Question

Calculate g at bottom of mine 8km deep at an altitude of 32km above earth's surface. radius of earth is 6.4x10^6m and gravitation on earth surface is 9.8 m/s^2please answer this question with proper steps◉‿◉​

Answer 1

Answer:

The ratio of gravity at position inside the Earth and at an altitude above the surface is 1.01

Explanation:

As we know that the gravitational field intensity inside the surface of Earth is given as

$g_{in} = g(1 - \frac{d}{R})$so we have

$g_{in} = 9.8(1 - \frac{8000}{6.4 \times 10^6})$

$g_{in} = (0.99875)g$

Now gravitational field at some height above the surface of earth is given as

$g_[up} = g(1 - \frac{2h}{R})$

$g_{up} = g(1 - \frac{2\times 32000}{6.4 \times 10^6})$

$g_{up} = 0.99g$

So the ratio of two gravitational field is given as

$\frac{g_{in}}{g_{up}} = \frac{0.99875}{0.99}$

$\frac{g_{in}}{g_{up}} = 1.01$

#Learn

Topic : Variation of gravity with height

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