Question

The percentage decreases in the acceleration due to gravity at a depth of 32 km below the surface of the earth is r=6400km

Answer 1

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

.