What height above the surface of the earth the value of acceleration due to gravity is reduced to one fourth of its value on the surface of the earth?

Answer

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Hint: In order to answer this question, we will let the new gravity be $g`$ in both cases i.e.. half of $g$and one-fourth of $g$ respectively. And then we will apply the formula to find the height above the earth’s surface in the terms of radius of the earth and the gravity to find the height in the given both cases.

Complete step by step answer:

Given that- Radius of the earth is, $R = 6400km$(i) Let g is the acceleration due to gravity.Let $g`$ be the half of $g$ .So, $g` = \dfrac{g}{2}$Now, for height, $h$ above the half of the earth’s surface, the formula is:$\because g` = g{(\dfrac{R}{{R + h}})^2}$where, $R$ is the radius of the earth.Now, we will put $\dfrac{g}{2}$ instead of $g`$ , as $g` = \dfrac{g}{2}$ .$ \Rightarrow \dfrac{g}{2} = g{(\dfrac{R}{{R + h}})^2}$Cancel out $g$ from both the sides:$\Rightarrow \dfrac{R}{{R + h}} = \dfrac{1}{{\sqrt 2 }} \\\Rightarrow \sqrt 2 R = R + h \\\Rightarrow h = \sqrt 2 R - R \\ $Now, we will take $R$ as common from the above equation and as we know, $\sqrt 2 = 1.414$ :$\Rightarrow h = (\sqrt 2 - 1)R \\\Rightarrow h = (1.414 - 1)R \\\Rightarrow h = 0.414 \times 6400 \\\Rightarrow h\, = 2649.6\,km \\\therefore h \approx 2650\,km \\ $

Hence, the height above the half of its value of the earth’s surface would cause the acceleration due to gravity to be $2650\,km$.

(ii) Again, $g`$ be the one- fourth of $g$ .So, $g` = \dfrac{g}{4}$For height, $h$ above the one-fourth of the earth’s surface:$\because g` = g{(\dfrac{R}{{R + h}})^2}$where, R is the radius of the earth.Now, we will put $\dfrac{g}{4}$ instead of $g`$ , as $g` = \dfrac{g}{4}$ .$\Rightarrow \dfrac{g}{4} = g{(\dfrac{R}{{R + h}})^2} \\\Rightarrow \dfrac{R}{{R + h}} = \dfrac{1}{{\sqrt 4 }} \\\Rightarrow R + h = 2R \\\Rightarrow h = 2R - R \\\Rightarrow h = R $We will put the value of $R$ that is given.$\therefore h = 6400\,km$

Hence, the height above the one-fourth of its value of the earth’s surface would cause the acceleration due to gravity to be $6400\,km$.

Note: The distance between the Earth's centre and a location on or near its surface is measured in Earth radius. The radius of an Earth spheroid, which approximates the figure of Earth, ranges from nearly \[6,378km\] to nearly \[6,357km\] . In astronomy and geophysics, a nominal Earth radius is sometimes used as a unit of measurement, with the equatorial value suggested by the International Astronomical Union.