Answer:
7.65 m
Explanation:
[tex]P_1[/tex] = Initial pressure = 0.03 atm
[tex]P_2[/tex] = Final pressure = 1 atm
[tex]r_1[/tex] = Inital radius = 21 m
[tex]V_1[/tex] = Intial volume of gas = [tex]\frac{4}{3}\pi r_1^3[/tex]
[tex]V_2[/tex] = Final volume of gas = [tex]\frac{4}{3}\pi r_2^3[/tex]
[tex]T_1[/tex] = Initial temperature = 200 K
[tex]T_2[/tex] = Final temperature = 323 K
From ideal gas law we have
[tex]\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}\\\Rightarrow \frac{P_1\frac{4}{3}\pi r_1^3}{T_1}=\frac{P_2\frac{4}{3}\pi r_2^3}{T_2}\\\Rightarrow \frac{P_1 r_1^3}{T_1}=\frac{P_2r_2^3}{T_2}\\\Rightarrow r_2=\frac{P_1r_1^3T_2}{T_1P_2}\\\Rightarrow r_2=\left(\frac{0.03\times 21^3\times 323}{200\times 1}\right)^{\frac{1}{3}}\\\Rightarrow r_2=7.65\ m[/tex]
The radius at liftoff is 7.65 m
