Question

A mass of 5Kg is 1st weighed on a balance at the top of a tower 20m high.The man is then suspended from a fine wire 20m long and reweivhed.find the difference in a weight ,assuming that the radius of the earth is 6400km,the mass of the earth is 6×10^24kg and G=6.673×-11 Nm^2/Kg-2.

Answer 1

Assume fine wire has zero mass
Newton law of gravitation
Weight=mg=GMm/r^2
where units are in m, kg, seconds
R1=6400000 (at top of tower)
R2=(6400000-20)=6399980 (at bottom)
M=6*10^24
G=6.673*10^(-11)

Difference in weight
=GMm(1/R1^2-1/R2^2)
Evaluate.

Hint: the difference is less than a mN

Answer 2

Change in weight =3.06×10 ^-4 N

Answer 3

Bekar