- #1

Haorong Wu

- 417

- 90

- 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:

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!

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!