# How high will a helium balloon rise?

1. Dec 9, 2015

### vetgirl1990

1. The problem statement, all variables and given/known data
A helium balloon of mass 0.2kg and volume 1m3 is attached to a long thin rope of mass density 1g/m. The coils of rope lay flat on the ground, so that as the balloon moves up the rope unwinds without resistance. How high will the balloon rise?
Air density = 1.204kg/m3
a) 1204m
b) 1004m
c) 602m
d) 502m
e) none of the above

2. Relevant equations
P = pgh

3. The attempt at a solution
I think the balloon will stop rising when the pressure inside the balloon equals the pressure of the air on the outside.

Pressure of balloon:
P(balloon) = F/A
V = 4/3πr3 --> r = 0.6204m
SA = 4πr2 --> A = 4.835m2
P(balloon) = F/A = 0.4053N/m2

P(balloon)=P(air)
P(balloon)=ρairgh
0.4053 = (1204)(9.8)h
h=0.0000343m

This answer is clearly wrong... Not sure where I'm going wrong.

2. Dec 9, 2015

### Buzz Bloom

Hi vetgirl:

I think your assumption is wrong, but I may be mistaken. I offer an alternative to consider.

The balloon will rise until the upward force of the air pressure on the bottom half of the balloon minus the downward force of the air pressure on the top half of the balloon equals the weight of the balloon plus the weight of the attached rope that is not still coiled on the ground, that is the height of the balloon times the rope's weight per meter.

Hope this helps.

Regards,
Buzz

3. Dec 9, 2015

### stockzahn

The balloon will stop rising when the net force is zero. What forces affect the balloon?

There is one solution, which doesn't take into account the decreasing density/pressure of air with increasing height - does the statement say something about this simplification?

4. Dec 9, 2015

### SteamKing

Staff Emeritus
Why do you think the pressure inside the balloon is different from the pressure outside when the balloon is released?

If the pressure inside is greater, the balloon keeps expanding, like when you inflate it. If the pressure inside is less, the balloon gets smaller, like when the helium runs out of it.

This is a simple problem in buoyancy, with an added twist.

5. Dec 9, 2015

### vetgirl1990

No it doesn't, which is what threw me off as well. I know air density decreases with altitude, so I suppose that simplification is assumed.

But I suppose it's a simple buoyancy problem then? The forces affecting the balloon are the gravitational force, and buoyancy force.
Fb = -Fg
ρairhAg = -mballoong
And then I think the Area that I use above, is the surface area of the balloon --> SA = 4πr2 --> A = 4.835m2
Then I would solve for height (h).

Does that make more sense?

6. Dec 9, 2015

### stockzahn

OK, then let's solve it like that.

The factor h⋅A in your equation ist the volume of the balloon - you can't calculate the height of the rising balloon from it. The buoyancy force Fb, as you've already stated correctly, is

Fbair⋅Vballoon⋅g

We've decided to neglect the decrease in pressure with increasing height, therefore the volume of the balloon stays constant. Till now you forgot to take the weight of the thin rope into account. So what forces are pointing downards (against the buoyancy) and which one is dependent on the height of the balloon?

7. Dec 9, 2015

### Staff: Mentor

This is incorrect. Have you learned about Archimedes principle? That gives you the upward pressure force of the air on the balloon, and it depends on displaced volume, not area. Also, you omitted the weight of the rope in the force balance equation.

Chet

8. Dec 9, 2015

### SteamKing

Staff Emeritus
No, it doesn't. The buoyant force on the balloon doesn't depend on the surface area of the balloon, but by the volume of air which is displaced by the less dense helium. That's why you're given the density of air and not the atmospheric pressure in the problem statement.

9. Dec 9, 2015

### Buzz Bloom

Hi SteamKing:

Good point! I had made the mistake of thinking that if the buoyancy force did not change with altitude, then then nothing would stop the balloon from rising. But the increase in the length of hanging rope stops the rising. The problem statement is intended to to make a not-quite-realistic simplifying assumption of a non-varying buoyancy force.

Regards,
Buzz

10. Dec 9, 2015

### Staff: Mentor

SteamKing: What do you think of this wording in Buzz Bloom's first post regarding the buoyant force of the air on the balloon?: "upward force of the air pressure on the bottom half of the balloon minus the downward force of the air pressure on the top half of the balloon."

In my judgement, it's technically correct if the directionality of the pressure force on the balloon surface is taken into account, but apparently, the OP interpreted it differently. Also, of course, the buoyant force is obtained much more easily than this.

Thoughts??

Chet

11. Dec 9, 2015

### SteamKing

Staff Emeritus
The buoyant force on any object can be obtained by integrating the pressure over the wetted surface (which is quite tedious) or by simply using the submerged volume to calculate the weight of displaced fluid (easy when objects are fully submerged; a little trickier when objects are only partially submerged). Hopefully, your calculations using each method will agree.

I don't want to quibble with Buzz on this point because it's irrelevant to solving this particular problem.