At what height above the surface of earth the value of g is same as that at a depth of 100 km

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Given,

Depth = d = 200 km

To find height (h)

We know,

At height h from the surface of the Earth, the value of acceleration due to gravity = \(g'_h=g(1-\frac{2h}{R})\)

At depth d from the surface of the Earth, the value of acceleration due to gravity = \(g'_d=g(1-\frac{d}{R})\)

Here R is the radius of the Earth

According to the question,

\(g'_h=g'_d\\ or, \; g(1-\frac{2h}{R})=g(1-\frac{d}{R})\\ or, \; \frac{2h}{R}=\frac{d}{R}\\ or, \; h=\frac{d}{2}=\frac{200}{2}=100 \; km\)

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If you compare the gravitational force on the earth due to the sun to that due to the moon, you would find that the Sun’s pull is greater than the moon’s pull. However, the tidal effect of the moon’s pull is greater than the tidal effect of the sun. Why?

Tidal effect is inversely proportional to the cube of the distance while gravitational force is inversely proportional to the square of the distance. 

The distance between moon and Earth is less as compared to the distance between Sun and Earth. Greater the distance, less is the pull of the tidal effect.