 2014 April 20.
It is travelling at 30 km/sec on an arc of 3.4%.
The asteroid should be within our solar system at 3.223 billion km. It is inside Neptune at 4.43 km.
The universe is a ellipsoid 22 ly x 2.997929=65.95 [1]
65.95-65=0.9544
0.9544*360 degrees=343.5977
343.5977-180-29.24=134.3577
135.3577-180=45.6 degrees
a X b= ||a|| ||b|| sin theta
=sqrt 3 * G * sin 1
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3. Paul T E Cusack et al. Galaxies, Asteroids, and Solar Flares
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=1.73 *0.6666* 0.866
=1
=t
The extinction asteroid could hit in Hollywood.
65,000,000 * 0.2422 days per year=15743000 days
15743000/365.2422=43403.9279 years
Remainder 0.9279 years =338.9155 days
0.9155 days = 21.97 hours
329.5710 degrees-180=149.5710 degrees
Continental drift
5 cm/year*65,000,000=2019 miles
2019miles /24860 miles=0.0812
0.0812 *360 degrees=29.24 degrees
149.5710-29.14=120.333 (cf Hollywood 118.33)
EXTINCTION ASTEROID:
The asteroid could be between Neptune and Jupiter. Refer to Figure 2 for a possible collision path
trajectory.
x^2/a +y^2=R2/c
x^2+y^2=91.445^3
x^2+y^2=1.3077
0.13077 cf 0.869
Figure 2: Possible Asteroid Collusion Path
SOLAR FLARES
Solar Flares may be able to be modeled by Bernoulli’s Equation from Fluid Mechanics. Refer to Figure 3
for the azimuth plot.
P/t=mgh +1/2 mv^2
P=i^2 R
i^2R]/.[mt]=gh * 1/2 v^2
1/3^2(R)/[m*0.4083]=9.81 (1000000) +1/2 (100000)^2
R/m=0.3141
m/R=B=31.839
B=31.8 (See above)
Figure 3: Solar Flares
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4. Paul T E Cusack et al. Galaxies, Asteroids, and Solar Flares
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© 2017, AJMS. All Rights Reserved.
So the flux on the solar flare is equal to the energy released by the gravitation of Earth. The pressure at
the Suns surface must be zero. It’s like a fire hose. p=0 at the hose end. Solar flare results.
Vol=4/3 Pi R^3 4/3 Pi (100000) ^3 4.1888 x 10^25
E rho=E/t=4/0.4083=9.7967
E rho /t=4/0.4083
E rho *Vol=41.000
B=R/M=31.8
31.8 /41=1.289=1.3 =di/dt
R/M=0.318
R/4.496=0.318
R=0.1426
1-0.1426=0.858 cf 0.86
CONCLUSION
So we see that there are still new consideration using basic mechanics to understand how our solar system
came to be the way it is.
REFERENCES
1. Cusack, P., Astrotheology, Cusack’s Universe, J. of Physical Mathematics., OMICS, Jan. 2016.
2. Beer, F. P., et al, Mechanics of Materials 7th Ed.. McGraw Hill, 2015.
3. Aris, R., Vectors, Tensors, and the Basic Equations of Fluid Mechanics. Dover.`990
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