Cosmic dust

Consider the diagram below:

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as the comet moves through the cloud, it presents a normal area, $\pi r^2$ , to the meteoroids and each area element on its surface, $dA$ , accumulates meteoroid matter in vertical stacks. By this mechanism, the comet grows in such a way that its two ends maintain their spherical character while the middle becomes a cylinder of height $L$ with diameter equal to the diameter of the original sphere. The rate of matter accumulation is therefore given by the normal area times the velocity times the density of the accumulating meteoroid matter ( $\dot{m} = \rho\pi r^2 v$ ).

By the end of the trip through the cloud, each normal area element $dA$ on the surface will pick up the mass $\rho L dA$ , and therefore the comet will have picked up the new mass $\rho\pi r^2 L$ .

The important idea here is that the comet maintains its original momentum throughout its transit of the cloud. Mathematically, we have:

$\displaystyle 0 =\dot{p} = \dot{m}v + m\dot{v} \rightarrow \dot{v} = -\frac{\dot{m}}{m}v$

Employing the rate of matter accumulation we found above, this becomes:

$\displaystyle \dot{v} = -\rho\pi r^2 \frac{v^2}{m}$

Multiplying the top and bottom by $v$ , we find

$\displaystyle \dot{v} = -\rho\pi r^2 \frac{v^3}{p}$

As the momentum, $p$ , is conserved throughout the process, it is a constant, $p_0 = Mv_0$ , in the above relation. We therefore have a simple equation governing the velocity of the comet. Solving, we have:

$\displaystyle v(t) = \frac{v_0}{\displaystyle\sqrt{1+\frac{2 \pi \rho r^2 t v_0}{M}}}$

The final velocity of the comet can be found by the conservation of momentum:

$\displaystyle \begin{aligned} M v_0 &= \left(M + \rho\pi r^2 L\right)v_f \\ v_f &= \frac{v_0}{\displaystyle 1 + \frac{\rho\pi r^2 L }{M}} \end{aligned}$

We can now solve for the time $t^*$ at which the comet reaches its final velocity (when it exits the cloud). We find:

$\displaystyle \boxed{t^* =\displaystyle L\frac{\pi L \rho r^2+2 M}{2 M v_0}}$

To summarize, the comet velocity goes as

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There are no external forces, so linear momentum is conserved. Furthermore, The comet experiences inelastic collisions with each particle in the cloud because we are given that each particle sticks onto the comet.

Let $M$ equal the mass of the comet and $v_{f}$ be the velocity at some time after the comet enters the gas cloud. Also, let $m$ be the mass gained by the comet at some time. Writing our momentum equation, we get $M \times v_{o} = (M+m) \times v_{f}$ . Solving for $v_{f}$ , we get

$v_{f}=\frac{M \times v_{o}}{(M+m)}$ $(1)$

Lets also imagine what happens when the comet plows through the gas cloud. It will leave in its wake a cylinder with formula $V=\rho A x$ , where $A= \pi \frac{D^{2}}{4}$ is the circle with the comet's radius.

Because $m=\rho V =\rho A x$ , we see that the comet gains mass linearly with distance. Plugging this into $(1)$ , we get that $v_{f}=\frac{M \times v_{o}}{(M+\rho A x )}$ . We see that this is a differential equation by setting $v_{f}=\frac{M \times v{o}}{(M+\rho A x )}=\frac{\mathrm{dx} }{\mathrm{d} t}$ . Integrating, $\int_{0}^{t_{f}} (M \times v_{o} )dt= \int_{0}^{x_{f}} (M+\rho A x )dx$ , with $x_{f}=500km$ and plugging in the other given constants, we get $t_{f}=5.628$

By the drag equation, we know that $\frac{dv}{dt}=kv^2$ , where $k=-\frac{\rho C_DA}{2m}$ . Integrating, we get $\frac{1}{v_0}-\frac{1}{v}=kt\implies v=\frac{dx}{dt}=\frac{v_0}{1-v_0kt}$ . Integrating again, we get $x=-\frac{1}{k}\ln(1-v_0kt)\implies t=\frac{1-e^{-kx}}{v_0k}$ . Plugging in our values, we get $k=-\frac{(10^9)(0.47)(4\pi(8/2)^2)}{2(10^{14})}=-4.725\times10^{-4}$ , and so $t=\frac{1-e^{4.725\times10^{-4}(500)}}{100(-4.725\times10^{-4})}=\boxed{5.64\text{s}}$

In this question we have to ignore the gravitational field due to the comic dust. Otherwise we cannot conserve linear momentum

This problem involves use of conservation of linear momentum, kinetics and geometry. With initial mass and velocity given, we must find the final mass to find out the final velocity. So:

Cross Sectional Area of comet = (pie)x(radius)^2 = (pie)x(4km)^2 = 16(pie) km^2 --> Be careful with units!! Volume of cloud = (pie)x(L)x(R^2) = 16(pie)x500 km^3 = 8000(pie) km^3 Mass gained by comet = density x volume of cloud = 10^9 x 8000(pie) kg = 8x10^12 x (pie) kg Final Mass= (10^14 + 8x10^12xpie) kg According to conservation of linear momentum, pi = pf, so

mi x vi = mf x vf (plug in the values!!)

10^16 = Vf x (10^14 + 8x10^12 x pie) Vf= 79.92 km/s

Using the kinetics formula, S=0.5x(u+v)t,

500=0.5(100+80)t, and t comes out to be 5.5555 (please note that the slight discrepancy with official answer results as rounded-off values)