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Acceleration Due Gravity Essay

Hi George,

From the information given in your question, Newton is the way to go on this.


Near the surface of the Earth, the acceleration due to gravity is ≈ 9.8 m/s2 (that's 9.8 meters per second per second). If you drop an apple from an airplane, skyscraper, or roof of your house, the acceleration due to gravity will always be ≈ 9.8 m/s2.


I respectfully disagree with Vivian:

  a) Galileo demonstrated (in his famous Leaning Tower of Pisa experiment in 1589) that heavy and light objects take the same time to fall to the ground. But for the friction effect of air, a feather and a cannon ball would take the same time to fall from the tower.


  b) The earth acceleration rate due to gravity is NOT the same universally. Earth's gravity decreases with altitude according to the "inverse square law" - the greater your altitude, the lesser is the earth's gravitational effect.


The inverse square law formula for gravity on earth is: g ≈ 9.8 (4000/(4000+h))2. where "h" is the altitude (in miles) above the surface of the earth (radius about 4,000 mi). Here are some examples:

  • In an airplane at 30,000 feet (5.7 miles), g ≈ 9.77 m/s2 (almost identical to surface of earth).
  • At about 235 miles (altitude of ISS satellite), g ≈ 8.74 m/s2.
  • At about 1,000 miles (inner space), g ≈ 6.3 m/s2.
  • At about 125,000 miles (half way to the moon), g ≈ 0.01 m/s2.


High school physics classes will never ask you about relativity, accounting for air friction, or high altitudes.

Don't over-engineer the problem and worry about all those vectors vectors.





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