The speed of a spaceship flying through dust

[Haorong Wu]

Homework Statement

A spaceship is flying through dust. The density of dust is $\rho$. Suppose the mass of the ship is $m_0$, and its initial speed is $v_0$. Since the dust will attach to the spaceship when it flies through it, the speed of the spaceship would change. Suppose the intersection of the ship is a cylinder with area of $S$, and the ship is flying along the axis of the cylinder, and all dust it passes through would attach to it. Calculate the relation between the speed of the spaceship and time.

Relevant Equations

None

The given solution is:

Suppose after time $dt$, the mass of the ship would increase $dm$, and the speed would become $v_0 - dv$.

According to the conservation of momentum, $\left ( m_0 + dm \right ) \left ( v_0 - dv \right ) = m_0 v_0$.

Omiting $dm \cdot dv$, we got $v_0 dm -m_0 dv =0$.

According to the problem, $dm=\rho S v dt$, so$\int_0^t \rho S v_0 dt = \int_{v_0}^v m_0 \frac {dv} v$

Thus, the result is $v=v_0 e^{\frac {\rho S v_0 t} {m_0}}$

However, I could not agree this solution, and my solotion is:

Suppose at time t, the mass of the ship is $m$, and its speed is $v$.

Also, suppose after time $dt$, the mass of the ship increases $dm=\rho S v dt$, so $\frac {dm} {dt}= v s \rho$.

Then according to the conservation of momentum, $mv=m_0 v_0$, or $m=\frac {m_0 v_0} {v}$

Then differentiate it with t, it becomes $\frac {dm} {dt}= \frac {-m_0 v_0} {v^2} \frac {dv} {dt}$

Substituting $\frac {dm} {dt}$, then $v s \rho= \frac {-m_0 v_0} {v^2} \frac {dv} {dt}$

Integrating again, then $t=\frac {m_0 v_0} {s \rho v^2} +c$. According to the initial conditions, $c=- \frac {m_0} {s \rho v_0}$.

so $v= \sqrt {\frac {m_0 {v_0}^2} {m_0+ s \rho v_0 t}}$

So which one is correct? it seems a exponential decay would be more convincing.

Thanks!

 

[PeroK]

Homework Statement: A spaceship is flying through dust. The density of dust is $\rho$. Suppose the mass of the ship is $m_0$, and its initial speed is $v_0$. Since the dust will attach to the spaceship when it flies through it, the speed of the spaceship would change. Suppose the intersection of the ship is a cylinder with area of $S$, and the ship is flying along the axis of the cylinder, and all dust it passes through would attach to it. Calculate the relation between the speed of the spaceship and time.
Homework Equations: None

The given solution is:
However, I could not agree this solution, and my solotion is:
So which one is correct? it seems a exponential decay would be more convincing.

Thanks!

The given solution doesn't look right. It uses $m_0$ in the differential equation, but that should be $m$.

Your solution looks correct to me.

 

[Haorong Wu]

The given solution doesn't look right. It uses $m_0$ in the differential equation, but that should be $m$.

Your solution looks correct to me.

Thanks, I found that the given solution is exponential increasing which is absurd.

 

[PeroK]

PS here's why it is NOT a negative exponential solution.

An exponential solution arises from $\frac{dv}{dt} = kv$. I.e. the acceleration must be proportional to the velocity. In this problem, as the velocity reduces, the amount of dust per second reduces in proportion. So, if the mass of the ship stayed the same, then the reduction in velocity would be proportional to the velocity. But, the ship is also gaining mass, which means its velocity is less affected by the dust over time. This means it must be slower than exponential.

 

[Orodruin]

First of all, please do not put the solutions in quotes, it makes it impossible to use the quote feature to quote them.

The given solution is wrong for several reasons reasons. First of all, the original assumption should be $(m+dm)(v+dv)= m_0 v_0$ in order to be valid at all times. Otherwise it is only valid at t=0. Also note the difference in the sign of dv. This change is crucial as it will make dv negative when v decreases and vice versa. If you use a negative sign you will not get v when you integrate dv! This is reflected in the given solution (which does this mistake) having an exponentially increasing solution, which is clearly unphysical.

Noting that $mv = m_0 v_0$ should then give you the same relation as you derived from differentiating that.