# Speed of a meteor before hitting a space station

1. Jul 2, 2013

### andreea_a

1. The problem statement, all variables and given/known data
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.

3. 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 )

Last edited: Jul 2, 2013
2. Jul 2, 2013

### Staff: Mentor

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.

3. Jul 2, 2013

### D H

Staff Emeritus
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 $||\vec v||^2 = GM(2/r-1/a)$.

4. Jul 2, 2013

### andreea_a

It's the vis viva equation.

5. Jul 2, 2013

### andreea_a

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

6. Jul 2, 2013

### Staff: Mentor

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

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

7. Jul 2, 2013

### D H

Staff Emeritus
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.

8. Jul 2, 2013

### Staff: Mentor

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...

9. Jul 2, 2013

### 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:
... should that be "trajectory"?
Does this mean that the meteor is travelling radially wrt the earth as it crosses the orbit of the satellite?
This is important to working out the vectors.

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

10. Jul 2, 2013

### D H

Staff Emeritus
That is the correct interpretation.

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

11. Jul 3, 2013

### andreea_a

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

12. Jul 3, 2013

### D H

Staff Emeritus
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.

13. Jul 3, 2013

### Staff: Mentor

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.

14. Jul 4, 2013

### andreea_a

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)

15. Jul 4, 2013

### Staff: Mentor

Right.
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.
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.

16. Jul 4, 2013

### 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)

17. Jul 4, 2013

### D H

Staff Emeritus
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.

18. Jul 4, 2013

### Staff: Mentor

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.

Last edited: Jul 4, 2013
19. Jul 4, 2013

### andreea_a

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...

20. Jul 4, 2013

### D H

Staff Emeritus
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.