Question

What is the acceleration of a hot air-balloon if the ratio of the air density outside the balloon to that inside is \[1.39\]? Neglect the mass of the balloon fabric and the basket.

Answer 1

Hint: Acceleration of an object occurs when there is a force applied on it in the direction in which the force is applied on it. The Archimedes’ principle and the concept of buoyancy is applied to determine the equations for the forces applied on the hot air-balloon. Newton's second law is applied as well to construct the relation between the forces that act on the balloon and hence the equation for the acceleration of the balloon is found out.

Complete step by step answer:
The above problem revolves around the concept of buoyancy and the Archimedes’ principle for an object said to be immersed in air. In order to find the acceleration of the balloon we first need to understand the concepts behind Archimedes’ principle and the meaning of buoyancy.

The concept of buoyancy comes into picture when a body is immersed in a fluid. When an object is said to be immersed in a fluid then the fluid is said to apply some amount of pressure on it. The body in fluid is subject to pressure in all directions, but however the pressure that is applied from the bottom of the body is said to be more that the other forces. This upward thrust or the upward force that is applied on the bottom surface of the body immersed in a fluid is called buoyancy.

Here, the body is said to be ‘immersed’ or surrounded by air which is said to be a gas. Hence, the buoyancy force applied by gases is applied because the balloon floating in air is also subject to pressure in all directions applied by the gas molecules. Thus, the Archimedes’ principle for gases is applicable in this case.

Let us now look into the concept of the Archimedes’ principle. The principle states that when a body is partially or wholly submerged in a fluid or a gas, it experiences an upward thrust equal to the weight of the gas displaced by it and its upward thrust acts along the center of gravity of the displaced gas.

The principle proves that the upward thrust or the buoyant force is equivalent to the weight of the gas that is displaced by the body immersed in gas which in this case is the hot air-balloon. Hence this force due to buoyancy is given by the equation:
${F_b} = {m_b}g$ ----------($1$)
Here, ${m_b}$ denotes the mass of the sir which applies the force.

We are asked to omit the mass of the balloon fabric and the basket of the hot air-balloon. But however there are two forces that are applied on the balloon. One force is the buoyant force that is applied by the air present outside the balloon that provides an upward thrust to the bottom surface of the balloon and one force will be applied by the air inside the hot air-balloon downwards back to balance out the applied buoyant force. This is in accordance with Newton's third law.
As per this concept the total mass over here will be the difference between the masses of the air outside the balloon and that inside the balloon. Hence the equation ($1$) becomes:
${F_b} = \left( {{m_o} - {m_i}} \right)g$ -----------($2$)

The air outside is said to be cooler as compared to the air present inside the hot air-balloon and hence the mass of the air outside will be having a greater magnitude that the mass of air inside the balloon. This is because the density of cooler air is more than the density of hot air since the volume of cool air is said to be occupying lesser volume since they are more packed than molecules that have kinetic energy due to their temperature.

 Since density is inversely related to volume the density of cooler air is more. Thus the mass-density relation is given by the equation:
$\rho = \dfrac{m}{V}$
By rearranging the terms we get:
$m = \rho V$
We substitute this in the above equation ($2$) to get:
${F_b} = \left( {{\rho _o} - {\rho _i}} \right)Vg$ -----------($3$)
The volume of both the gases are said to be taken as constant.
We now apply Newton's second law of motion which states that there is an acceleration that is produced due to the force applied on a body which is inversely dependent on its mass. Hence the equation for Newton’s second law is given by:
${F_g} = {m_i}a$
Here, ${m_i}$ denotes the mass of air inside the balloon.

The hot air-balloon is said to accelerate in the upward direction due to the gravitational force that is applied on it and for this force which considers only the balloon and its acceleration then only the mass of the balloon is considered. This mass will be with respect to the gas present inside it
By similarly applying mass-density relation we get:
${F_g} = {\rho _i}Va$ -----------($4$)
As per the Newton’s third law that the every action has an equal and opposite reaction and by applying the Archimedes’ principle that the weight of the air displaced by the body is equivalent to the buoyant force we can say that:
${F_b} = {F_g}$
Now, from equations ($3$) and ($4$) we have:
${\rho _o}Vg - {\rho _i}Vg = {\rho _i}Va$
The common terms are taken out to get:
$\left( {{\rho _o} - {\rho _i}} \right)Vg = {\rho _i}Va$
The common terms are cancelled out to get:
$\left( {{\rho _o} - {\rho _i}} \right)g = {\rho _i}a$

We now divide the equation by ${\rho _i}$ on either sides to get:
$\dfrac{{({\rho _o} - {\rho _i})g}}{{{\rho _i}}} = \dfrac{{{\rho _i}a}}{{{\rho _i}}}$
By splitting up the terms we get:
$\dfrac{{{\rho _o}g}}{{{\rho _i}}} - \dfrac{{{\rho _i}g}}{{{\rho _i}}} = \dfrac{{{\rho _i}a}}{{{\rho _i}}}$
By cancelling out the common terms we get:
$\dfrac{{{\rho _{out}}g}}{{{\rho _{in}}}} - g = a$
Taking the common term $g$out we get:
$\left( {\dfrac{{{\rho _{out}}}}{{{\rho _{in}}}} - 1} \right)g = a$
$a = \left( {\dfrac{{{\rho _{out}}}}{{{\rho _{in}}}} - 1} \right)g$ ---------($5$)
Hence this is the equation for acceleration of the balloon
We are given the ratio of the densities of air outside and inside the hot air-balloon in the question. We are asked to find the acceleration of the balloon.

Given, $\dfrac{{{\rho _o}}}{{{\rho _i}}} = 1.39$ [Since, density of air outside is more that density of air inside]
By substituting this in equation ($5$) we get:
$a = \left( {1.39 - 1} \right)\left( {9.8} \right)$
By further solving this out we get:
$a = \left( {0.39} \right)\left( {9.8} \right)$
$ \Rightarrow a = 3.822m/{s^2}$
By approximating to $3$ significant figures we get:
$ \Rightarrow a = 3.82m/{s^2}$
This acceleration of the balloon is said to be upwards since the upward force or the buoyant force is greater than the gravitational force and hence the acceleration is in the direction to the buoyant force which is upwards.

Note: There is often a misconception that Archimedes’ principle can only be applied to objects that are immersed in a fluid which is wrong. As we can see in the solution of the given problem, we can apply Archimedes’ principle to any object that is said to be partially immersed in not only water but the law is also applicable to gases (or air) as well since there is said to be some kind of buyout force applied on the object by the gases as well.