# Making a Lead Balloon Float

1. Nov 23, 2008

### Ronerin

1. The problem statement, all variables and given/known data
So the question is what volume is required to make a lead spherical balloon float. It is filled with helium, the P=10^5 Pa, the Temp is 300K and the thickness of the lead balloon shell is .1mm

Any ideas on sending me in a direction with this one?

2. Relevant equations

3. The attempt at a solution

2. Nov 23, 2008

### LogicalTime

the system has to be less dense than the air that surrounds it

3. Nov 23, 2008

### Ronerin

I'm looking to determine what Volume will overcome the weight of the balloon itself minimally, but not sure about what steps to take to find it. Any clues?

4. Nov 23, 2008

### LogicalTime

$Density = \frac{ M_{lead \ balloon}+M_{helium \ in \ balloon}} {V_{balloon}}$

so set that equal to the density of air and solve for the radius of the sphere

Last edited: Nov 23, 2008
5. Nov 23, 2008

### Ronerin

how using that do I solve for the radius? Sorry I need so much help and thanks for the interest!

Im not sure of the weight of the lead, just the density of it, and the thickness...so lost.

6. Nov 23, 2008

### mgb_phys

For the balloon to float you need.
Mass lead + mass helium = mass air
(volume envelope * density lead) + (volume balloon * density he) = (volume balloon * density air)
Or:
Area surface * thickness * density lead = volume balloon * ( density air - density he)
Now you just need the formulae for the surface area and volume of a sphere.

7. Nov 23, 2008

### Phrak

For starters look up the area of a sphere of radius r.

8. Nov 23, 2008

### LogicalTime

no problem! I just like to error on the side of too little info so I don't spoil the problem for you, or treat you like you don't know stuff which could offend some people. I'm very happy to help.

the amount of lead we need for the balloon is a spherical shell, and the surface area of a sphere:
$SA_{sphere} = 4 \pi r^3$

since the balloon thickness is very small a good approximation is just taking this surface area and multiplying it by the thickness to get the volume of lead we are using

$V_{lead} = 4 \pi r^3 * (Thickness)$

from there to find the mass of this volume of lead using the density

the volume of the helium is just the volume of the sphere, and then you use the density of helium to find the mass of all the helium

9. Nov 23, 2008

### Ronerin

So if I haven't been given the radius, how should I determine the Surface Area in order to find the Volume?
Thx for All the Interest! :D

10. Nov 23, 2008

### LogicalTime

basically we have to keep it all in theoretical form till the end and then solve for our unknown r
$M_{lead\ balloon} = (Density\ of\ Lead)(4\pi r^2)(Thickness)$
$M_{helium\ in\ balloon} = (Density\ of\ Helium)(\frac{4}{3} \pi r^3)$
$V_{balloon} =\frac{4}{3} \pi r^3$

11. Nov 23, 2008

### Ronerin

Thanks a Ton! I think I can come up with an answer now :D. I didn't realize it was a theoretical R, and was getting very frustrated over my own ignorance! Thanks again!

12. Nov 23, 2008

### LogicalTime

yeah, have to work with it abstractly for a while before you can put numbers in. If you get used to the abstract it's really nice because at the end you can plug in any values you want, like if you wanted to make a titanium balloon etc. just plug in a different density and voila.

13. Nov 24, 2008

### cindyrachelle

I am having trouble with this problem still. Can you please explan in greater detail how to get the volume needed?

What I have so far is that the
density of air= density of lead(SA)(thickness) + Denisty of He (Volume of a sphere)/ volume of a sphere

my volume of a sphere cancel out I solve for radius in the SA and get a very small number of .2821 m

Once I plug that into the volume of a sphere formula I get .0940 m3 or 94 mm3. The density of lead given to us by our instructor was 10000

These numbers seem really small to me, but the thickness is very small. Am I doing this right?

14. Nov 24, 2008

### Ronerin

Are you at ETSU, Dr. Raz?!

Basically its like they say

Mass Air = Mass He + Mass Pb

(air density)(V)=(He density)(V) + (Pb density)(4pi r^2)(.1/1000m)

just use some algebra to move the stuff around and solve for r

15. Nov 24, 2008

### LogicalTime

$Density\ of\ Air = \frac{M_{lead\ balloon} + M_{helium\ in\ balloon}}{\frac{4}{3} \pi r^3}$
substituting and simplifying
$Density\ of\ Air = \frac{3(Density\ of\ Lead)(Thickness)}{r} + (Density\ of\ Helium)$

from here just make sure your units match up, density should be in kg/m^3, make sure thickness is in meters and when you solve r will be in meters

16. Nov 24, 2008

### mgb_phys

I'm not sure why you are trying to solve for the density of air.
You need the equations I wrote in #6 and the volume of a sphere = 4/3 pi r^3

Density of Helium is roughly 0.18 kg/m^3 and air 1.2 kg/m^3

17. Nov 24, 2008

### LogicalTime

the idea is to find the point at which the
$density\ of\ air \geq density\ of\ the\ balloon$

if the balloon is less dense than the air it will float up. Finding where they are equal is the balancing point between the balloon falling down and floating up.

both of our ways work I believe, what do you think mgb_phys?

18. Nov 24, 2008

### LogicalTime

19. Nov 24, 2008

### cindyrachelle

I see where I went wrong. My algebra failed me! Thank you!
I forgot to divide SA by 4/3 pi r^3.

My numbers look a whole lot better now!!