The percentage decrease in the weight of a rocket, when taken to a height of 32 km above the surface of earth will, be: (Radius of earth =6400 km )

Question

The percentage decrease in the weight of a rocket, when taken to a height of 32 km above the surface of earth will, be: (Radius of earth =6400 km ) (A) 1 % (B) 3 % (C) 4 % (D) 0.5 %

Tardigrade Admin · Accepted Answer

Acceleration due to gravity at a height h << R is g prime= g (1-(2 h / R )) ∴ (Δ g / g )=(2 h / R ) ⇒ (Δ g / g ) × 100=(2 h / R ) × 100 =2 × (32/6400) × 100=1 %

Q. The percentage decrease in the weight of a rocket, when taken to a height of $32 km$ above the surface of earth will, be :
(Radius of earth $= 6400 km$ )

$1%$

$3%$

$4%$

$0.5%$

Acceleration due to gravity at a height $h << R$ is
$g^{'} = g (1 - \frac{2 h}{R})$
$∴ \frac{Δ g}{g} = \frac{2 h}{R}$
$\Rightarrow \frac{Δ g}{g} \times 100 = \frac{2 h}{R} \times 100$
$= 2 \times \frac{32}{6400} \times 100 = 1%$

Q. The percentage decrease in the weight of a rocket, when taken to a height of 32km above the surface of earth will, be : (Radius of earth =6400km )

1%

3%

4%

0.5%