New weight at 1 mile in the sky

[pwhite4541]

1. 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 600N, to the top of the building.
2. (1) F = GmM/R^2 and (2) F= GmM/(R+h)^2, where h is the height in meters above the earth.
3. I figured that if I could calculate the acceleration due to gravity at 1 mile above Earth's surface, I could figure out the new weight. I used the equation above that includes h and got the answer wrong. The solution divides formula 1 by 2. I don't understand why my method did not work or why this method does work. Help!

 

[Chestermiller]

Show us the details of what you did. Both methods are equivalent, so you should have gotten the same result.

 

[pwhite4541]

Determined the mass to be 61.22 kg, then m = 61.22.

F = GMm/(R+h)^2 = (6.67e-11)(5.98e24)(61.22)/(6.37e6+1609.34)^2 = 601.48 N

601.48 N is the new weight at 1 mile above the earth? Shouldn't it be slightly less than the original of 600 N?

(61.22)(g) = 601.48, g= 9.82 < -- Gravity is stronger at 1 mile above the earth? I got this result and stopped the problem to re-examine my arithmetic and my understanding of the equations. I'm not sure what I am doing wrong.

 

[nasu]

Well, you take g=9.80 m/s^2 to calculate the mass and then use some values for radius and mass of the Earth to calculate the weight.
The problem is that these values are not compatible. With these values for radius and mass, you get g=9.83 m/s^2.
It's a small difference but the effect you are looking for is also small.
As the problem does not give you the radius and mass of the Earth, it is better to solve without assuming specific values, when possible.
If you estimate the ratio, even if the value for the radius of the Earth is a little off the result will be in the right direction.