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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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Answer:

As follows,

Explanation:

KE=1/2mv^2

In 1st question,

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In 3rd question,

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For 4th,

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b.Speed/velocity gained

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