Conservation of momentum of spacecraft and asteroid

[songoku]

Homework Statement

Please see below

Relevant Equations

Conservation of Momentum

By "DART will have a relative speed of 6250 ms^-1 when it collides with the asteroid", I assume it is the relative speed of the DART with respect to the asteroid.

Using that assumption, I can answer question (a)

For question (b), I don't understand the solution from the teacher. He did it like this:
$$\tan \theta=\frac{\Delta v_{\text{asteroid}}}{\text{initial velocity of asteroid}}$$
$$=\frac{4 \times 10^{-4}}{0.16}$$
$$\theta = 0.14^o$$

What's the logic behind that working?

I tried using conservation of momentum but got stuck because I don't know the initial speed of the DART.

Thanks

 

[Bandersnatch]

What's the logic behind that working?

Draw the velocity vectors for the asteroid - the initial, the final, and the change. Remember that the change in velocity can only happen along the direction of the collision. This should nett you a right triangle, to which you can then apply trigonometry.

 

[songoku]

Draw the velocity vectors for the asteroid - the initial, the final, and the change. Remember that the change in velocity can only happen along the direction of the collision. This should nett you a right triangle, to which you can then apply trigonometry.

Maybe you mean that triangle, where:
a = initial velocity of asteroid
b = change in velocity of asteroid
c = final velocity of asteroid

But why don't the triangle in the shape like this:

What does it mean by "along the direction of collision" ? Is it direction of x in the picture below?

Thanks

 

[PeroK]

View attachment 298435
Maybe you mean that triangle, where:
a = initial velocity of asteroid
b = change in velocity of asteroid
c = final velocity of asteroid

But why don't the triangle in the shape like this:
View attachment 298441

What does it mean by "along the direction of collision" ? Is it direction of x in the picture below?
View attachment 298445

Thanks

It says in the question to assume that the collision is at $90$ degrees to the direction of the asteroid's velocity. Conservation of momentum (in all directions) tell you the rest.

 

[Bandersnatch]

But why don't the triangle in the shape like this:

Because you are told in the problem that the collision with DART is at 90 degrees to the velocity of the asteroid. So any velocity the asteroid may have gained ($\Delta V$) will be in the same direction as DART was moving along. Otherwise, the asteroid would have gained some momentum 'sideways', even though nothing ever hit it sideways. That would run afoul of conservation of momentum, so it's disallowed.

Maybe you mean that triangle, where:
a = initial velocity of asteroid
b = change in velocity of asteroid
c = final velocity of asteroid

Yes. Along the direction of collision would mean 'like b', since that's the direction DART hits the asteroid towards. That's the direction of the push, of the impulse.
It is, by the way, the same picture as the last one you made. The velocity vector c in the former is the one you labelled 'x' in the latter. The x is the sum of the other two.
There is a potential false intuition that might creep in here and confuse you, to the effect of: since x is longer than initial velocity, the body must have accelerated sideways as a result of the collision. But in reality, it is only longer in the way they hypotenuse in a right triangle is longer than its base. All the extra length is gained at the right angle.

 

[songoku]

Because you are told in the problem that the collision with DART is at 90 degrees to the velocity of the asteroid. So any velocity the asteroid may have gained ($\Delta V$) will be in the same direction as DART was moving along. Otherwise, the asteroid would have gained some momentum 'sideways', even though nothing ever hit it sideways. That would run afoul of conservation of momentum, so it's disallowed.Yes. Along the direction of collision would mean 'like b', since that's the direction DART hits the asteroid towards. That's the direction of the push, of the impulse.
It is, by the way, the same picture as the last one you made. The velocity vector c in the former is the one you labelled 'x' in the latter. The x is the sum of the other two.
There is a potential false intuition that might creep in here and confuse you, to the effect of: since x is longer than initial velocity, the body must have accelerated sideways as a result of the collision. But in reality, it is only longer in the way they hypotenuse in a right triangle is longer than its base. All the extra length is gained at the right angle.

I understand this.

It says in the question to assume that the collision is at $90$ degrees to the direction of the asteroid's velocity. Conservation of momentum (in all directions) tell you the rest.

I have written the conservation of momentum in all directions but I can't see from the equation that the change in velocity of asteroid will be 90 degree to its initial direction of motion.

Let asteroid moving to the right collides with DART moving upwards, where:
m_a = mass of asteroid
m_d = mass of DART
u_a = initial velocity of asteroid
u_d = initial velocity of DART
v = final velocity of asteroid and DART
θ = angle between v and horizontal

From conservation of momentum in x - direction:
m_a . u_a = (m_a + m_d) . v cos θ ... (1)

From conservation of momentum in y - direction:
m_d . u_d = (m_a + m_d) . v sin θ ... (2)

Dividing (2) by (1): tan θ = (m_d . u_d) / (m_a . u_a) ... (3)

Change in velocity of asteroid in x - direction: Δv_x = v cos θ - u_a
Change in velocity of asteroid in y - direction: Δv_y = v sin θ

Change in velocity of asteroid:
$$\Delta v = \sqrt{(\Delta v_x)^2 + (\Delta v_y)^2}$$
$$=\sqrt{v^2-2v u_a \cos \theta +u_a^2} ~... (4)$$

From all of those equations, how can I know the change in velocity of the asteroid will have a direction of vertically upwards? I expect Δv_x to be zero so v cos θ = u_a but how to get this relation?

Thanks

 

[Tom.G]

but I can't see from the equation that the change in velocity of asteroid will be 90 degree to its initial direction of motion.

The wording there is critical to understanding. There are two main ideas:
1) That the spacecraft hits the asteroid perpendicular to the astroid's path. That means the asteroid originally had no velocity in that direction.
2) The question asks for the CHANGE in travel direction of the asteroid caused by the spacecraft impact.

Your first sketch in post #3 shows the result, with the length 'b' being the CHANGE in asteroid direction, not its new path.

If the spacecraft approached and hit the asteroid from perhaps the 7 o'clock position in the diagram, then your second diagram would almost show the result. "Almost" because the impact would be at the point where sides 'a' and 'b' meet and the new path would be side 'b'. Side 'c' would not exist because nothing happened where 'a' and 'c' meet

Hope this helps!
Tom

 

[haruspex]

That means the asteroid originally had no velocity in that direction.

That it originally had no velocity in DART's direction.

with the length 'b' being the CHANGE in asteroid direction

Theta is the change in the asteroid direction. In the diagram, b represents the change in its velocity vector.

 

[Delta2]

@Bandersnatch is it enough to change an asteroid speed by 0.4mm/s so that is collision course with Earth is altered enough to prevent collision? I guess maybe when the asteroid is far away from Earth right?

 

[jbriggs444]

@Bandersnatch is it enough to change an asteroid speed by 0.4mm/s so that is collision course with Earth is altered enough to prevent collision? I guess maybe when the asteroid is far away from Earth right?

Suppose that the projected Earth impact was not dead center but at a point some 2000 km into the Earth's disc. Obtaining a 2000 kilometer deflection at 0.4 mm/second would require 2000 * 1000 * 1000 * 2.5 = 5 billion seconds. That is about 158 years.

If we could arrange for an impact one year in advance then that would suggest the need for 158 DART sized impactors.

There is some uncertainty about the effect of the impact. It might not be a simple merger. Chasing down a reference from Wikipedia to a somewhat better than average journalistic source...

There's a degree of uncertainty over how Dimorphos will respond to the impact, in part because its interior structure isn't known. If Dimorphos is relatively solid inside, rather than full of spaces, it might produce lots of debris - which would give the object an extra push.

That said, there is little reason for optimism about a real life scenario.

 

[songoku]

Thank you very much for all the help and explanation Bandersnatch, PeroK, Tom.Gm haruspex, Delta2, jbriggs444