Find the percentage decrease in the weight of the body when taken to a height of 16km above the surface of earth.radius of earth is 6400km
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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
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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0
Answer 0.5% decrease
Explanation:
Explanation:
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