If you weighed 100 lbs at sea level, how much would you weigh flying at 30,000 ft? (not mass.. weight)?

Answer 1

Your weight = W = GMm/r^2; where G = a constant, M = mass of Earth, m = your mass after Holiday dinner, and r = the distance between you and the center of Earth's mass. r ~ 24,000/2pi ~ 4,000 miles, which is Earth's radius.

Your weight on the surface of Earth = W0 = GMm/r0^2. Your weight at 30K ft = W = GMm/r^2; where r = r0 + 30K ft ~ 4,000 + 6 miles, assuming a mile is about 5,000 feet.

Thus W = W0 (r0/r)^2 by taking the ratio W/W0 and everything but the r and r0 cancel out. As you can see, since r > r0, your weight at 30K will be (r0/r)^2 of the original weight of 100 pounds. Therefore, your 100 pounds at 30K would be (4000/4006)^2 of that 100 pounds, or W = W (.997) = 99.7 pounds...a cool way to shed those post-Holiday pounds.

Answer 2

Assuming the plane that you are in is not accelerating or you are not in free fall, your weight would follow by the equation from the unversal law of gravitation where F=GMm/d^2. Where G is the gravitational constant, M is the mass of the earth, and m is the mass of the object (you), and d is the distance between the centers of the masses.

Usually weight is represented by F=mg. But the g is actually dervived from the law of universal gravitation and is short hand for the value GM/d^2 and is represented in metric units as 9.8 meters per second square. As seen from the gravitation equation The actual value of g will vary depending on the distance of your mass from the earth. Thus the closer you are to the surface the more g will be valued, and the thus your weight as well. And the farther away from the surface of the earth the less g and your weight is will be valued.

Thus you expect your weight to be slightly less at high altitudes. The determine more precisely use the gravitational formal. Use G as 6.67 * 10 ^ (-11) N*m^2/kg^2 and the M (earth) for 6 * 10^24 kg.
and d in meters. To figure out your mass, m, divide your weight (scale weight) on the surface of the earth by 9.8 m/s^2

Convert your current weight to metrics first!

Answer 3

You appropriate. At 5"0 you opt to be between ninety 8 and 108. So your on the backside area of standard. I advise you paintings on preserving this weight- until of direction you strengthen then you definitely would be wanting to benefit weight.

Answer 4

I forgot how high it is to be classified in outer space, but you would only weigh a tad bit less, maybe like around 95-98lbs since your still pretty close to earth & the magnetic field is still somewhat real strong.

Answer 5

A scale would show the same amount. 30,000 ft just isn't enough to make a difference.

Answer 6

Less

Good Luck!!!