Question

Acceleration due to gravity at earth's surface is $g\, ms^{-2}$
Find the effective value of gravity at a height of $32\, km$ from sea level $(R_e = 6400\, km)$

• A. $0.5\, g\, ms^{-2}$
• B. $0.99\, g\, ms^{-2}$
• C. $1.01\, g\, ms^{-2}$
• D. $0.90\, g\, ms^{-2}$

Answer 1

Answer: (B)

For height h above the earth's surface
$g' = g \left(1 - \frac{2h}{R_e}\right)=g \left(1 - \frac{64}{6400}\right)$
$(\because R_e = 6400\, km)$
$= g(1-0.01)$
$= 0.99\, g\, ms^{-2}$