How Can Inflatable Balloons Enable Mars Exploration?

AI Thread Summary
Inflatable balloons could facilitate Mars exploration by utilizing the low density of the Martian atmosphere, which is 0.0154 kg/m^3. To achieve buoyancy, calculations indicate that a balloon with a radius of approximately 0.974 meters and a mass of 0.0596 kg would float just above the Martian surface. When considering the same balloon size on Earth, the initial acceleration was incorrectly calculated, yielding an unrealistic value of 754 m/s^2. The calculations for buoyancy and mass displacement were discussed, highlighting the need for clarity in force balance equations. Overall, the exploration concept hinges on understanding the physics of buoyancy in different atmospheric conditions.
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Fluid Mechanics Question! Help needed :(

Homework Statement



It has been proposed that we could explore Mars with inflatable balloons. the density of the Martian atmosphere is 0.0154kg/m^3 (ignore that this varies with temperature). Suppose we make the balloons out of a thin tough plastic where density is such that each square meter has mass 5.00g, we inflate them with light gas of negligible mass.
(a) what's the radius and mass of the balloons such that they float immediately above the martian surface?
(b) If balloons of the same size were used on Earth, find the initial acceleration. the density of the Earth's atmosphere is 1.20kg/m^3.
(c) If on Mars these balloons have 5X the radius they currently do, how heavy an instrument package could they carry?

Homework Equations



density=mass/volume
volume sphere=(4/3)πr^3
area of sphere =4πr^2
f=ma

The Attempt at a Solution



(a) density of martian atmos. same as balloon's if it floats just above surface.
mass of balloon=(5g * 4πr^2)=0.02πr^2 kg
dens.=mass/volume

0.0154=(0.02πr^2)/((4/3)πr^3)

r=0.974 m and mass=0.0596 kg (both 3sf)

(b) here's where I became unstuck! Volume of balloon= 3.87 m^3 (3sf) therefore mass of air displaced =4.64 kg(3sf). using F=ma.

Mass of air displaced(buoyancy) - Mass of balloon = (mass of balloon)*acceleration.

plugging in my numbers, got a ridiculous answer for a, a=754 m/s^2 , clearly I have made some error.
I would greatly appreciate any help at all.

many thanks,

S.
 
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I got about the same answer (769 vs 754). What you did looks OK, although I certainly would have had the g's explicitly in the force balance equation.
 
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