How Much Less Do You Weigh at the Top of the World Trade Center?

[robertmatthew]

Homework Statement

You weigh 518 N at sidewalk level outside the World Trade Center in New York City. Suppose that you ride from this level to the top of one of its 410 m towers. Ignoring Earth's rotation, how much less would you weigh there (because you are slightly farther from the center of Earth)?

Homework Equations

F = (GMm)/r²
Mass of Earth: 5.97x10²⁴ kg
Radius of Earth: 6378100 m

The Attempt at a Solution

518 = ((6.67x10^-11)(5.97x10²⁴)(m))/(6378100²)
m = 52.92 kg

F = (GMm)/(r+h)²
F' = ((6.67x10^-11)(5.97x10²⁴)(52.80))/(6378510²)
F' = 517.94 N

So my calculated difference is .0575 N, but apparently that's not right. Not sure where I've gone wrong here, any help is much appreciated.

 

[SammyS]

Homework Statement

You weigh 518 N at sidewalk level outside the World Trade Center in New York City. Suppose that you ride from this level to the top of one of its 410 m towers. Ignoring Earth's rotation, how much less would you weigh there (because you are slightly farther from the center of Earth)?

Homework Equations

F = (GMm)/r²
Mass of Earth: 5.97x10²⁴ kg
Radius of Earth: 6378100 m

The Attempt at a Solution

518 = ((6.67x10^-11)(5.97x10²⁴)(m))/(6378100²)
m = 52.92 kg

F = (GMm)/(r+h)²
F' = ((6.67x10^-11)(5.97x10²⁴)(52.80))/(6378510²)
F' = 517.94 N

So my calculated difference is .0575 N, but apparently that's not right. Not sure where I've gone wrong here, any help is much appreciated.

Look at $\displaystyle \ \frac{1}{r^2}-\frac{1}{(r+h)^2} =\frac{r^2+2rh+h^2-r^2}{r^2(r+h^2)}=\frac{2rh+h^2}{r^2(r+h^2)}\approx\frac{2h}{r^3}\ .$

 

[PeterO]

Homework Statement

You weigh 518 N at sidewalk level outside the World Trade Center in New York City. Suppose that you ride from this level to the top of one of its 410 m towers. Ignoring Earth's rotation, how much less would you weigh there (because you are slightly farther from the center of Earth)?

Homework Equations

F = (GMm)/r²
Mass of Earth: 5.97x10²⁴ kg
Radius of Earth: 6378100 m

The Attempt at a Solution

518 = ((6.67x10^-11)(5.97x10²⁴)(m))/(6378100²)
m = 52.92 kg

F = (GMm)/(r+h)²
F' = ((6.67x10^-11)(5.97x10²⁴)(52.80))/(6378510²)
F' = 517.94 N

So my calculated difference is .0575 N, but apparently that's not right. Not sure where I've gone wrong here, any help is much appreciated.

Gravity is an inverse square relation, so I would add 410 to the Earth radius you had to find the ratio of increase [I get that r is now 1.000064 times the original. Square that 1.000129
so the inverse says the weight force is reduced by that factor.

518 / 1.000129 gave me about 0.0666 N less than 518, rather than your .0575 answer.

 

[robertmatthew]

Thank you both very much, got the right answer and I actually understand how I got there!

 

[Nickie]

Your calculation is correct, but your answer is slightly off due to rounding errors. I would suggest using more significant figures in your calculations to get a more accurate answer. Also, make sure to include units in your final answer (N for force).

The correct answer is 517.944 N, which is a difference of 0.056 N from your original weight at sidewalk level. This small decrease in weight is due to the decrease in gravitational force as you move away from the center of the Earth.