Question

the height above the earth's surface at which the weight of a person becomes 1/4th of his weight on the surface of earth is (R is the radius of earth) ​

Answer 1

Answer:

At height r or 6400km, the acceleration due to gravity will be 1/4th of its value at the surface of earth

Explanation:

here,

$g'=\frac{1}{4} (g)$

we know that,

$\frac{g'}{g} = (\frac{r}{r+r'})^{2} \\\frac{\frac{1}{4} (g)}{9} =(\frac{r}{r+r'})^{2} \\\frac{1}{4} =(\frac{r}{r+r'})^{2} \\ (\frac{r}{r+r'})=\frac{1}{2}$

$hence,\\r+r'=2r\\r'=2r-r=r\\therefore, \\r'=r$

ie. r'=6400km

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