Question

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 757 N, to the top of the building.

Answer 1

According to Physics, the weight (w) depends on the mass of the body and the acceleration of gravity as the following equation:

w = mg

The earth is always constant in the same object, regardless the place or planet where they find. As we increase the height, g is decreasing, so that the weight (w) also will do it.

To calculate more accurately the weight variation according to the distance between it and the Earth's surface, let´s use this equation:

w = G (m (object) x m (Earth)) / (r(Earth´s radius) + (height)) ^ 2

Where   G = 6.67384 X 10 ^-11 N m^2 / Kg^2, the mass of the Earth is 6 X 10 ^24 Kg and radius of the Earth is 6470 000 meters.

To calculate mass of the object and height let´s use the data given in the problem:

weight at street level = 757 N = m object g using g= 9.81 m/s2 and clearing out m object:

m object = 757 N / 9.81 m/s^2 = 77.16615698 Kg

height = 1 mile = 1 609. 344 meters

So let´s occupy all the previous data in the proposed formula:

w = G (m (object) x m (Earth)) / (r(Earth´s radius) (height)) ^ 2

w = (6.67384 X 10 ^-11 N m^2 / Kg^2) ( ( 6 X 10 ^24 Kg) ( 77.16615698 Kg) / (( 6470 000 m + 1 609. 344 m ))^2

Solving the last expression:

w = 737.7841591 N   We can observe weight is reduced and it fulfills our statment established above whe we said:

"As we increase the height, g is decreasing, so that the weight (w) also will do it" The weight was reduced about 20 Newtons that is about 2.6% of the weight registered at street level.