Atwhat altitude above Earth’s surface is free fall acceleration equal to three quarters of its value at surfac
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Atwhat altitude above Earth’s surface is free fall acceleration equal to three quarters of its value at surfac

[From: ] [author: ] [Date: 11-05-25] [Hit: ]
g{ 0 } = g_e (surface) ≈ 9....g{ h } ⁄ g{ 0 } = (Re)² ⁄ (Re + h)² = 0. (Re)² ⁄ (Re + h)² = 0.......
The force of gravity is determined by the equation:

 Fɢ = G • (Me) • (Mo) ⁄ (Re + h)² = (Mo) • g

 ... where: mass of object = Mo

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~…

The acceleration due to gravity or "g" is:

  g{ h } = G • (Me) ⁄ (Re + h)²  ...  equation 1
  g{ 0 } = G • (Me) ⁄ (Re)²  ...  g{ 0 } = g_e (surface) ≈ 9.8 m/sec²

 ... where: h = height or altitude of object

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~…

 g{ h } ⁄ g{ 0 } = (Re)² ⁄ (Re + h)² = 0.75

        (Re)² ⁄ (Re + h)² = 0.75

    (0.75)(Re)² + (1.5)(Re)h + (0.75)h²  = (Re)²

     (0.75)h² + (1.5)(Re)h − (0.25)(Re)² = 0  ... Re = 6.37 ×10^6 m

 (0.75)h² + (9.555 ×10^6)h − 1.014 ×10¹ ³ = 0

  h = [ -b ± √ ( b² − 4ac ) ]   ⁄   (2a)
  h = [ -9.555 ×10^6 ± 11.03 ×10^6  ]   ⁄   1.5
  h = 985 ×10³ m
  h = 985 Km  ...  altitude (only positive solution)

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~…

Check of Solution:
     g{ h } = G • (Me) ⁄ (Re + h)² ... equation 1

  g { 985 ×10³ } = (6.673 ×10 ‾ ¹ ¹ ) • (59.8 ×10² ³ ) ⁄ [ (6.37×10^6) + ( 985×10³ ) ] ²

  g { 985 ×10³ } = 7.38 m/sec² ... which is about 75% of g{ 0 } ≈ 9.8 m/sec²
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