# Time to impact in freefall: mass dependant?

1. Oct 18, 2005

### Staff: Mentor

My point is this: though the earth accelerates the feather and the ball bearing equally, the feather and the ball bearing do not accelerate the earth equally. As a result, the time to impact for the ball bearing will be less than for the feather. For objects that small, it may not be noticeable, but it is still there.

To notice it, you really need bigger masses. Any object arbitrarily low in mass will hit the earth in about two seconds when dropped from 9.8m. But lets say you could drop another earth from 9.8m above this one. Since each is being accelerated toward the other at 9.8m/s^2, impact would occur in 1s.

2. Oct 18, 2005

### ZapperZ

Staff Emeritus
We have seen something similar like this a few times on here. Perhaps some day we'll have a definitive explanation and put it up as an FAQ or something.

There is an implicit assumption of a "fixed" earth when we try to explain the drop of an object. What this means is that the center of mass of the whole system is fixed, and fixed right at the center of the earth. The object then is moving, and falls to the earth (remember, we're dealing with classical mechanics here). Now, in most cases, this is perfectly valid since most objects that we deal with terrestrially is waaaaaaay smaller and lighter than the earth itself.

Now, as you have alluded to, what happen if the object is larger and of significant fraction of the earth? Then the center of mass of the system would change, and may in fact no longer be right at the center of the earth. When you have two objects such as these, and you let go, the earth will start moving towards the other object, while the other object will start moving towards the earth. How fast this happens certainly depends on the force of gravity between the two AND where they meet! The latter also explicitly depends on the masses of both of them since this depends on the center of mass.

So two very light objects will have the same acceleration, and travels the same distance to hit the earth surface, But a VERY heavy object, comparable to the earth's mass, will have a different distance to travel before it hits the earth, since the earth is also moving siginificantly towards it.

Zz.

3. Oct 18, 2005

### pervect

Staff Emeritus
Basically, if you are dropping a heavy enough mass that the Earth moves, I have to also question the auxillary assumption that the Earth is acting as a rigid body.

This assumption is being made when one says that the Earth "moves" a certain distance. It's quite likely that different parts of the Earth move a different distance.

There will be significant and time-varying tidal forces on the Earth due to the falling large mass. I assume that the time is being computed when the falling mass reaches the surface of the Earth that's underneath the mass. Since the Earth is probably not holding it's shape, one would have to include all the dynamics of how the Earth responds to the time-varying tidal forces to answer this question.

4. Oct 18, 2005

### Danger

Okay, I see what you were on about. I was indeed referencing the Earth as a fixed gravity source. When given the reality that it can also be accelerated by the gravity of another body, your argument is correct.
There's another factor which would totally screw up the experiment if one of the bodies is extremely massive. The comparison object would be attracted to both it and Earth, and thus would no longer have a normal 1g acceleration.

5. Oct 18, 2005

### Staff: Mentor

Yeah, we were just using different simplifying assumptions. We're cool - well, I am, anyway...

6. Oct 18, 2005

### Danger

Roger that.

7. Oct 20, 2005

### pallidin

My, what an interesting look at the standard problem! Great job!

8. Oct 20, 2005

### Danger

We've never embraced the KISS principle.