Question

(3) At what distance below the surface of the
Earth, does the acceleration due to gravity
decrease by 10% of its value at the surface,
given the radius of Earth is 6400 km. ​

Answer 1

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Data:- gd = g -0.1g = 0.9g, R = 6400 km

$\frac{gd}{g} = 0.9 \\ gd = g(1 - \frac{d}{r} ) \\ \frac{gd}{g} = 1 - \frac{d}{r} \\ \frac{d}{r} = 1 - \frac{gd}{g} = 1 - 0.9 = 0.1 \\ d = 0.1r = 0.1 \times 6400km$

= $\boxed{\boxed{640 km}}$

This is the required distance.

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Answer 2

Answer:

• The Distance below (d) the Earth's Surface is 640 Km.

Given:

1. Earth's Radius (R) = 6400 Km.
2. Acceleration decreases by 10 %

Explanation:

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From the Formula, We Know,

$\large \bigstar\; \boxed{\tt g_d = g\Bigg(1 - \dfrac{d}{R}\Bigg)}$

$\mathfrak{Here}\begin{cases}\text{g Denotes Acceleration due to gravity at surface} \\ \sf{g_d}\text{ Denotes Acceleration due to gravity at depth d} \\ \text{R Denotes Radius of Earth}\end{cases}$

We Know, The acceleration due to gravity  decrease by 10% of its value at the surface of the earth,

Therefore,

$\displaystyle \dashrightarrow \tt g_d = g- 10 \; \% \; of \; g \\\\ \dashrightarrow\tt g_d = g-\dfrac{10}{100} \times g \\\\\dashrightarrow\tt g_d = g-0.1 \times g\\\\\dashrightarrow\tt g_d = g-0.1 \;g \\\\\dashrightarrow\tt \large \underline{\blue{g_d = 0.9 \; g}}$

Now,

$\large \boxed{\tt g_d = g\Bigg(1 - \dfrac{d}{R}\Bigg)}$

Substituting the values,

$\displaystyle \dashrightarrow\tt 0.9\;g=g\Bigg(1-\dfrac{d}{6400}\Bigg)\\\\\\\dashrightarrow\tt 0.9\;\cancel{g}=\cancel{g}\Bigg(1-\dfrac{d}{6400}\Bigg)\\\\\\\dashrightarrow\tt 0.9=\Bigg(1-\dfrac{d}{6400}\Bigg)\\\\\\\dashrightarrow\tt 0.9-1=-\dfrac{d}{R}\\\\\\\dashrightarrow\tt -0.1=-\dfrac{d}{R}\\\\\\\dashrightarrow\tt 0.1=\dfrac{d}{R}\\\\\\\dashrightarrow\tt d=0.1\times R\\\\\\\dashrightarrow\tt d=0.1\times 6400\ \because[R=6400\;Km]\\\\\\\dashrightarrow \large \underline{\boxed{\red{\tt d=640\;Km}}}$

∴ The Distance below (d) the Earth's Surface is 640 Km.

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