The percentage decreases in the acceleration due to gravity at a depth of 32 km below the surface of the earth is r=6400km
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Explanation:
Percentage decrease in weight of body is 1%.
Solution:
Given: Height = 32 km
Radius of the earth = 6400 km.
The formula to find the gravitational pull at a height above surface of earth is given by \bold{g=g_{0}\left(1-\frac{2 h}{R}\right)}g=g
0
(1−
R
2h
)
Rearranging the equation we get \frac{2 h}{R}=g_{0}-g
R
2h
=g
0
−g
Therefore percentage decrease =\frac{2 h}{R} \times 100 \%=
R
2h
×100%
On evaluating we get the percentage decrease in acceleration due to gravity is 2 \times \frac{32}{6400} \times 100 \%=1 \%2×
6400
32
×100%=1%
As g is 1% less than \bold{g_{0}}g
0
, weight mg is also 1% less than \bold{m g_{0}}mg
0
.
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