# Asteroid impact speed

1. Jun 3, 2014

### spinnaker

Hi there,

I have a question here:

"An asteroid is measured to have a speed relative to the Earth of 15 km/s while it is still far away. What is the speed of impact of the asteroid with the Earth?"

I'm stumped. I don't even know where to start with this one -- it doesn't talk about how far "far" means, and you have nothing on the mass so I can't even calculate the new velocity.

Does anyone have at least a hint of where to start with this one?

2. Jun 3, 2014

### Nathanael

You just have to estimate the answer. If you're far away, it may be the case that going even farther has little affect on the change in velocity.
Try and see the total change in velocity (as a function of the starting distance away) approaches a constant as you go farther and farther. You could use that to estimate the impact speed of this object that is "far".

(Hope that was clear)

3. Jun 3, 2014

### Nathanael

Also,
Does the mass matter?

4. Jun 3, 2014

### ehild

Think of the potential energy of the asteroid. Approximately, what is it very far away? What is the total energy of the asteroid?
The mass of the asteroid is m. Write the equations in terms of m.

ehild

5. Jun 8, 2014

### spinnaker

So, what I've done is:

(K+U)i = (K+U)f

I assumed "far" to mean, for example, 1 AU away.

.5mv_i^2 + G M_earth m/r1^2 = .5mv_f^2 + 0 (since there is no potential energy at impact)

Divided through by mass of asteroid to eliminate it as it's not relevant.
But that gives me an impact speed that's actually faster if U is positive, and barely decreases the speed if U is negative (Ui = 1.79e-8)

That would give me an impact velocity of 15km/s which still seems fast.
Am I missing something?

6. Jun 8, 2014

### D H

Staff Emeritus
Yes, you are missing a number of things.

One is units. $\frac{GMm}{r^2}$ has units of force, not energy.

Another is "since there is no potential energy at impact". A better way to look at it is that it has zero potential energy when it is "far, far away". Potential energy involves an arbitrary constant, but typically the arbitrary constant is taken such that potential is zero at infinity. This means gravitational potential at a finite distance is negative.

Do this right and you will get a velocity larger than the initial velocity of 15 km/s. Why?

7. Jun 8, 2014

### spinnaker

Or am I approaching this all wrong? If I take the initial speed and add the escape velocity (basically the effect of gravity), I get

v_f = v_i + sqrt(2GM/r)
= 15km/s + 11.18km/s
= 26.183 km/s

Does that make more sense?

8. Jun 8, 2014

### D H

Staff Emeritus
You can't do that.

What you can do is to use $E = \frac 1 2 m v_i^2 + \frac{GMm}{R}$, where $E$ is the energy delivered to the atmosphere and planet surface and $R$ is the radius of the Earth, or $v_i = \sqrt{2 \bigl(\frac E m - \frac {GM} r\bigr)}$. But with E=18 MJ and m=1 kg, this leads to an imaginary initial velocity.

9. Jun 8, 2014

### ehild

Why 1 AU? very far is infinity....

The potential energy of an asteroid of mass m in the gravitational field of the Earth is : -GMearthm/r. It is negative as the gravitational potential is attractive: The work of Earth is negative when the asteroid is moved from a certain distance r to infinity. The potential at infinity is considered zero.

The potential energy at impact is -GmM/R where R is the radius of the Earth.

Conservation of energy means that the energy of the asteroid at very far away is the same as that at the impact. (assuming the asteroid moves towards the Earth)

1/2 mvi2=1/2 mvf2-GmM/R

ehild