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let there be a spaceship which is floating in space with a surface size of A. The spaceship floats in a constant speed of $v_0$, dust with density of $\rho$ gets stuck on the (only) surface of the spaceship. The starting mass of the spaceship is $m_0$

Find the velocity as a function of time

So we now that the momentum is preserved as there is no outer force acting on the spaceship:

$$\frac{d(mv)}{dt}=0$$

Because the mass is a function of time the derivative is:

$$\dot{m}v+m\dot{v}=0$$

At an instant of time when the spaceship is in the dust the change in the mass is:

$$dm=\rho Avdt$$

Using again the fact the the momentum is preserved we get:

$$mv=m_0v_0$$

So:

$$\frac{m_0v_0}{v}\dot{v}+v^2\rho A=0$$

How did we arrive to the last equation?

If we take:

$$dm=\rho Avdt\iff \frac{dm}{dt}=\rho Av$$

And:

$$mv=m_0v_0\iff m=\frac{m_0v_0}{v}$$

taking the derivative:

$$\frac{dm}{dt}=\frac{d}{dt}\frac{m_0v_0}{v}\Rightarrow \rho Av=-\frac{m_0v_0\dot{v}}{v^2}$$?

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  • $\begingroup$ You want to get to the final answer as a function of time? $\endgroup$
    – Karthik
    Commented Apr 8, 2019 at 13:01
  • $\begingroup$ @KV18 correct a function of time $\endgroup$
    – newhere
    Commented Apr 8, 2019 at 13:22

1 Answer 1

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I am not entirely sure about this one. But I used the control volume approach and ended up with a first order differential equation which I tried to solve. I will be posting it below, and suggestions are welcome. I, probably, might have got it wrong.

Velocity of spaceship as a function of time

Here,

  v = Instanteneous velocity of spaceship
  v_{d} = Absolute Velocity of dust striking the spaceship 
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