The percentage decrease in the weight of a rocket, when taken to a height of $32 \mathrm{~km}$ above the surface of earth will, be :
$($ Radius of earth $=6400 \mathrm{~km})$
Solution
<p>$\because$ $g = {{GM} \over {{r^2}}}$</p>
<p>$\Rightarrow {{\Delta g} \over g} = 2{{\Delta r} \over r}$</p>
<p>$$ \Rightarrow {{\Delta g} \over g} \times 100 = 2 \times {{32} \over {6400}} \times 100\% = 1\% $$</p>
<p>$\Rightarrow$ % decrease in weight = 1%</p>
About this question
Subject: Physics · Chapter: Gravitation · Topic: Gravitational Field and Potential
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