Question
How much below the surface of the earth does the acceleration due to gravity become $70 \%$ of its value at the surface of the earth? Radius of the earth is 6400 km .
✓
Answer
Here, it is given that the acceleration due to gravity becomes $70 \%$ of the value on surface of the Earth at a depth d .
So, $70 \%$ value of g means $\frac{70}{100} g=0.7 g$
Hence, it is clear that g at depth d becomes $g ( d )=0.7 g$
Now, we know the formula for $g$ at depth as
$
g(d)=g\left(1-\frac{d}{R_E}\right)
$
Here, $R _{ E }$ is the radius of the Earth and g is the acceleration due to gravity on the surface of the Earth.
So, we get
$
\begin{aligned}
& 0.7 g=g\left(1-\frac{d}{R_E}\right) \\
& 0.7=1-\frac{d}{R_E} \\
& \frac{d}{R_E}=1-0.7=0.3 \\
& \therefore d=0.3 R_{E}=0.3 \times 6400=1920 km
\end{aligned}
$
Hence, we can see that at a depth of 1920 km from the surface of the Earth, the value of $g$ becomes $70 \%$ as that on surface.
So, $70 \%$ value of g means $\frac{70}{100} g=0.7 g$
Hence, it is clear that g at depth d becomes $g ( d )=0.7 g$
Now, we know the formula for $g$ at depth as
$
g(d)=g\left(1-\frac{d}{R_E}\right)
$
Here, $R _{ E }$ is the radius of the Earth and g is the acceleration due to gravity on the surface of the Earth.
So, we get
$
\begin{aligned}
& 0.7 g=g\left(1-\frac{d}{R_E}\right) \\
& 0.7=1-\frac{d}{R_E} \\
& \frac{d}{R_E}=1-0.7=0.3 \\
& \therefore d=0.3 R_{E}=0.3 \times 6400=1920 km
\end{aligned}
$
Hence, we can see that at a depth of 1920 km from the surface of the Earth, the value of $g$ becomes $70 \%$ as that on surface.
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