Answer:
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
