Question

calculate acceleration due to gravity which is at height above the earth surface equal to 2/3 of radius of the earth .​

Answer 1

Answer:

$14.7 \frac{m}{ {s}^{2} }$

Explanation:

$\frac{g'}{g} = \ { (\frac{r}{r'} )}^{2}$

Since,

$r' = \frac{2}{3} r$

substitute it in the first equation,

$hence \\ \frac{g'}{g} = \frac{r}{ \frac{2r}{3} } \\ \frac{g'}{g} = \frac{3}{2} \\ g' = \frac{3}{2} \times g \\ ie. \: g' = 1.5 \times g$

$therefore \: g' = 14.7 \frac{m}{ {s}^{2} }$

hoped this helped you!!

please like and vote ((:

Answer 2

Answer:

3.528 ms^-2

Explanation:

here let's assume,

$r'=\frac{2}{3} r$

$\frac{g'}{g} =(\frac{r}{r+r' } )^{2} \\\\\frac{g'}{g} =(\frac{r}{r+\frac{2r}{3} } )^{2} \\hence,$

$\frac{g'}{g} =(\frac{r}{\frac{5r}{3} } )^{2} \\g'=(\frac{3r}{5r})^{2} (g)\\\\g'=(\frac{3}{5})^{2} (g)$

$here, \\g=9.8 ms^{-2}\\hence,\\g'=(\frac{3}{5})^{2} (g)$

$g'=(0.36)^{2}(9.8)\\g'=3.528 m s^{-2}$