# Momentum conservation of asteroid in a dust cloud

• LANS
In summary: You can use the substitution form for v(t) to get v(t) = -kv^2*t + k*v^1. Plugging this into the equation for dm gives you dm = -kt.
LANS
Note: this is one of the suggested practice problems for my second-year classical mechanics course.

Homework Statement
A spherical asteroid of mass $$m_{0}$$ and radius R, initially moving at speed $$v_{0}$$, encounters a stationary cloud of dust. As the asteroid moves through the cloud, it collects all the dust that it hits, and slows down as a result. Ignore the increase in radius of the asteroid, and its gravitational effect on distant dust grains. Asume a uniform average density D (mass per unit volume) in the dust cloud.

a)show that $$\frac{dv}{dt} = -kv^{3}$$ and evaluate k.
b) find $$v(t)$$

The attempt at a solution

Let $$A_{c} = 2*\pi*R$$ be the cross-sectional area of the asteroid.

Conservation of momentum:
$$m_{0}v_{0} = m(t) v(t)$$
$$dm = (\pi R^2)*(v(t)dt)*D$$
dm is from mass of dust which the asteroid hits in time dt. Cross-sectional area * distance traveled in time dt * dust density.

I'm not sure where to go from here. Any help is greatly appreciated.

Thanks

Last edited:
LANS said:
$$m(t) = m_{0} - dm*t$$
Where did this equation come from? It doesn't really make sense (and besides, the units are inconsistent, so at least that needs to be fixed).
LANS said:
$$dm = (2 \pi R)*(v(t)dt)*D$$
dm is from mass of dust which the asteroid hits in time dt. Cross-sectional area * distance traveled in time dt * dust density.
What's the cross-sectional area of a sphere? Remember that it has to have the proper units for area

Fixed first post.

diazona said:
Where did this equation come from? It doesn't really make sense (and besides, the units are inconsistent, so at least that needs to be fixed).

$$m(t) = m_{0} - dm*t$$

Thanks, didn't quite think about that one. Forgot to consider that the amount of mass the asteroid picks up in a set amount of time is dependant on its velocity, and is thus not a constant term.

You should respond in your new post rather than going back and changing what you previously posted - not only because it makes it possible for others to follow the discussion, but also because editing your original post after it's been replied to is a violation of the PF guidelines.

Anyway, what is the value of this expression?
$$\frac{\mathrm{d}(mv)}{\mathrm{d}t}$$
The resulting equation will be useful to you. Together with the two that are currently in your original post, it will allow you to solve the problem.

Alternatively (really the same as from diazona in the end, but slightly different initially), note that m = m0v0/v, so what's dm in terms of dv?

From part a) onwards you just have to solve the simple ODE.

## 1. What is momentum conservation?

Momentum conservation is a fundamental law of physics that states that the total momentum of a closed system remains constant over time, regardless of any internal or external forces acting on the system.

## 2. How does momentum conservation apply to asteroids in a dust cloud?

In the context of asteroids in a dust cloud, momentum conservation means that the total momentum of the asteroids and the dust particles remains constant as they interact with each other.

## 3. What factors affect momentum conservation in this scenario?

The momentum conservation of asteroids in a dust cloud can be affected by the masses and velocities of the objects involved, as well as any external forces acting on the system, such as gravity or collisions with other objects.

## 4. Can momentum be transferred between asteroids and dust particles in a dust cloud?

Yes, momentum can be transferred between objects in a dust cloud through collisions or other interactions. However, the total momentum of the system will still remain constant.

## 5. How does momentum conservation impact the overall motion of the dust cloud?

Momentum conservation plays a crucial role in determining the overall motion of a dust cloud. As the various objects within the cloud interact and exchange momentum, the cloud's center of mass will remain constant, resulting in a predictable and stable motion of the entire system.

Replies
2
Views
3K
Replies
2
Views
2K
Replies
2
Views
2K
Replies
14
Views
1K
Replies
3
Views
1K
Replies
335
Views
10K
Replies
3
Views
2K
Replies
1
Views
2K
Replies
1
Views
3K
Replies
4
Views
4K