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Second PROOF for GRAVITY EQUATION

 

We know that for an object in space not to be affected by earth's gravity, it must be at an infinite distance, "r_a", from earth's surface, "r_e". To fall to earth, this object must be pushed slightly. It will eventually strike the earth with an impact velocity, "v_e", which can be found because we know its kinetic energy, "E_k", on impact, is equal to its original potential energy, "E_p". Average gravity during the fall is "g_ave". Gravity at earth's surface is "g_e". Gravity at the infinitely remote point is "g_a". "F" is force.

E_k = E_p
E_k = (1/2)mv_e²
E_p = Fr_a
F = ma or mg_ave

(1/2)mv_e² = Fr_a

(1/2)mv_e² = mg_ave r_a

m divides out of the equation.

(1/2)v_e² = g_aver_a

v_e² = 2g_aver_a

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From equation on page 8:

B = 1 / [(r_a/r_e) + 1]

When r_a is infinite, r_a/r_e is so
great that "+1" is negligible, so

B = 1 / (r_a / r_e) = r_e / r_a

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From equation on page 8:

g_ave = Bg_e + (1 - B)g_a

At distance r_a, g_a is zero, so

g_ave = Bg_e = (r_e / r_a)g_e

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From above:

v_e² = 2g_aver_a

v_e² = 2(Bg_e)r_a

v_e² = 2[(r_e / r_a)g_e]r_a

v_e² = 2r_eg_e

v_e = (2r_eg_e) ^1/2

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The energy needed for a rocket to escape from earth is the same as its kinetic energy from falling from an escape distance, so v_e is its escape velocity.

Since the rocket must have fallen at the same rate as the nether surrounding it would have fallen, the velocity of the nether at the earth's surface would also be v_e.
The equation v_e = (2r_eg_e) ^1/2 can be generalized as
v = (2rg) ^1/2 or g = v ² / 2r.

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So the escape velocity for any celestial body at any point is the same as the instantaneous velocity of the nether moving past at that point.
 

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