91Ó°ÊÓ

To hover on Mars, the buoyant force should equal the weight of the balloon. The buoyant force is equal to the weight of the displaced Martian atmosphere. Using the formula \( F_b = \rho_{atm} \cdot V \cdot g_{Mars} \), where \( \rho_{atm} = 0.0154 \; \mathrm{kg/m^3} \) and \( g_{Mars} = 3.71 \; \mathrm{m/s^2} \), and knowing that the balloon is weightless when hovering, we deduce that the volume \( V \) must be such that the weight of the atmosphere equals the weight of the balloon. Volume \( V = \frac{m_{balloon}}{\rho_{atm}} \).The mass of the plastic of the balloon is \( 5 \; \mathrm{g/m^2} \) which converted gives \( 0.005 \; \mathrm{kg/m^2} \). The weight of this shell is \( 4 \pi r^2 \cdot 0.005 \) with \( 4\pi r^2 \) being the sphere's surface area. By setting both forces equal, we find the required radius \( r \): \( \frac{4}{3} \pi r^3 \cdot \rho_{atm} \cdot g_{Mars} = 4 \pi r^2 \cdot 0.005 \cdot g_{Mars} \Rightarrow r = \sqrt{\frac{2 \cdot 0.005}{3 \cdot \rho_{atm}}} = \sqrt{\frac{2 \cdot 0.005}{3 \cdot 0.0154}} = 0.58 \; \mathrm{m}. \)

On Earth, we use the equation for the net force, \( F_{net} = F_b - m_{balloon} \cdot g_{Earth}\), where \( \rho_{Earth} = 1.20 \; \mathrm{kg/m^3} \) and \( g_{Earth} = 9.81 \; \mathrm{m/s^2} \).The changed buoyant force is \( F_b = \rho_{Earth} \cdot V \cdot g_{Earth} \) and initial volume \( V = \frac{4}{3} \pi r^3 = \frac{4}{3} \pi (0.58)^3 \approx 0.82 \; \mathrm{m^3}\).\[ F_b = 1.20 \cdot 0.82 \cdot 9.81 = 9.65 \; \mathrm{N} \]The mass of balloon plastic remains, \( m_{balloon} = 4 \pi r^2 \cdot 0.005 = 4 \pi (0.58)^2 \cdot 0.005 \approx 0.021 \; \mathrm{kg} \).The weight is \( m_{balloon} \cdot g_{Earth}= 0.021 \cdot 9.81 = 0.21 \; \mathrm{N} \).Since \( F_b \) > \( m_{balloon} \cdot g_{Earth} \), net force is upward, yielding:\[ a = \frac{F_{net}}{m_{balloon}} = \frac{9.65 - 0.21}{0.021} \approx 449 \; \mathrm{m/s^2} \]

We increase the radius by five times on Mars, hence \( r = 5 \times 0.58 = 2.90 \; \mathrm{m} \).The new volume is \( V = \frac{4}{3} \pi r^3 \), \[ V = \frac{4}{3} \pi (2.90)^3 \approx 102.11 \; \mathrm{m^3} \]The increased buoyant force is \( F_b = \rho_{Mars} \cdot V \cdot g_{Mars} \). \[ F_b = 0.0154 \cdot 102.11 \cdot 3.71 \approx 6.07 \; \mathrm{N} \]Accounting the balloon's mass, \( m_{balloon} = 4 \pi (2.90)^2 \cdot 0.005 \approx 0.531 \; \mathrm{kg} \).The weight of the balloon is \( m_{balloon} \cdot g_{Mars} = 0.531 \cdot 3.71 = 1.97 \; \mathrm{N} \)The extra force available for lifting the package is: \[ F_{package} = F_b - m_{balloon} \cdot g_{Mars} \approx 4.10 \; \mathrm{N} \].Considering \( F_{package} = m_{package} \cdot g_{Mars} \), the mass the balloon can lift is: \[ m_{package} = \frac{4.10}{3.71} \approx 1.10 \; \mathrm{kg} \]