Question

the acceleration due to gravity at the height R(radius of the earth) from the earth's surface is....

Answer 1

$\huge{\underline{\bf{Answer:-}}}$

$\large{\bigstar{\underline{\sf{Given:-}}}}$

→ Gravity g = G M / R²

→ Gravity above earth's surface = g', at altitude h.

→ g'/g = R² / (R + h)² = 1 - 2 h/R

→ Gravity below Earth's surface = g" at a depth d and radius R"

$\large{\bigstar{\underline{\sf{Explainationtion:-}}}}$

= M" / M =  R"³ / R³

= g" / g = (R² / R"² ) × (R"³ / R³)

=  R" / R =  (R - d) / R = 1 - d/R

= g" / g'  =  (R - d) / (R - 2 h)

If, d = h = R/20  then:-

g " / g'   =  (19R /20) / (18R/20)  = 19/18

g" = 19/18 × 9 m/sec²

= 76/9

= 9.5 m/sec²

________________________________________________________________________