Static condition for the formation of earth’s surface under gravity

doi:10.15406/oajmtp.2018.01.00004

Open Access Journal of

eISSN: 2641-9335

Mathematical and Theoretical Physics

Opinion Volume 1 Issue 1

Static condition for the formation of earth’s surface under gravity

Paul TE Cusack

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Park Ave, Saint John, Canada

Correspondence: Paul TE Cusack, BScE, Dule 23 Park Ave,Saint John, NB E2J 1R2, Canada

Received: February 02, 2018 | Published: March 2, 2018

Citation: Cusack PTE. Static condition for the formation of earth’s surface under gravity. Open Acc J Math Theor Phy. 2018;1(1):19-20. DOI: 10.15406/oajmtp.2018.01.00004

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Opinion

When comparing the centripetal acceleration of the Earth which tends to spin the Earth’s mass outward, with the gravitational forces pulling it together, we see an interesting result.

centripetal force

$F_{C} = M a = M_{C} (ω ² r)$ (1)

$ω ² = (d θ / d t) ² = (40, 075 k m / 24 h o u r s) ² = 1669 ² k m / h r$

gravitational forces

$F_{G} = g G M_{1} M_{E} / r ²$ (2)

Set the two equal to determine the static conditions for the surface of the planet¹

$F = g G M_{1} M_{E} / r ² = F_{G} = M a = M_{C} (ω ² r) C$ (3)

$(N o t e : G = 6.52 n o t 6.67 = L n π / 1.618)$

$6.52 (1) (5.972) / (6371) ² C = (5.972) (1669.79) ² (6371)$

$6.52 C = (1669.79) ² (6371) ³ = 721.07$

$C = 3.007 ~ c = s p e e d o f l i g h t$ (4)

$2.9979 / 3.007 = 99.69 %$ Accurate

Now Using $G = 6.67$ (5)

$g 6.67 (1) (5.972) / (6371) ² = (5.972) (1669.79) ² (6371)$

$6.67 g = (1669.79) ² (6371) ³ = 721.078$

$g = 1.081$ Atomic Mass of Hydrogen (6)

And

$6.52 C = 721.078$

$C = 3.007 ~ c = t ²$ (7)

$1.081 \times 3.007 / 6.52$

$= 0.498 ~ 0.5$

$= 1 / 201$

$= 1 / Y f r o m A T M a t h$ (8)

$= t$

Continuing,

$E = 1 / t$ (9)

$t = 1 / E = 1 / Y$

$1 / 2.01 = 0.498$

And $E = M c ²$ (10)

$1 / E = 1 / (M c ²)$

Let $M = 1$

$E / 1 / c ²$

Now, the distance D.E.:

$d = v_{i} t + 1 / 2 a t ²$ (11)

Let $v = a$

$C = c = v = 3 = a$

$t ² = 3$

$d = 1 / 2 (3) (3)$

$= 9 / 2$

$= 4.5$

Figure 1

Figure 1 Static Equilibrium of acceleration to gravity with a hollow core.

Now Circumference=Area

$2 π R = π R ²$ (12)

$R = 2$

$R_{E} = 6371$

$6.371 - 2 = 4.371$

Let $s = d = 4.371$

$d = 4.371 - 4.5 = v_{i} t + 1 / 2 a t ²$ (13)

$v_{i} t = 0.271 = e = A T$ Math Energy

$v_{i} (\sqrt t) = 2.718 - 1.73$

$= 1.0$

$d = 1.0 + 4.5 = 5.5$ (14)

$4.371 - 5.5$

$= 1.271$

$~ 4 / π$

$= ρ$

AT Density

We see that the constant is very close to $c$ , the speed of light, within the marginal error of significant digits. The $\sqrt c = e i g e n v e c t o r$ from AT Math.