In: Physics

# What would be its initial acceleration assuming it was the same size as on Mars

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 0.0154 (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 4.60 . We inflate them with a very light gas whose mass we can neglect. So far I found the following: What should be the radius of these balloons so they just hover above the surface of Mars? Radius of the balloon = /896 m What should be the mass of these balloons so they just hover above the surface of Mars? Mass of balloon = 4.64*10^-2 kg If we released one of the balloons from part A on earth, where the atmospheric density is 1.20 , what would be its initial acceleration assuming it was the same size as on Mars? If on Mars these balloons have five times the radius found in part A, how heavy an instrument package could they carry?

## Solutions

##### Expert Solution

The density of the Martian atmosphere, ρ = 0.0154 kg/m3

A)

The volume of the sphere, V = (4/3)πr3

The area of the sphere, A = 4πr2

The mass of the balloon is

m = (4.60 g/m2)A

= [4.60x10-3 kg/m2][4πr2]

The buoyant force is

F = ρVg

mg = ρVg

Then,

[4.60x10-3 kg/m2][4πr2] = ρ(4/3)πr3g

Therefore, the radius of the balloon is

r = 3[4.60x10-3 kg/m2]/ρ

= 3[4.60x10-3 kg/m2]/[0.0154 kg/m3 ]

= 0.896 m

The mass of the balloon is

m = [4.60x10-3 kg/m2][4πr2]

= [4.60x10-3 kg/m2][4π][0.896 m]2

= 4.64x10-2 kg

B)

The density of the air on earth, ρ = 1.20 kg/m3

The volume of the balloon is

V = (4/3)(π)(0.896 m)3

= 3.01156 m3

In this case, the net force acting on the balloon is

ΣF = ρVg - mg = ma

Then, the initial acceleration of the balloon is

a = [ρVg - mg ]/m

= [(1.20 kg/m3)(3.01156 m3)(9.8 m/s2) - (4.64x10-2 kg)(9.8 m/s2)]/(4.64x10-2 kg)

= 753.474 m/s2

= 753.5 m/s2        (nearly)

= 754 m/s2            (nearly)

Direction: Upward

C)

The volume of the total system is

V' = (4/3)π(5r)3

= (4/3)π(5)3(0.896 m)3

= 376.446 m3

The mass of the total system is

m = [4.60x10-3 kg/m2][4π(5r)2]

= [4.60x10-3 kg/m2][4π(5r)2]

= [4.60x10-3 kg/m2][4π](0.896 m)2]

=1.159587 kg

The buoyant force is equals to the weight of the total mass (balloon+load) is

F = (m + m')g

ρV'g = (m + m')g

ρV' = (m + m')

Therefore, the required mass is

m' = ρV' - m

= (0.0154 kg/m3)(376.446 m3) - (1.159587 kg)

= 4.64 kg

Part A

The mass of the balloon is  4.64x10-2 kg.

Part B

Direction: Upward

Part C

The required mass is 4.64 kg