JEE MAIN - Physics (2022 - 26th July Morning Shift - No. 14)
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 $$=6400 \mathrm{~km})$$
1%
3%
4%
0.5%
Vysvětlení
$$\because$$ $$g = {{GM} \over {{r^2}}}$$
$$ \Rightarrow {{\Delta g} \over g} = 2{{\Delta r} \over r}$$
$$ \Rightarrow {{\Delta g} \over g} \times 100 = 2 \times {{32} \over {6400}} \times 100\% = 1\% $$
$$\Rightarrow$$ % decrease in weight = 1%