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Eng Dule-1641, Sandy Point Rd, Saint John NB, Canada

- *Corresponding Author:
- Paul TE Cusack

Eng Dule-1641, Sandy Point Rd

Saint John NB, Canada

**Tel:**(506) 214-3313

**E-mail:**St-michael@hotmail.com

**Received date:** October 10, 2016; **Accepted date:** January 02, 2017; **Published date:** January 10, 2017

**Citation: **Cusack PTE (2017) A Revised Solar System. J Astrophys Aerospace Technol 5:134. doi:10.4172/2329-6542.1000134

**Copyright:** © 2017 Cusack PTE. This is an open-access article distributed under the terms of the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original author and source are credited.

**Visit for more related articles at** Journal of Astrophysics & Aerospace Technology

I hereby suggest that there may have been plants that were moons of other planets. Mercury may have been a moon of Venus; Pluto may have been a second moon to Earth. We also see an explanation as to where the new planetV113 may have come from. Finally, we see the calculation for the inclination of Venus.

Moons; Venus; Earth; V113; Pluto; Mercury

I hereby suggest a possible new model for the Planets in our Solar System [1,2]. I calculated that:

• V-113 the new Pluto was the second Moon to Earth.

• Pluto was the second Moon to Earth.

• Mercury was a moon to Venus.

• The inclination of Venus.

**V113 - New Planet**

Refer to **Figure 1** for showing the moon as the new planet V113 and
the forces acting upon it.

F = ks

0.1525 = k (0.0655)

s = 0.0655 = 6.55%

The tilt on the elliptic orbit of Mars = about 7%

V113 is 1.239 billion kilometers from Earth. So, 1239+149.9 = 1388.6 (6.78 degrees tilt).

V113 was the Moon planet to Mars.

F = GM1M2/R^2 = 0.666 (58.9) (0.639)/(0.129)^2 = 0.1508

tan 24.59 degrees = 0.4526/30 km/s = 0.015 = 0.1508

Refer to **Figure 2** for the trajectory of the possible path of how the
Earth’s second moon is Pluto.

F = GM1M2/R^2 = 0.666(0.736)(5.98)/384000^2) = 1.9825

F = GM1M2/R^2 = 0.666(1.219)(5.98)/332000^2 = 3.0120

F T. = 1.9825+3.0120 = 4.9945 = 5

FE = GM1M2/R^2 = 0.666(5.98)(1.989)/ 149.6^2 = 6.67

F Net = 5-6.7 = 1.7

s = 0.3+0.4+2^n = 1.7

F = ks, k = 1

M Pluto = 1.3 cf 0.1334 of moon of Venus

Earth Inclination = 23.6 degrees.

Tan 23.6 degrees = 0.4327

M Earth = 5.98 Mt

0.4327*5.98 = 2.5878 = P

P1 = M1V1 = P2M2V2

P1 = 2.5878 = 5.98v

v = 0.4327

**Now Pluto**

P = MV

0.5522 = 1.29 (0.4327)

P 1 = 1.29(30 km/s) = 38.7

P2 = 38.7*0.5532 = 6.995 = 7

v2 = 4.66 P = 1.29(4.66) = 6.0

delta MV = 1

The energy goes into the rotation (original rotation neglected as it is small).

Mv = 1 = 1.29v

v = 0.77

**Rotation**

Omega = 48.7

48.7/0.77 = 0.01590

So, 1/6.99 -0.01590 = 0.01430-0.1590 = 1.618 Golden Mean

Considering Pluto was a Second Moon of Earth,

F = G M1M2/R^2 F = 1.29(5.98)/R^2 = 7.7142

F = 1.29(1.989)/R^2 = 256452

R = 256452/7.7142 = 332441; cf Moon = 384406 km

Refer to **Figures 3** and **4** for the illustration of the forces acting on
the moon.

F = ks = GM1M2/R^2

F = s

k = 1

(1)s = F = G k = 2/3 (1/1.3) = 0.5128

k = 0.5

0.5 s = G = 6.67

s = 0.1334

cf F net = m 0.1334 for Venuses' Moon Mercury z

0.4+0.3 +2^n the elongation of Venus averages 47 to 45 degrees. So 0.1420-0.1334 = 86

86 *7.14/7 degrees = 1.02 = 45.6 degrees = 46 degrees

Using Energy from the dampening of the Magma inside Mercury, we have:

Dampened cosine Curve

E = Y = e^-t cos theta

= 0.2016 = e^-t cos theta

0.2016 = e^-t cos 60 degrees

E^-t = 0.4032

We also know,

Work = E = F*d

F = Ma

If a = v

F = Ma = M sin theta = [e^-t cos theta]/d

M cos theta = e^-t cos theta/d

M = e^-t/d = 0.4032/d

Mass of Mercury = 3.301

So,

3.301 = 0.4032/d

D = 1.2214

Since the D = 1.2214

Since the push and pull can be modelled as cos = sin,

1/0.8415 = 1.1884

Delta d = 3.300%

Now the tile is 0.6 degrees/ 29.124 = 2.062%

2.0602 ×3.300 = 6.7985 degrees

Tilt = 6.776 degrees

6.7985/6.776 = 1.0033 = 1+ delta d

So, the dampening motion of the magma inside Mercury was enough to cause the planet to move to a different orbit about the Sun.

Refer to **Figure 5** for an illustration of the title on Venus.

Venus is at 6 degrees inclination on its own axis. Conservation of P = M1V1 = M2V2

P1 = (3.3+48.7) = 52.0

P2 = 48.7

P1/P2 = 52/48.7 = 1.06776

52-48.7 = 3.3

3.3/48.7 = 0.06776

0.06776*90 degrees = 6.089 degrees = 6 degrees true

The Moon is on a 5-degree ecliptic. Pluto is on a 17-degree elliptic. Added together it is 22 degrees. This is off the Earth's tilt by 1.4 degrees. I suggest Pluto is at (17+1.4) = 18.4-degree tilt to the elliptic or the moon has dropped by 1.4 degrees from 6.4 degrees to 5 degrees which is more likely. Perhaps the swell of the Earths oceans is slowing down the Moon's elliptic? It absorbs energy lost in heat and friction to move the oceans on earth- 28% over 5 billion years.?

28%*0.834 = 23.4 degrees-tilt of Earth

5.6 × 10^-11

G = 6.67 ×10^-11

0.839G = sin 1 * G

sin 45 degrees * 1/sin 1 = 1/sqrt2*1/0.839 = 0.8428 = sin 1

The 2^{nd} moon was 90 degrees off the lunar moon. Sin 1 = Cos 1
shifted by 90 degrees (1 quasackian = 45 degrees)

Pluto is inclined at 66.66 degrees. 66.66-17 = 49.66 degrees 1-49.66 = 50.34 degrees cf 51.28 1.4 degrees (see above paragraph) (1) s = F = G k = 2/3 (1/1.3) = 0.5128.

So, that's it. Pluto is Earth's second moon gone astray.

We see that there are possible 3 planets that may have been moons of other planets.

- Wikipedia (2010) List of unsolved problems. Astronomy and Astrophysics.
- Cusack PTE (2010) Conservation with Wm. J. Cusack RE: Plant 113.

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