How high does a rocket have to go above earth's surface before its weight is half what it would be on Earth?
Ok, before "Justin" or any other idiot gets on here and says "omg this is cheating ,blah blah" it's not cheating my class is different from her/his and we already have all the answers, but I really want to know how to do it! Mrs Luca explained it to us but I didn't get it, so I'd like someone to clean it up for me step by step please???
2007-12-03 12:46:05 · 2 answers · asked by E 2 in Science & Mathematics ➔ Physics
Weight = force of gravity on a mass
F = GMm/r^2 where M = mass of earth, G = Newton's constant
Let W = GMm/Re^2 Re = radius of earth and W = weight.
Now for W/2 = GMm/r^2 ---> r ^2= 2GMm/W = 2GMm/(GMm/Re^2)
r ^2= 2Re^2 ---> r = sqrt(2)* Re = 9020 km
h = height above surface = r -Re = 2642 km
2007-12-03 12:52:43 · answer #1 · answered by nyphdinmd 7 · 0⤊ 0⤋
the rockets distance from the middle of the earth (to experience purely 0.5 its weight) might must be sqrt(2) cases the radius of earth which could be 2.6426*10^3 km above the earths floor
2016-12-10 11:44:23 · answer #2 · answered by ? 4 · 0⤊ 0⤋