How much lighter are you on the observatory (top floor) of 1 World Trade Center?

2 Answers

  • 8 months ago
    Favourite 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

    ~0.0385523 lbs

    ~0.616836 oz

    ~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).

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  • Anonymous
    8 months ago

    Virtually zero.

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