Applying Newton's Universal Law of Gravitation

AI Thread Summary
To determine the height (h) at which a person's weight is reduced to one-third of their weight on Earth's surface, Newton's Law of Gravitation is applied. The calculations show that the radius (r) at this height is approximately 11,085,125.17 meters. Subtracting Earth's radius (6,400,000 meters) yields a height of 4,685,125.17 meters. Additionally, the acceleration due to gravity (g) at this height is calculated to be approximately 9.37 X 10^-3 m/s². The solution confirms the correct application of gravitational equations to find both h and g.
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Homework Statement


This problem only requires to find h, but I want to make sure that I also found Me and g correctly.
Starting with Newton's Law of Gravitation, determine the height h one person has to go from the surface of the Earth in order for that person's weight to be reduced to one-third of his/her weight at the surface of the earth. The only information you have is that the radius of the Earth is 6.4X106m. Knowing that the acceleration due to gravity at the surface of the Earth is 9.8m/s2, determine the acceleration due to gravity at the point were the weight of the person is reduced to 1/3rd.

Homework Equations


GMem/r2 = 1/3GMem/Re2
r2 = 3Re2
r = √3(6,400,000)
h = r - Re = 11,085,125.17m - 6,400,000m = 4,685,125.17m
Me = gRe2/G
g = G9.0 X 10-9/6.4 X 106

The Attempt at a Solution


GMem/r2 = 1/3GMem/Re2 Cross out the GMem, leaving: r2 = 3Re2
r = √3(6,400,000) = 11,085,125.17m
h = r - Re = 11,085,125.17m - 6,400,000m = 4,685,125.17m
6400km X 1000m/1km = 6.4 X 106m
h = 4,685,125.17m,
Me = gRe2/G = 9.8m/s2 X (6.4 X 106)2/6.67 X 10-11 = 9.0 X 10-9
Acceleration due to gravity at the point were the weight of the person is reduced to 1/3rd = g = G9.0 X 10-9/6.4 X 106 = 9.37 X 10-3m/s2
 
Last edited:
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science_rules said:

Homework Statement


This problem only requires to find h, but I want to make sure that I also found Me and g correctly.
Starting with Newton's Law of Gravitation, determine the height h one person has to go from the surface of the Earth in order for that person's weight to be reduced to one-third of his/her weight at the surface of the earth. The only information you have is that the radius of the Earth is 6.4X106m.

Homework Equations


GMem/r2 = 1/3GMem/Re2
r2 = 3Re2
r = √3(6,400,000)
h = r - Re = 11,085,125.17m - 6,400,000m = 4,685,125.17m
Me = gRe2/G
g = G9.0 X 10-9/6.4 X 106

The Attempt at a Solution


GMem/r2 = 1/3GMem/Re2 Cross out the GMem, leaving: r2 = 3Re2
r = √3(6,400,000) = 11,085,125.17m
h = r - Re = 11,085,125.17m - 6,400,000m = 4,685,125.17m
6400km X 1000m/1km = 6.4 X 106m
h = 4,685,125.17m,
Me = gRe2/G = 9.8m/s2 X (6.4 X 106)2/6.67 X 10-11 = 9.0 X 10-9
g = G9.0 X 10-9/6.4 X 106 = 9.37 X 10-3m/s2
 
Last edited:
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