Question:

How high does a rocket have to go above the Earth's surface so that its weight is reduced 78.2%???

by Guest61437  |  earlier

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How high does a rocket have to go above the

Earth's surface so that its weight is reduced

to 72:8% of its weight at the Earth's sur-

face? The radius of the Earth is 6380 km

and the universal gravitational constant is

6:67 x 10^-11 N m2/kg2. Answer in units of

km.

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2 ANSWERS


  1. the weight is due to acceleration of gravity (g), at sea level

    g = 9.81

    g = [  (G)(M)  ] / ( R ^ 2 )

    G = gravitational constant 6.673E - 11

    M = mass of earth 5.98E + 24 kg

    R = radius of earth 6.37E + 6 m

    at height (H) the acceleration of gravity is reduced to (0.782)(g), and the new equation is

    0.782(g) = [  (G)(M)  ] / (R + H) ^ 2

    H = ??? height in m

    (R + H) ^ 2 = [ (G)(M) ] / [0.782(g)]

    (R + H) ^ 2 = [ (G)(M) ] / [0.782(9.81)]

    (R + H) ^ 2 =  5.215E + 13

    (R+ H) =  7.22E + 6

    H =  (7.22E + 6) / (6.37E + 6)

    H = 851,532 m or 851 km

    (above stratosphere)


  2. Already plugged into u-texas...

    Use this equation..

    Re x ((Sqrt (1- (72.8/100)) - 1)

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