Question

at what distance above the surface of earth the acceleration due to gravity decrease by 10 of its value at the surface​

Answer 1

Question:-

At what distance above the surface of earth, the acceleration due to gravity $\displaystyle\sf {g}$ decreases by 10% of its value at the surface?

Answer:-

At a height $\displaystyle\sf {\underline {\underline {h=346.2\ km}}}$ from the surface of the earth.

Solution:-

The variation of the value of $\displaystyle\sf {g}$ due to the height $\displaystyle\sf {h}$ from the surface of the earth is given as,

$\displaystyle\longrightarrow\sf {g'=g\left [1+\dfrac {h}{R}\right]^{-2}}$

where

• $\displaystyle\sf {g'=}$ the acceleration due to the gravity at that height $\displaystyle\sf {h.}$

• $\displaystyle\sf {R=}$ radius of the earth.

Here, the value of the acceleration due to gravity decreases by 10% of its actual value at the height $\displaystyle\sf {h.}$ So,

$\displaystyle\longrightarrow\sf {g'=g-\dfrac {10g}{100}}$

$\displaystyle\longrightarrow\sf {g\left [1+\dfrac {h}{R}\right]^{-2}=\dfrac {9g}{10}}$

$\displaystyle\longrightarrow\sf {\left [1+\dfrac {h}{R}\right]^{-2}=\dfrac {9}{10}}$

$\displaystyle\longrightarrow\sf {\left [1+\dfrac {h}{R}\right]^{2}=\dfrac {10}{9}}$

$\displaystyle\longrightarrow\sf {1+\dfrac {h}{R}=\sqrt{\dfrac {10}{9}}}$

$\displaystyle\longrightarrow\sf {\dfrac {h}{R}=\sqrt{\dfrac {10}{9}}-1}$

$\displaystyle\longrightarrow\sf {h=R\left [\sqrt{\dfrac {10}{9}}-1\right]}$

Well, $\displaystyle\sf {R=6400\ km.}$ Then,

$\displaystyle\longrightarrow\sf {h=6400\left [\sqrt{\dfrac {10}{9}}-1\right]}$

$\displaystyle\longrightarrow\sf {\underline {\underline {h=346.2\ km}}}$