Question

determine the percentage decrease in the weight of the body when taken to a height of 16km above the surface of the earth. radius of the earth is 6400 km​

Answer 1

Answer:

Weight of body = mg

radius = 6400 km = 64×10^5 m

mass off earth = M = 5.98×10^24 kg

G = 6.673×10^-11 N.m^2/kg^2

g( near earth surface) = GM/(R+h)^2

g = {5.98× 10^24 × 6.673×10^-11} / {64×10^5}^2

= 9.8 m/s^2

hence

since h= 16 km and R = 6400km

then

R+h = 6416km = 6416×10^3 m

g = {5.98×10^13×6.67} / (6416 ×10^3)^2

= 9.7 m/s^2

hence % of change of weight at 16 km height

= [9.7×100] / 9.8

= 98.97 %

hence % of decrease in weight =100-981.92.97

= 1.02 % decrease

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Answer 2

Answer:

0.5%

Explanation:

Here d= 32 km R= 6400 km

weight of body at depth d is mg`

=mg(1-d/R)%

decrease in weight =( mg- mg`)/mg × 100= 32 / 6400×100 = 0.5%.