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?

the radius of the balloon r = 0.896 m and mass m = 4.64 × 10⁻² kg

the initial acceleration is a = 753.47 m/s²

the  an instrument package could they carry a required mass of 4.64 kg

Explanation:

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

Given that :

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

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

m = (4.60×10⁻³ kg/m²)(4πr²)

The formula for the buoyant force is expressed as :

F = ρVg

m×g = ρ×V×g

m = ρ×V

Now;

(4.60×10⁻³ kg/m²)(4πr²)= ρ(4/3)πr³

r = 3(4.60×10⁻³ kg/m²)/ ρ

r = 3(4.60×10⁻³ kg/m²)/ 0.0154 kg/m³

r = 0.896 m

Thus; the radius of the balloon r = 0.896 m

The mass of the balloon is (4.60×10⁻³ kg/m²)(4πr²)

m = (4.60×10⁻³ kg/m²)(4π×0.896²)

m = 4.64 × 10⁻² kg

b) what would be its initial acceleration assuming it was the same size as on Mars?

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

V = 3.01156 m³

Considering the net force acting on the balloon ; we have

ΣF = ρVg - mg = ma

However; making the initial acceleration  a of the balloon the subject ; we have:

a = (ρVg - mg)/m

a = (1.20 kg/m³)(3.01156 m³)(9.8 m/s²) - (4.64×10⁻² kg)(9.8 m/s²)]/(4.64×10⁻² kg)

a = 753.47 m/s²

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

The volume of the total system is V' = (4/3)π(5r)³

V' =  (4/3)π(5)³(0.896 m)³

V' =376.446 m³

The mass of the total system is m = (4.60×10⁻³ kg/m²) (4π(5r)²)

m = [4.60×10⁻³ kg/m²][4π][25](0.896 m)²

m = 1.159587 kg

We can then say that  the buoyant force is equals to the weight of the total mass (balloon+load) and is expressed as:

F = (m + m')g

ρV'g = (m + m')g

ρV' = (m + m')

Thus; the required mass m' is =  ρV' - m

m'  =  ρV' - m

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

m' = 4.64 kg

Thus; the  an instrument package could they carry a required mass of 4.64 kg

multiply 2kg by latent heat of vap

beta is electron from nucleus

something from nucleus has changed, but charge must be same for cons of charge