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Murrr4er [49]
4 years ago
8

At what distance above the surface of the earth is the acceleration due to the earth's gravity 0.785 m/s2 if the acceleration du

e to gravity at the surface has magnitude 9.80m/s2?
Physics
1 answer:
cestrela7 [59]4 years ago
8 0
The acceleration due to gravity falls off as the inverse of the square of the distance from the center of the Earth. Thus the distance of interest is
  1/(d/R)^2 = 0.785/9.8
  d = R·√(9.8/0.785) ≈ R·3.53328
or R·2.53328 above the surface of the earth.

Wikipedia says the radius of the earth is 6371 km, so your distance is
  (6371 km)·2.53328 ≈ 16,140 km
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I₂² / I₁²   =  1/ 0.3049

I₂² / I₁²   =  3.2797

√  (I₂² / I₁² )   =  √3.2797

b) I₂ / I₁  =  1.81

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P₁´  =  V₂² / R₁

P₁  =  (127)²/R₁      and     P₁´ = (230)²/ R₁

Then the increasing factor is:

c) (230)² / ( 127)² =  3.28

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