Look at Newton's law of universal gravitation:
F = G·m·M / r²
This says that the FORCE of gravity is proportional to 1/r² (an "inverse square law").
Now, you also know, from Newton's 2nd law, that F = m·a. So set these 2 expressions equal to one another:
G·m·M / r² = m·a ──► a = G·M / r²
This shows that the ACCELERATION due to gravity also goes as 1/r². So, for example, if you double your distance from the center of the earth (that is, you travel 6.37x10^6 m straight up), the acceleration you would feel would only be a quarter that of what you feel on the surface.
A consequence of this result is that as you go higher and higher in the atmosphere, the velocity required to stay in orbit decreases. To stay in orbit, your linear velocity must be such that your centripetal acceleration and the acceleration due to gravity are equal:
a = v² / r = G·M / r² ──► v² = G·M / r
v = √(G·M / r)
So the velocity goes as 1/√(r).
2007-07-25 04:40:21
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answer #1
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answered by Anonymous
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Gravity field is defined as weight per unit mass.
If you are moving away from the Earth against the field you would experience a decrease in the gravity field.
The reason is that the gravity field is directly proportional to the mass which experiences the gravity field and inversely proportionl to the square of the distances between the two center of masses of the two interacting masses of which the field is born.
So Acceleration is a description of the gravity field.
At high jet plane altitude or on top of a high mountain a heavy person would weight less that at Sea level
2007-07-25 11:51:21
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answer #2
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answered by goring 6
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Blue is right..and that equation is right too.
basically what he's trying to say is the strength of gravity is inversely squared to the distance from the earth.
So it is the force of gravity divided by the [X^2]
Where X is the distance from the earth
SO..it varies by decreasing by the power of 2 as you go further from earth
2007-07-25 11:48:34
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answer #3
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answered by Anonymous
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It decreases slightly because of the separation from the center of mass.
2007-07-25 11:33:50
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answer #4
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answered by Pfo 7
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