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