Are they any tables or charts that show the relative affect of gravity vs distance from sea level?

Question

Ideally, I am looking for something that will state:

0 sea level = 100 lbs
1 mile above sea level = X lbs
5 miles above sea level = Y lbs
50 miles = Z
etc.

I'm not overly concerned in the increments or type of measurement (miles, kilometers, pounds, kilograms, etc.) What I am specifically looking for is the ratio at which perceived weight at sea level decreases as distance from the Earth increases.

Answer 1

No, but you can easily calculate it yourself. Gravity obeys the inverse square law, which means double the distance is a quarter the gravity. The trick is that it is distance from THE CENTER OF THE EARTH and not from sea level. Earth is about 8,000 miles in diameter, so at sea level you are 4,000 miles from the center. Skipping all the steps to arrive at the final equation, the formula for the weight at different altitudes (of an object that weighs 100 pounds at sea level) is:
W=100*4,000^2/(4,000+H)^2

For example, 4,000 miles up (double the distance from the center of the Earth) you would expect 25 pounds, and:
W=100*4,000^2/(4,000+4,000)^2
W=100*4,000^2/8,000^2
W=100*16,000,000/64,000,000
W=100*0.25
W=25
works out!

Now at 1 mile it is:
W=100*4,000^2/(4,000+1)^2
W=100*4,000^2/4,001^2
W=100*16,000,000/16,008,001
W=100*0.9995
W=99.95

At 100 miles you get:
W=100*4,000^2/(4,000+100)^2
W=100*4,000^2/4,00^2
W=100*16,000,000/16,810,000
W=100*0.9518
W=95.18

You can easily calculate any other altitude you like.

Answer 2

Although this should be easy enough to calculate, the answer probably wouldn't be accurate in practice. The variation in gravitational attraction due to these differences in altitude would be pretty small and would probably be overshadowed by variations in the density of materials making up the ground in a given area.

Highly sensitive instruments can actually detect subsurface cavities such as caves and tunnels due to the tiny variations in gravity caused by the "missing" mass in those areas. Likewise a large deposit of something like lead would probably increase the gravitational force in an area by a tiny amount.

Also, you may need to take into account the slight variation in centripetal force caused by the rotation of the Earth. At higher elevations, this force should be a tiny bit higher. The exact amount of variation would depend on how close you were to the equator or poles, however.

Answer 3

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Answer 4

http://physics.webplasma.com/physics10.html

Answer 5

(A constant called G) x (mass of first object) x
(mass of second object)

Force =
---------------------------------------------------

(divided by the square of the distance between
them)