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 (Re = 6400 km)

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 (Re = 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

Tardigrade Admin · Accepted Answer

For height h above the earth's surface g' = g (1 - (2h/Re))=g (1 - (64/6400)) (∵ Re = 6400 km) = g(1-0.01) = 0.99 g ms-2

Q. Acceleration due to gravity at earth's surface is $g m s^{- 2}$
Find the effective value of gravity at a height of $32 km$ from sea level $(R_{e} = 6400 km)$

$0.5 g m s^{- 2}$

$0.99 g m s^{- 2}$

$1.01 g m s^{- 2}$

$0.90 g m s^{- 2}$

For height h above the earth's surface
$g^{'} = g (1 - \frac{2 h}{R _{e}}) = g (1 - \frac{64}{6400})$
$(∵ R_{e} = 6400 km)$
$= g (1 - 0.01)$
$= 0.99 g m s^{- 2}$

Q. Acceleration due to gravity at earth's surface is gms−2 Find the effective value of gravity at a height of 32km from sea level (Re​=6400km)

0.5gms−2

0.99gms−2

1.01gms−2

0.90gms−2