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.

 

[Nickie]

I would like to clarify that the weight of an object is determined by its mass and the acceleration due to gravity at a particular location. Therefore, the weight of an object will change as it moves to a different height on Earth's surface. In this case, if the mile-high building was constructed, there would be a change in your weight as you ride the elevator from the street level to the top of the building.

To calculate the change in weight, we need to consider the equation for gravitational force (F) between two objects, which is given by F = GmM/R^2, where G is the universal gravitational constant, m is the mass of the object, M is the mass of the Earth, and R is the distance between the object and the center of the Earth.

At the street level, the distance (R) between you and the center of the Earth is the radius of the Earth (RE). Therefore, your weight (W) at the street level is given by W = GmM/RE^2.

At the top of the building, the distance (R+h) between you and the center of the Earth will be RE + h, where h is the height of the building. Therefore, your weight (W') at the top of the building is given by W' = GmM/(RE+h)^2.

To find the change in your weight, we can subtract the two equations: W' - W = GmM/(RE+h)^2 - GmM/RE^2. We can simplify this to get W' - W = GmM(1/RE^2 - 1/(RE+h)^2).

Now, we can use the fact that h is equal to 1 mile (1609.34 meters) and the radius of the Earth (RE) is approximately 6,371,000 meters. Plugging in these values, we get W' - W = GmM(1/6,371,000^2 - 1/6,372,609.34^2) = GmM(1/40,654,828,000 - 1/40,657,176,000).

Using the value of G (6.67 x 10^-11 Nm^2/kg^2), the mass of the Earth (5.97 x 10^24 kg), and your mass (m), we can calculate the