## Mars balloon , forces, pressure

1. The problem statement, all variables and given/known data
We have a balloon that is need to stay buoyant in Mars atmosphere, to make it hover.
Density given for Mars's atmosphere is 0,0154 kg/m^3. Lets assume that we have to make balloons out of thin firm plastic, with surface density of 5 g/m^2. Balloons would be inflated with a very light gas that we can neglect.

What radius(spherical balloon) would balloon have to have, to make it hover above Mars surface.

When we would put the same balloon in Earths atmosphere, with air density of 1,2 kg/m^3, what would happen? Would the balloon go up or down? Find starting acceleration.

2. Relevant equations

G=mg
F=rho*g*v (Buoyant force)
V=4/3*r^3* pi
A=4pi*r

3. The attempt at a solution

I solved the first part. U equalize mg and rho*g*v. g of mars cancels. Mass u get from surface density * surface it self. etc etc.

u get effective radius of: r = 0,974 m.

BUT second part is a bit tricky.

we have newtons first law, ma= rho(air)*g*V - mg.

m we calculated before, and we get that ma=44,97 N

m was 0,059 kg. And from these statements we get that

a=762,34 m/s^2 which is physical absurd imho.

Help? Does now air inside balloon matter or?

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 I can find no fault in your calculation. Of course the moment it started to move, air-resistance would come into effect and the balloon would very quickly reach terminal velocity.
 Very non intuitive! That is my problem. I almost get the speed of sound in a second an a half. But i think real-case scenario is much more complicated. Thanks

## Mars balloon , forces, pressure

 Quote by Bassalisk Very non intuitive! That is my problem. I almost get the speed of sound in a second an a half. But i think real-case scenario is much more complicated. Thanks
But it will not accelerate at this rate for any appreciable length of time. When it starts to move you have to consider air resistance.

Using the drag equation the terminal velocity of your balloon is 7.3m/s, so it will not get any faster than this.

 Quote by Mr.A.Gibson But it will not accelerate at this rate for any appreciable length of time. When it starts to move you have to consider air resistance. Using the drag equation the terminal velocity of your balloon is 7.3m/s, so it will not get any faster than this.
Can you explain this equation? I found it on wiki, but components are not as clear.

Thanks

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 You calculate the drag force from the eqation, it's components are: ρ the density of air, v the velocity of the balloon, when the forces are balanced this is the terminal velocity A cross-sectional area of the balloon, calculate from the radius C_d Drag coefficient, this changes for different shapes for a sphere it is 0.47

 Quote by Mr.A.Gibson You calculate the drag force from the eqation, it's components are: ρ the density of air, v the velocity of the balloon, when the forces are balanced this is the terminal velocity A cross-sectional area of the balloon, calculate from the radius C_d Drag coefficient, this changes for different shapes for a sphere it is 0.47
K but i still don't get how did u end up with 7,3 m/s ?

 bump
 When at terminal velocity the forces are in balance Bouancy = weight + drag Put in the formula you have and then rearrange for v. Then show an attempted solution.