Question

It has been proposed that we could explore Mars using inflated balloons to hover just above the surface. The buoyancy of the atmosphere would keep the balloon aloft. The density of the Martian atmosphere is 0.0154kgm30.0154kgm3(although this varies with temperature). Suppose we construct these balloons of a thin but tough plastic having a density such that each square meter has a mass of 5.00 g. We inflate them with a very light gas whose mass we can ignore.
(a) What should be the radius and mass of these balloons so they just hover above the surface of Mars?
(b) If we released one of the balloons from part (a) on earth, where the atmospheric density is 1.20kgm31.20kgm 3, what would be its initial acceleration assuming it was the same size as on Mars? Would it go up or down?
(c) If on Mars these balloons have five times the radius found in part (a), how heavy an instrument package could they carry?
(d) Would it go up or down?
(e) If on Mars these balloons have five times the radius found in part A, how heavy an instrument package could they carry?

Answer 1

Answer:

a) mb = 0.0596 kg ; r = 0.974 m

b) a = 754 m/s^2 .. (Upward)

c) mL = 5.96 kg

Explanation:

Given:-

- The density of Mars atmosphere , ρ = 0.0154 kg/m^3

- The surface density of ballon, σ = 5.0g/m^2

Solution:-

(a) What should be the radius and mass of these balloons so they just hover above the surface of Mars?

- We will first isolate a balloon in the Mar's atmosphere and consider the forces acting on the balloon. We have two forces acting on the balloon.

- The weight of the balloon - "W" - i.e ( Tough plastic weight + Gas inside balloon). Since, the balloon is filled with a very light gas we will assume the weight due to gas inside to be negligible. So we have:

                            W = mb*g

Where,  mb: Mass of balloon

             g: Gravitational constant for Mars

- The mass of the balloon can be determined by using the surface density of the tough plastic given as "σ" and assuming the balloon takes a spherical shape when inflated with surface area "As".

                           As = 4πr^2

Where,  r: The radius of balloon

So,                      mb =  4σπr^2

- Substitute the mass of balloon "mb" in the expression developed for weight of the balloon:

                         W = 4*σ*g*πr^2    ......... Eq1

- The weight of the balloon is combated by the buoyant force - "Fb" produced by the volume of Mars atmosphere displaced by the balloon acting in the upward direction:

                        Fb = ρ*Vs*g

Where,    Vs : Volume of sphere = 4/3 πr^3

So,                    Fb = ρ*g*4/3 πr^3   ....... Eq 2        

- Apply the Newton's equilibrium conditions on the balloon in the vertical direction:

                       Fb - W = 0

                       Fb = W

                       ρ*g*4/3 πr^3 = 4*σ*g*πr^2        

                       r = 3σ / ρ

                       r = 3*0.005 / 0.0154

                       r = 0.974 m           .... Answer            

- Use the value of radius "r" and compute the "mb":

                       mb =  4σπr^2

                       mb =  4*0.005*π (0.974)^2  

                       mb = 0.0596 kg   ... Answer  

(b) If we released one of the balloons from part (a) on earth, where the atmospheric density ρ = 1.20kg/m^3, what would be its initial acceleration assuming it was the same size as on Mars? Would it go up or down?

- The similar analysis is to be applied when the balloon of the same size i.e r = 0.974 m and mass mb = 0.0596 kg is inflated on earth with density  ρ = 1.20kg/m^3.

- Now see that the buoyant force acting on the balloon due to earth's atmosphere is different from that found on Mars. So the new buoyant force Fb using Eq2 is:

                       Fb = ρ*g*4/3 πr^3

Where,   g: Gravitational constant on earth = 9.81 m/s^2

                       Fb = (1.20)*(9.81)*(4/3)* π*(0.974)^3

                       Fb = 45.5 N

- Apply the Newton's second law of motion in the vertical direction on the balloon:

                      Fb - W = mb*a

Where,          a: The acceleration of balloon

                     a = (Fb - W) / mb

                     a = Fb/mb - g

                     a = 45.5/0.0596 - 9.81

                    a = 754 m/s^2  (upward) ..... Answer

c), d) If on Mars these balloons have five times the radius found in part (a), how heavy an instrument package could they carry?

- The new radius of the balloon - "R" -is five times what was calculated in part (a):

- Apply the Newton's equilibrium conditions in the vertical direction on the balloon with the addition of downward weight of load "WL":

                     Fb - W - WL = 0

                     WL = Fb - W

                     mL*g = ρ*g*4/3 πR^3 - 4*σ*g*πR^2      

Where,          mL : The mass of load due to instrument package

                     mL =  ρ*4/3 πR^3 - 4*σ*πR^2

                     mL = 0.0154*4/3*π*(5*0.974)^3 - 4*(0.005)*π*(5*0.974)^2    

                     mL = 7.45 - 1.45

                     mL = 5.96 kg   ..... Answer                      

Answer 2

Answer: a) radius = 0.974m

Mass = 0.0596 kg

b) acceleration = - 754m/^2

c) weight = 58.5 N

d) it will go up

e) 5689.6 N

Explanation: Please find the attached files for the solution