How Does Height Affect Weight in a Hypothetical Mile-High Building?

[dominus96]

Homework Statement

In 1956, Frank Lloyd Wright proposed the construction of a mile-high building in Chicago. Suppose the building had been constructed. Ignoring Earth's rotation, find the change in your weight if you were to ride an elevator from the street level, where you weigh 520 N, to the top of the building.

Homework Equations

a = Gm/R^2

The Attempt at a Solution

g = Gm/R^2, so I used 6.67E-11 for G, the mass of the Earth (5.97E24) for m, and the radius of the Earth plus the distance from Earth's surface (1 mile, which is about 1609 meters) for R. I calculated all that and got about 9.77 for g and then found weight and subtracted it from 520, but it was wrong. What's the deal?

 

[borgwal]

You don't need the mass of the Earth, nor the value of G: the fractional change in g (which gives you the fractional change in weight) equals the fractional change in 1/r^2, with r the distance to the center of Earth (so, that number you do need).

The fractional change in g that you get is way too large (rounding errors?).