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 &#160; km above the surface of earth will, be (Radius of earth = 6400 &#160; km )

Aicharya · Accepted Answer

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

The percentage decrease in the weight of a rocket, when taken to a height of 32   km above the surface…

Solution

The percentage decrease in the weight of a rocket, when taken to a height of 32 &#160; km above the surface…

Solution

The percentage decrease in the weight of a rocket, when taken to a height of 32 km above the surface…