Question

What will be the value of g at the bottom of sea 64km deep?

Answer 1

Given:

$Acceleration due to gravity, g =9.8 m/s^{2}\\\\Depth, D = 64km = 64\times 10^{3} m\\\\ Radius of Earth = 6370km = 6370\times 10^{3} m$

Explanation: We have to calculate the value of the acceleration due to gravity at the bottom of sea

Step by Step Solution:

We have the expression to calculate the acceleration due to gravity at the bottom of sea,

$gd = g(1-\dfrac{d}{r} )\\\\gd = 9.8(1-\dfrac{64\times 10^{3} }{6370\times 10^{3}} ) \\\\gd = 9.8\times (1-0.0100)\\\\gd = 9.8\times 0.99\\\\gd = 9.702 m/s^{2}$

Answer 2

Answer:

Given:

Accelerationduetogravity,g=9.8m/s2Depth,D=64km=64×103mRadiusofEarth=6370km=6370×103m\begin{lgathered}Acceleration due to gravity, g =9.8 m/s^{2}\\\\Depth, D = 64km = 64\times 10^{3} m\\\\ Radius of Earth = 6370km = 6370\times 10^{3} m\end{lgathered}

Accelerationduetogravity,g=9.8m/s

2

Depth,D=64km=64×10

3

m

RadiusofEarth=6370km=6370×10

3

m

Explanation: We have to calculate the value of the acceleration due to gravity at the bottom of sea

Step by Step Solution:

We have the expression to calculate the acceleration due to gravity at the bottom of sea,

gd=g(1−dr)gd=9.8(1−64×1036370×103)gd=9.8×(1−0.0100)gd=9.8×0.99gd=9.702m/s2\begin{lgathered}gd = g(1-\dfrac{d}{r} )\\\\gd = 9.8(1-\dfrac{64\times 10^{3} }{6370\times 10^{3}} ) \\\\gd = 9.8\times (1-0.0100)\\\\gd = 9.8\times 0.99\\\\gd = 9.702 m/s^{2}\end{lgathered}

gd=g(1−

r

d

)

gd=9.8(1−

6370×10

3

64×10

3

)

gd=9.8×(1−0.0100)

gd=9.8×0.99

gd=9.702m/s

2