Speed of a meteor before hitting a space station

[andreea_a]

Homework Statement

Hello! Suppose a meteor was approaching the Earth along a distance that passes through the Earth's center.I have a space station that moves around the Earth in a circular orbit (radius R)The meteor hit the space station and becomes incorporated.After the impact the space station moves in an elliptical orbit so that the minimum distance relative to the center of the Earth is R/2. Now my question is:what was the speed of this meteor before hitting the space station?
I know M=Earth's mass,m1 = meteor's mass,m2 = space station's mass and K=gravitational constant.

The Attempt at a Solution

Momentum is conserved,right?So I wrote m1 v1 + m2 v2 = (m1+m2) v where v = speed of the meteor after the impact. I should get from here v1. v2 should be √KM/R(circular orbit) and v = √KM(2/r-1/a)
Substituting I get m1 v1 + m2 √KM/R=(m1+m2)√KM(2/r-1/a)
But can I write in another way the expression for the speed after the impact?Should I replace r by R/2 (velocity at perihelion)?
The final answer should be v = (m1+m2)/m1 √2*K*M/R*(3*m2^2 / 2*(m1+m2)^2 -1 )
Thanks in advance!

 

[mfb]

v = √KM(2/r-1/a)

Where does that equation come from? r and a should be explained there.A strange setup. Such an asteroid would just destroy the space station.

 

[D H]

Substituting I get m1 v1 + m2 √KM/R=(m1+m2)√KM(2/r-1/a)

You are implicitly assuming that you can treat v1, v2, and v as scalars. You cannot do this in general because velocity is a vector. Treating those vectors as scalars is only valid if they are all parallel to one another. They aren't. Draw a picture.

BTW, a better way to write the vis viva equation is $||\vec v||^2 = GM(2/r-1/a)$.

 

[andreea_a]

Where does that equation come from? r and a should be explained there.


A strange setup. Such an asteroid would just destroy the space station.

It's the vis viva equation.

 

[andreea_a]

You are implicitly assuming that you can treat v1, v2, and v as scalars. You cannot do this in general because velocity is a vector. Treating those vectors as scalars is only valid if they are all parallel to one another. They aren't. Draw a picture.

BTW, a better way to write the vis viva equation is $||\vec v||^2 = GM(2/r-1/a)$.

Then if I can't treat them as scalars how can I derive that answer ?

 

[mfb]

It's the vis viva equation.

But how do you get the semi-major axis then?

Then if I can't treat them as scalars how can I derive that answer ?

Treat them as vectors (or use radial/tangential velocity as independent scalar values).

 

[D H]

Then if I can't treat them as scalars how can I derive that answer ?

Draw a picture. At what angle must the meteor and station meet?

You're going to need more than the vis viva equation here. Angular momentum might help.

 

[gneill]

I agree with D H. Both specific energy and specific angular momentum will be required. Also, I think that the suggested solution could benefit from a few parentheses to make the order of operations clear. That square root symbol doesn't cover enough ground...

 

[Simon Bridge]

I'm going to support the others in conservation of energy and momentum: draw a picture, work the vectors ... I just have a few observations about the question:

... a meteor was approaching the Earth along a distance

... should that be "trajectory"?

that passes through the Earth's center.

Does this mean that the meteor is traveling radially wrt the Earth as it crosses the orbit of the satellite?
This is important to working out the vectors.

I have a space station that moves around the Earth in a circular orbit (radius R). The meteor hit the space station and becomes incorporated.

The space-station and the meteor end up stuck together? Got it!

After the impact the space station moves in an elliptical orbit so that the minimum distance relative to the center of the Earth is R/2. Now my question is: what was the speed of this meteor before hitting the space station?

 

[D H]

Does this mean that the meteor is traveling radially wrt the Earth as it crosses the orbit of the satellite?

That is the correct interpretation.

andreea_a: This is very important. It is why you cannot treat those velocities as scalars.

 

[andreea_a]

That square root symbol doesn't cover enough ground...

That's my mistake,sorry. That square root doesn't cover just the 2KM term but all that expression with 3*m2^2/2(m1+m2)^2 - 1

 

[D H]

Have you figured out the geometry of the problem yet?

What made you use the vis viva equation? This is going to make you have to chase down the semi-major axis and the eccentricity. There are easier ways to attack this problem. Big hint: Conservation laws.

 

[gneill]

That's my mistake,sorry. That square root doesn't cover just the 2KM term but all that expression with 3*m2^2/2(m1+m2)^2 - 1

You might want to look into the built-in LaTeX capabilities for rendering the more complicated math. Sure beats sorting out the order of operations of ascii-rendered formulas.

If I'm not mistaken, I believe that your suggested solution would go like this:

$$v = \frac{m_1 + m_2}{m_1}\sqrt{2\frac{\mu}{R}\left( \frac{3}{2}\frac{m_2^2}{(m_1 + m_2)^2} - 1 \right) }$$
where $\mu$ is your KM.

 

[andreea_a]

Have you figured out the geometry of the problem yet?

What made you use the vis viva equation? This is going to make you have to chase down the semi-major axis and the eccentricity. There are easier ways to attack this problem. Big hint: Conservation laws.

So I have the radial velocity of the meteor - the vector is pointing to the center of the Earth,right?
If I understand correctly the reason why I can't apply the conservation of momentum is because that radial velocity of the meteor is not parallel to the space station's velocity. I should apply conservation of angular momentum for the space station?(initially moving in a circular orbit and then in an elliptical one)

 

[gneill]

So I have the radial velocity of the meteor - the vector is pointing to the center of the Earth,right?

Right.

If I understand correctly the reason why I can't apply the conservation of momentum is because that radial velocity of the meteor is not parallel to the space station's velocity.

Conservation of momentum applies just fine. The problem is, you were treating the collision as a 1-dimensional one where the velocities are aligned, but in this case the initial velocities are at right angles to each other. You need to consider a 2-D collision geometry and apply conservation of momentum accordingly. The resultant velocity will have components in both radial and tangential directions.

I should apply conservation of angular momentum for the space station?(initially moving in a circular orbit and then in an elliptical one)

You'll need that, too. It's always a good idea to take note of the all the conserved items
So determine the specific angular momentum and specific energy immediately after the collision. They'll come in handy.

 

[andreea_a]

Before the collision the angular momentum of the space station is $$L_i = m_2 v R$$ where v is the circular velocity $$v = \sqrt{\frac{KM}{R}}$$
I have one question. How about the final angular momentum of the space station?I mean after the collision the meteor and the space station end up stuck together. For $$L_f$$(final angular momentum) what is the final velocity of the space station moving in an elliptical orbit?(this is one of the reasons I thought about using the vis-viva equation)

 

[D H]

Before the collision the angular momentum of the space station is $$L_i = m_2 v R$$ where v is the circular velocity $$v = \sqrt{\frac{KM}{R}}$$
I have one question. How about the final angular momentum of the space station?I mean after the collision the meteor and the space station end up stuck together.

The collision event is presumably very short in duration, which means that linear momentum is conserved during that event. Use conservation of linear momentum to find the velocity of the station+meteor object. Then find the angular momentum using the definition of angular momentum. Remember that angular momentum is a vector quantity.

Alternatively, angular momentum is also conserved across the collision. The angular momentum of the station+meteor object is the vector sum of the angular momenta of the station and meteor just prior to collision. In fact, angular momentum is conserved not just across the collision event but conserved, period, since the only force is the radial gravitational force.

 

[gneill]

Before the collision the angular momentum of the space station is $$L_i = m_2 v R$$ where v is the circular velocity $$v = \sqrt{\frac{KM}{R}}$$
I have one question. How about the final angular momentum of the space station?I mean after the collision the meteor and the space station end up stuck together. For $$L_f$$(final angular momentum) what is the final velocity of the space station moving in an elliptical orbit?(this is one of the reasons I thought about using the vis-viva equation)

Rather than using the angular momentum, L, you may find it more convenient to use the specific angular momentum, h. That is, the angular momentum per kilogram. It's calculated in the same way as the angular momentum only the mass is left out of the equation. It's handy because it is conserved just as angular momentum is and you don't need to worry about the particular mass of the object. Remember that bodies follow the same orbits regardless of their mass, provided that their mass is much smaller than that of the central body.

The same goes for the specific mechanical energy; It's the energy per kilogram. Again, the mass of the orbiting object is left out. So the specific KE is $\frac{v^2}{2}$ and the specific PE is $-\frac{\mu}{r}$ (with $\mu$ being your KM). Thus
$$\xi = \frac{v^2}{2} - \frac{\mu}{r}$$

Once the the collision happens and two bodies become one, the mass of the resulting object does not change again, and nor does the mass affect the shape and size of the orbit! Conservation of linear momentum will tell you the velocity of the object that results from the collision.

Find both the specific angular momentum and the specific energy for the object immediately after the collision.

 

[andreea_a]

The collision event is presumably very short in duration, which means that linear momentum is conserved during that event. Use conservation of linear momentum to find the velocity of the station+meteor object. Then find the angular momentum using the definition of angular momentum. Remember that angular momentum is a vector quantity.

But didnt'I do that?Using conservation of linear momentum was my first idea and it is written in my first post but you said I can't treat them as scalars...

 

[D H]

But didnt'I do that?Using conservation of linear momentum was my first idea and it is written in my first post but you said I can't treat them as scalars...

Right. You cannot treat velocities as scalars. So treat them as the vectors that they are. Momentum is a vector, also. Use conservation of momentum as a vector equation and you'll be fine.

 

[andreea_a]

This is what I've got so far:
Applying the conservation of linear momentum for the space station and the meteor:
$$\vec{p_i} =\vec{p_f} => m_1\vec{v1}+m_2\vec{v2}=(m_1+m_2)\vec{v_f}$$From this I should derive the relationship for v1-the speed of the meteor before hitting the space station but first I need that v_f.
So from conservation of angular momentum for the space station:$$L_i=L_f ⇔ m_2R\sqrt{\frac{KM}{R}} = (m_1+m_2)v_f r_f$$ But how do I get that r_f?How do I get that final velocity(meaning the velocity of the space station+meteor - velocity in an elliptical orbit)?Even applying conservation of energy it seems that there's something I'm missing.The initial energy of the space station is the energy in that circular orbit and the final one is the energy in the elliptical orbit,right?So$$E_i=E_f ⇔\frac{-KMm_2}{2R} = \frac{-K(m_1+m_2)M}{2r_f}$$
Please tell me what is wrong !

 

[voko]

But how do I get that r_f?

Is $r_f$ different from $r_i$ in a collisions?

 

[D H]

So from conservation of angular momentum for the space station:$$L_i=L_f ⇔ m_2R\sqrt{\frac{KM}{R}} = (m_1+m_2)v_f r_f$$

Once again, no. You need to stop treating vectors as scalars. This is your number one problem with this problem.

The correct formula for angular momentum is $\vec L = \vec r \times (m\vec v)$. Note the cross product. One way to calculate the magnitude is via $L = mrv \sin\theta$, where θ is the angle between the position vector and the velocity vector. You don't want to do that. Another way to calculate the magnitude is via $L=mrv_{\perp}$, where v_⊥ is the part of the velocity vector that is perpendicular to the position vector. You can get this from conservation of linear momentum. You didn't use the fact that the pre-collision velocities are normal to one another when you applied conservation of linear momentum.

But how do I get that r_f?

The same way you got the post-collision velocity. Conservation of linear momentum only applies in systems in which there are no external forces. There obviously is an external force here, the Earth's gravitational force. So what makes it valid to use the conservation of momentum? The answer is that the collision takes place very quickly. The impact forces, which are internal forces in the station+meteor system, will overwhelm the gravitational force by many, many orders of magnitude. This in turn means that conservation of linear momentum is approximately true during the collision event. Conservation of linear momentum is exactly true in the limit of the collision event being instantaneous.

If Δt, the duration of the collision event, is extremely small, what does that say about the change in position (vΔt)? In the limit of an instantaneous collision event (Δt → 0), what is the change in the position, exactly?

 

[andreea_a]

Once again, no. You need to stop treating vectors as scalars. This is your number one problem with this problem.

The correct formula for angular momentum is $\vec L = \vec r \times (m\vec v)$. Note the cross product. One way to calculate the magnitude is via $L = mrv \sin\theta$, where θ is the angle between the position vector and the velocity vector. You don't want to do that. Another way to calculate the magnitude is via $L=mrv_{\perp}$, where v_⊥ is the part of the velocity vector that is perpendicular to the position vector. You can get this from conservation of linear momentum. You didn't use the fact that the pre-collision velocities are normal to one another when you applied conservation of linear momentum.

The velocity (space station+meteor) after the impact wouldn't be the perigee velocity?Isn't angular momentum conserved(initial angular momentum in the circular orbit-final angular momentum at perigee)? For the circular orbit θ would be 90 . The same happens at perigee,right? Can't I find this way the perigee velocity ? $v_max =\frac{ 2m_2}{m_1+m_2}\sqrt{\frac{KM}{R}}$
But I couldn't derive the formula for the initial speed of the meteor.

 

[D H]

The velocity (space station+meteor) after the impact wouldn't be the perigee velocity?

No. The space station+meteor is not at perigee immediately after collision. That should be obvious. Perigee distance is R/2. That's a given in the problem statement. Since the collision is essentially instantaneous, the post-collision distance from the center of the Earth is still R.

Isn't angular momentum conserved

Yes. So use the correct formula for computing angular momentum or add the pre-collision angular momentum of the meteor and the pre-collision angular momentum of the station to get the post-collision angular momentum of the station+meteor object.

But I couldn't derive the formula for the initial speed of the meteor.

Try. Show some work, even some partial work, and we'll be able to help you.

 

[andreea_a]

Apart from the speed of the meteor I have to calculate the perigee velocity. The answer for the perigee velocity is $V_max = \frac{2m_2}{m_1+m_2}\sqrt{\frac{KM}{R}}$This is the velocity calculated by me in the previous post. I used the conservation of angular momentum but I am not sure if that's correct. I said initial angular momentum in the circular orbit equals to the final angular momentum which is angular momentum at perigee. From here I get that $v_max$ at perigee which is the same as in my answer book.

 

[D H]

You have the correct perigee velocity. This is not the velocity immediately after impact.

 

[voko]

Are you still unable to calculate the speed of the meteor?

 

[andreea_a]

Are you still unable to calculate the speed of the meteor?

I need $v_f$ - velocity after impact in order to calculate the speed of the meteor

 

[D H]

What is it, symbolically, in terms of the pre-collision velocities of the meteor and space station?

Use conservation of momentum. The total momentum, as a vector quantity, is conserved across the collision event.

 

[andreea_a]

What is it, symbolically, in terms of the pre-collision velocities of the meteor and space station?

Use conservation of momentum. The total momentum, as a vector quantity, is conserved across the collision event.

But I've done that...in the previous posts. The problem is that I have 2 unknowns. In order to get $v_f$(velocity station+meteor after impact) I must know $v_1$(initial velocity of the meteor)

 

[Simon Bridge]

Would it be easier to be more mechanical about this? i.e.

There are three stages.
I. just before the collision
II. just after the collision
III. on reaching perigee

You should have written an explicit statement for the tangential and radial velocities of each object at each stage.

You should have an expression relating each stage by conservation of energy and conservation of momentum. This gives three sets of equations.

Which quantities are conserves between each of the cases?This should give you something like 8 equations and 8 variables depending on your notation convention. The next step is simplification and cancellation.

i.e. Most of the variables will have a trivial relationship to each other or known values ...
eg. At stage I there is one unknown out of four velocities:
1. tangential velocity of satellite: given;
2. radial velocity of satellite: infer from "circular orbit";
3. tangential velocity of meteor: infer from it's trajectory;
4. radial velocity of satellite: unknown
etc.

When you get used to this sort of calculation, you get used to anticipating some of the results by your choice of notation and which quantities you write an expression for. When you are still struggling, it can be a big help to write down everything, be pedantic. It can help to use a large window and a dry-erase marker as you get to see all the setup relations in one place, and you can rub out mistakes and when you want to make substitutions.

I shall now return you to your regular programming...

 

[D H]

The problem is that I have 2 unknowns. In order to get $v_f$(velocity station+meteor after impact) I must know $v_1$(initial velocity of the meteor)

You are making this problem much, much harder than it is.

In post#21 you wrote

This is what I've got so far:
Applying the conservation of linear momentum for the space station and the meteor:
$$\vec{p_i} =\vec{p_f} => m_1\vec{v1}+m_2\vec{v2}=(m_1+m_2)\vec{v_f}$$

What happens when you divide both sides of this expression by $m_1+m_2$?