The value of g will become 1% of its value at the earth's surface at the height ________ above the surface of the earth. Take the radius of the earth as 6400km

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Solution

Here acceleration due to gravity at height $h$ above the surface of Earth is $g_{h} = 1 % o f g$
$g_{h} = \frac{1}{100} g . . . . . . . . . . . . . . . (1)$
2. As we know $g_{h} = g {(R / R + h)}^{2} . . . . . . . . . . . . . . (2)$
where $g$ is an acceleration due to gravity on the surface of Earth and $R$ is a radius of Earth.
By putting the value of the equation $(1)$ in $(2)$ we get
$\Rightarrow \frac{1}{100} = {(R / R + h)}^{2}$
$\Rightarrow 10 R = R + h$
$\Rightarrow 9 R = h \Rightarrow h = 9 \times 6400 \Rightarrow h = 57, 600$
Hence, we will get that value of g will become 1% of the earth's surface at the height of 57,600 above the surface of the earth.

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