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COINCIDENTALLY

In the first edition of Book Two of this series, I postulated dropping an object for a distance equal to the earth's radius within a constant gravity equal to that found on earth's surface. The impact velocity of such an object exactly equalled the velocity of nether passing through the earth's surface. I wondered why this was so.

The following explains why the correct equation can be obtained in this manner. The information is not exactly necessary, but it does constitute an easy way to remember the gravity versus the instantaneous velocity for a planetary surface.

The E_p of an object dropped to the center of the earth as if there were no impediment (as is the case with the nether), and in a constant gravity equal to g_e, has the equation

E_p = Fr_e.         F = mg.

So:     E_p = mg_er_e

        E_k = E_p         E_k = (1/2)mv_i²         where v_i is impact velocity

(1/2)mv_i² = mg_er_e

  (1/2) v_i² = g_er_e

          v_i² = 2g_er_e

          v_i = (2g_er_e)^1/2

The impact velocity is the same velocity as that of the nether at the surface of the earth. The reason this works is as follows.

B = 1/[(r_e/r_c) +1]           Because r_c is nearly 0,   r_e/r_c is nearly
                                      infinite, +1 is negligible, and B = r_c/r_e.

g_ave = Bg_e + (1 - B)g_a           Here g_a is g_e,   so

g_ave = Bg_e + (1 - B)g_e

g_ave = Bg_e + g_e - Bg_e

r_c/r_e is nearly zero.     Usually g_c would be used in lieu of the first g_e,   and g_c is much larger than g_e,   but g_c was not used.     So Bg_e is essentially zero.

g_ave = 0 + g_e - 0

g_ave = g_e

(1/2)v_e² = g_aver_e

v_e² = 2g_er_e

v_e = (2g_er_e)^1/2       which can be generalized to v = (2gr)^1/2

This explains why the method works that was used to find equation 6 on page 18 of the first edition of Book Two of the BLI series (on Gravity).
 

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