# space ship problem

## space ship problem

An spaceship has a shape of cylinder of length 54 km and internal radius 16 km. It is filled inside by air. The ship has its own gravity, which comes from rotating around its axis once per every four minutes. On inside surface the pressure of air is 1 bar.

The entrance to ship is an orifice in the centre of one of its basis. Before removing the space suit, on axis is breathable air, what is its density comparing to density on cylinder surface. Assume the temperature is the same everywhere.

I have been pondering over this problem for the last three days and couldn't land up with a solution

The entrance to ship is an orifice in the centre of one of its basis. Before removing the space suit, on axis is breathable air, what is its density comparing to density on cylinder surface. Assume the temperature is the same everywhere.

I have been pondering over this problem for the last three days and couldn't land up with a solution

**Pinnaka**- Admin
- Posts : 36

Join date : 2008-05-20

Age : 31

Location : India

## Re: space ship problem

Hi everyone,

I don't think the responses so far quite answer the question, so I'll have a go at it.

First, the original problem states that the air is at a uniform temperature, so we can forget about the adiabatic assumption, although I agree that the adiabatic atmosphere model is more physically realistic.

The type of spaceship mentioned in the problem is a fictional interstellar design, intended to cruise slowly between stars for many hundreds of years, while providing a livable environment and apparent "gravity" via rotation. (It is similar to the spaceship in "Rendezvous with Rama" by Arthur C. Clarke, RIP.) Therefore, it is reasonable to assume that the air has had plenty of time to come to rest relative to the rotating floor of the ship. The problem is therefore one of a rotating cylinder in which the air rotates along with it. While it is true that there will be winds, the velocity of the winds will likely be no more than an average day on earth, while the rotation velocity of the air at the floor of the cylinder is very high. Therefore, the winds are negligible compared to this rotational velocity. They also average away (after all, in the long term, there is nowhere for the air to go, so its average velocity relative to the spaceship must be zero).

You can look at this problem in two ways, which both give the same answer. First, you could use a rotating coordinate system that turns along with the ship. This is the point of view of an observer standing on the floor inside the ship. This observer cannot see that the ship is rotating, but will still feel the effect of the rotation as if it were gravity. In this case, the ship, observer, and air within it are subject to a "fictitious force" pointing outward from the axis of the ship toward the floor. This force acts like gravity, with an artificial strength g' = rw^2, where r is the radial distance from the axis and w is the angular velocity of this ship. Since the entire ship and the air within it are all turning together, w is constant everywhere, so g is proportional to r. alternatively, you could use a non-rotating coordinate system, in which case the air is accelerating due to its circular motion, and has an acceleration a = rw^2 towards the axis of the cylinder. One way of looking at things gives F - mg' = 0 and the other way gives F = ma, but these are algebraically the same equation.

Now the gradient of the pressure is the net force that it exerts per unit volume, and in cylindrical coordinates, the pressure gradient is dP/dr, because the pressure doesn't vary along the z direction or the angular direction. (You can derive this by looking at an infinitesimal cube in cylindrical coordinates, but it's a bit messy, so I'll skip the details. It's a well-known formula anyway.)

Now consider a small "cubical" parcel of air at a distance r from the axis in cylindrical coordinates. It must satisfy Newton's second law, so F = ma gives

F/V = (m/V)a,

where V is the volume. By the ideal gas law, m/V = mP/nGT = MP/GT, where M is the molar mass in kg/kmol and G is the gas constant 8314J/kmol. (Usually the gas constant is called R, but that is already being used for the radius of the spaceship, so I'm calling it G instead). Putting all this into the equation above gives

dP/dr = (MP/GT)*w^2r.

Let's abbreviate by setting A = Mw^2/GT, so that dP/dr = APr. This rearranges into dP/P = Ardr, which integrates to ln(P) = (A/2)r^2 + C. Since we know that at the floor, P = P_0, C must satisfy ln(P_0) = (A/2)R^2 + C, so C = (A/2)R^2 - ln(P_0). Plugging this back into the previous equation and using properties of logarithms gives

ln(P/P_0) = (A/2)(r^2 - R^2).

The left side is dimensionless, so the right side must be also. That makes it convenient to make EVERYTHING dimensionless, by putting back the definition of A and factoring out an R:

ln(P/P_0) = (M(Rw)^2/2GT)((r/R)^2 - 1).

This new constant B = M(Rw)^2/2GT is dimensionless and can be computed directly from known values of the air's molar mass, earth gravity, temperature, radius of spaceship, and gas constant. I get a value of (29kg/kmol)*(16000m*2pi/240s)^2/2*(8314J/kmolK)*T = 306.0111K/T. Assuming you want the spaceship to be at room temperature (this wasn't given in the problem), you can set T = 300K, so B = 1.02.

Therefore, ln(P/P_0) = 1.02((r/R)^2 - 1). Setting r=0 in this finds the pressure at the axis, where the space traveler enters: ln(P/P_0) = 1.02(0 - 1) = -1.02, so the pressure there is

P = P_0*exp(-1.02) = 0.3606 bars.

That is enough to be breathable, but it is equivalent to a very high mountain altitude on earth.

Hope that helps, and good luck on your interstellar journey!

--Pinnaka

noTE: THIS QUESTION HAS BEEN ANSWERED BY OUR EXPERT PINNAKA

I don't think the responses so far quite answer the question, so I'll have a go at it.

First, the original problem states that the air is at a uniform temperature, so we can forget about the adiabatic assumption, although I agree that the adiabatic atmosphere model is more physically realistic.

The type of spaceship mentioned in the problem is a fictional interstellar design, intended to cruise slowly between stars for many hundreds of years, while providing a livable environment and apparent "gravity" via rotation. (It is similar to the spaceship in "Rendezvous with Rama" by Arthur C. Clarke, RIP.) Therefore, it is reasonable to assume that the air has had plenty of time to come to rest relative to the rotating floor of the ship. The problem is therefore one of a rotating cylinder in which the air rotates along with it. While it is true that there will be winds, the velocity of the winds will likely be no more than an average day on earth, while the rotation velocity of the air at the floor of the cylinder is very high. Therefore, the winds are negligible compared to this rotational velocity. They also average away (after all, in the long term, there is nowhere for the air to go, so its average velocity relative to the spaceship must be zero).

You can look at this problem in two ways, which both give the same answer. First, you could use a rotating coordinate system that turns along with the ship. This is the point of view of an observer standing on the floor inside the ship. This observer cannot see that the ship is rotating, but will still feel the effect of the rotation as if it were gravity. In this case, the ship, observer, and air within it are subject to a "fictitious force" pointing outward from the axis of the ship toward the floor. This force acts like gravity, with an artificial strength g' = rw^2, where r is the radial distance from the axis and w is the angular velocity of this ship. Since the entire ship and the air within it are all turning together, w is constant everywhere, so g is proportional to r. alternatively, you could use a non-rotating coordinate system, in which case the air is accelerating due to its circular motion, and has an acceleration a = rw^2 towards the axis of the cylinder. One way of looking at things gives F - mg' = 0 and the other way gives F = ma, but these are algebraically the same equation.

Now the gradient of the pressure is the net force that it exerts per unit volume, and in cylindrical coordinates, the pressure gradient is dP/dr, because the pressure doesn't vary along the z direction or the angular direction. (You can derive this by looking at an infinitesimal cube in cylindrical coordinates, but it's a bit messy, so I'll skip the details. It's a well-known formula anyway.)

Now consider a small "cubical" parcel of air at a distance r from the axis in cylindrical coordinates. It must satisfy Newton's second law, so F = ma gives

F/V = (m/V)a,

where V is the volume. By the ideal gas law, m/V = mP/nGT = MP/GT, where M is the molar mass in kg/kmol and G is the gas constant 8314J/kmol. (Usually the gas constant is called R, but that is already being used for the radius of the spaceship, so I'm calling it G instead). Putting all this into the equation above gives

dP/dr = (MP/GT)*w^2r.

Let's abbreviate by setting A = Mw^2/GT, so that dP/dr = APr. This rearranges into dP/P = Ardr, which integrates to ln(P) = (A/2)r^2 + C. Since we know that at the floor, P = P_0, C must satisfy ln(P_0) = (A/2)R^2 + C, so C = (A/2)R^2 - ln(P_0). Plugging this back into the previous equation and using properties of logarithms gives

ln(P/P_0) = (A/2)(r^2 - R^2).

The left side is dimensionless, so the right side must be also. That makes it convenient to make EVERYTHING dimensionless, by putting back the definition of A and factoring out an R:

ln(P/P_0) = (M(Rw)^2/2GT)((r/R)^2 - 1).

This new constant B = M(Rw)^2/2GT is dimensionless and can be computed directly from known values of the air's molar mass, earth gravity, temperature, radius of spaceship, and gas constant. I get a value of (29kg/kmol)*(16000m*2pi/240s)^2/2*(8314J/kmolK)*T = 306.0111K/T. Assuming you want the spaceship to be at room temperature (this wasn't given in the problem), you can set T = 300K, so B = 1.02.

Therefore, ln(P/P_0) = 1.02((r/R)^2 - 1). Setting r=0 in this finds the pressure at the axis, where the space traveler enters: ln(P/P_0) = 1.02(0 - 1) = -1.02, so the pressure there is

P = P_0*exp(-1.02) = 0.3606 bars.

That is enough to be breathable, but it is equivalent to a very high mountain altitude on earth.

Hope that helps, and good luck on your interstellar journey!

--Pinnaka

noTE: THIS QUESTION HAS BEEN ANSWERED BY OUR EXPERT PINNAKA

**Pinnaka**- Admin
- Posts : 36

Join date : 2008-05-20

Age : 31

Location : India

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