Calculating Meteoroid Speeds on a Collision Course with Earth

[Winegar12]

Homework Statement

Two meteoroids are heading for earth. Their speeds as they cross the moon's orbit are 2.0 km/s. The first meteoroid is heading straight for earth. What is its speed of impact?
The second misses the Earth by 5000 km. What is its speed at its closest point?

Homework Equations

I figured out the first part using the equation Vf=sqrt(2GM((1/Rmoon+Rearth)-(1/r(distance between moon&earth))) and got 11.3 which is the right answer.

The Attempt at a Solution

I'm assuming that I would use the same equation, but I'm not sure. In all honesty I have no idea how to figure out the 2nd part or how to set it up. Do I use the same equation, if so where does the 5,000 km (which I change to m) go in the equation?

 

[tiny-tim]

Welcome to PF!

Hi Winegar12! Welcome to PF!

(have a square-root: √ and try using the X² tag just above the Reply box )

I figured out the first part using the equation Vf=sqrt(2GM((1/Rmoon+Rearth)-(1/r(distance between moon&earth))) and got 11.3 which is the right answer.

Your equation is weird.

What does R_moon have to do with it? The Moon isn't anywhere near this asteroid.

The two distances should be the distances from the centre of the Earth to the initial and final positions.

(and that should help with the second part also)

 

[Winegar12]

the Rmoon is the radius of the moon...I didn't really know how to do the first part of the problem, but I looked through the book and followed a problem that was similar to the first part and got the right answer and understand the first part now, but there was nothing really how to do the second...hence why I'm a little confused.

 

[Winegar12]

So for the second part would I do the same equation, but instead of having (1/Rmoon+Rearth)-1/r(distance between moon&earth), would I change the Rmoon+Rearth to the initial position t+final position and then subtract from the center of the earth?

 

[tiny-tim]

Hi Winegar12!

… I looked through the book and followed a problem that was similar to the first part and got the right answer and understand the first part now

No you don't understand the first part, all you've done is find a problem that's not the same, but somehow gives you approximately the right answer.

As I said before, R_moon has nothing to do with this problem … the asteroid is crossing the Moon's orbit, but it isn't anywhere near the Moon itself.

Do you understand why the equation v = √2GM(1/R - 1/r) works, and what R and r are supposed to be?

 

[Winegar12]

Apparently not...the book was showing the R was the radius and r was the distance between the moon and the earth

 

[tiny-tim]

(just got up :zzz: …)

Apparently not...the book was showing the R was the radius and r was the distance between the moon and the earth

You mean R was the radius of the Earth?

Then the radius of the moon doesn't come into it.

Anyway, your equation doesn't include the 2.0 km/s speed given in the question, so it can't be right.

Start again. Call the speed v at distance r, and V at distance R.

You need to use conservation of energy, ie ∆KE = -∆PE.

What is ∆KE in this case (the change in kinetic energy)?

And what is ∆PE?
​

 

[Winegar12]

The book was doing the same type of problem, but with the sun instead of the moon and it said the R in 1/R was the radius of the sun+the radius of the Earth and then you subtracted 1/r (from 1/R) and r being the distance between the sun and the earth. Then you multiplied that answer by 2GM and sqrt the whole thing...sorry I think this is all confusing me more

Thanks for your help anyway, but I did get the second part figured out last night. I'm probably more confused now then when I first started...

 

[tiny-tim]

… Thanks for your help anyway, but I did get the second part figured out last night. I'm probably more confused now then when I first started...

Perhaps it would help if you show us what you've done …

we can see if anything doesn't work, and if it does work, we can explain why.