How much lighter are you on the observatory (top floor) of 1 World Trade Center?
- TStoddenLv 78 months agoFavourite answer
Top of 1 World Trade Center (excluding the top spire), is 1,776 ft above ground & NYC, ON AVERAGE is 33 ft above sea level... placing your current altitude at 1,809 ft above sea level.
According to Wikipedia & using VERY ROUGH figures... you experience a 0.29% decrease in gravity at ~30,000 ft (or ~9 km) above sea level.
SO (1809 / 30000) * 0.0029 = ~0.00017487 or ~0.017487% difference (99.982513% of Earth's Gravity)...
This means that if you weighed in at 220.000 lbs (100.000 kg) at sea level. You would weigh in at ~219.9615286 lbs (~99.982513 kg), which is a difference of
~17.487 g (= 17,487 mg = 0.017487 kg)
This would be due to the minute gravitational difference that you may not perceive unless you have an extremely high-precision scale & this difference would be negated when you descend back to ground floor (as the difference between ground floor & sea level would be a ~0.00029% difference... which gets down to the scale of milligrams or smaller in sensitivity to detect).
- Anonymous8 months ago