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The Question

(Submitted January 06, 1997)

My son, as a requirement for his 7th grade science class, is to have a working model science fair project. While going through some magazines he came upon the idea to build a "Trash Bag Hot-Air Balloon", which would show and demonstrate one type of lighter than air vehicle.

While this seemed to be a great idea for his project, his teacher is not too enthused as "everyone knows how they work." As a result of his comments we have been trying to find the explanation of the scientific principles that are being demonstrated in this project and, hopefully, prove his project more "science project" worthy to his teacher.

Is there any help, information, or advice you can possibly offer that will help him with this project?

Thank You for your time.

The Answer

Our expertise is in designing and building detectors to collect X-rays and gamma-rays from astrophysical objects, and then to interpret the data. It might seem like hot air balloons would be a little outside of our area of interest, but actually, since the radiation we are most interested in observing is absorbed by the Earth's atmosphere, many of our high energy astrophysics experiments are flown on balloons. This way they can get above a substantial fraction of the absorbing atmosphere. The balloons used for scientific payloads are helium filled, but the principle of employing a balloon filled with a lighter gas to get and remain airborne is basically the same.

The basic physics behind hot air balloon travel is the effect of increased temperature on the motions of molecules of a gas, and thereby on the density of the gas. In order to understand this, you'll need a little algebra, and one of the basic ideas of thermodynamics, called the Ideal Gas Law.

A hot air balloon stays afloat in the cooler air surrounding it due to the buoyant force on it. This is the same force that acts on you when you are in a pool of water. You have probably noticed it is much easier to lift someone if you are both in a pool of water, and this is due to the partial support the water is offering: the buoyant force. This force was studied by the Greeks before 200 B.C. and can be understood by Archimedes' principle: any body completely or partially submerged in a fluid is buoyed up by a force equal to the weight of the fluid displaced by the body. If B represents the buoyant force and W the weight of the displaced fluid, then B=W.

We need to consider the sum of the forces acting on the balloon, which is totally submerged in the air around it. The buoyant force B acts upward on the balloon, and gravity acts downward. The weight of the balloon, w, is the same as the gravitational force downward. Since these two forces act in opposite directions (buoyant force up, gravity down), the total force on the balloon is F(total)=B-w.

Now we need to represent each force in terms of things we can measure. These are: the density of the fluid and the volume of the balloon. Remember that density is defined as the mass of an object divided by its volume. The weight of the displaced air equals the buoyant force, and weight is always equal to the mass times the acceleration due to gravity (W=Mg). The mass of the displaced air, M, is just the density of the displaced air multiplied by the volume of the balloon (that's the volume being displaced). So, if D=density of the cool, surrounding air, then W=D*V*g, where V is the volume of the balloon. And since B=W, B=D*V*g. For the weight of the balloon, w=mg. Assuming the basket attached to the balloon can be ignored for now, we need only get the mass of the air inside the balloon. That will be equal to the density of the air inside, d, times the volume. w=d*V*g. Then F(total)=B-w=D*V*g-d*V*g=(D-d)*V*g. When this is a positive number, the force is in the upward direction. That occurs when the density of the air inside the balloon, d is less than the density of the surrounding, cooler air, D.

How can the density of the air inside become less that the air outside? Since the gas inside is the same as outside the balloon, we can use the Ideal Gas Law to study what happens to the density as the temperature is increased. One way of stating this law is: for a gas with a constant molecular weight, the pressure is proportional to product of the density and the temperature (P=K*D*T). Here, K is just a constant, and T is the temperature. That means for a gas at constant temperature, the density is given by: D=P/(K*T). As the temperature increases, the density decreases. At this point the story is almost complete. If you and your son are still interested in following this line for a science project, then I will leave it to you to discover what is happening to the molecules of gas inside the balloon that causes the density to drop. You can find a discussion of the ideas of buoyancy, and how gases are affected by temperature, in any introductory high school or college physics book. The books will have lots of pictures and worked-out examples, and will give you a more detailed description than the one here.

I spoke with a high school physics teacher this weekend about your question, and she told me that there is a kit available to students for making a hot air balloon. Perhaps your son's teacher was concerned that if he used a kit to build the balloon he might not learn as much as if he started a project from scratch. I would encourage you to talk to the teacher some more, and find out what they want the kids to get out of their project. If your son comes away from this project with a better understanding of the forces acting on a balloon, and on the effects of heat on a gas, then it sounds like a good learning experience. It the teacher is aiming for something different in the class, perhaps you and your son can build the balloon on your own in your spare time!

I hope this helps. Good luck with the balloon and the science fair.

Regards,
Padi Boyd
for Imagine the Universe!

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