# Law of Gravitation problem

1. Apr 2, 2013

### robertmatthew

1. The problem statement, all variables and given/known data
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)?

2. Relevant equations
F = (GMm)/r2
Mass of Earth: 5.97x1024 kg

3. 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.

Last edited: Apr 2, 2013
2. Apr 2, 2013

### SammyS

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

3. Apr 2, 2013

### PeterO

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.