Re: Calculate how far the earth moves when a pair of keys are dropped?

Hi Jane,

It's no bother; and I'm glad you asked.

The formula you offered is fine as long as you know the two speeds and the
time. Those can be determined, but they are all variables : that is to say
that while the distance is determined by the speed and the time of fall,
the speed and time are also determined by the distance. Just as the keys
will fall faster and take longer from a high point, the earth will "rise
higher" and take longer to get there.

We can use your values for the masses of the earth and your keys; but you
need to change your acceleration to read 9.8 m/s/s. (Remember first
that an acceleration must express a speed change divided by a time change.
So m/s per s = m/s/s -or- m/s^2.) And since the precision of the keys and
acceleration are only to 2 digits, I'll use 6.0 * 10^24 kg for the earth.

While there are lots of variables in a free fall situation, we can count on
two equalities: The force between the earth & keys and the time of fall.
Newton's Third Law states that every force comes as a pair. The two forces
are equal in magnitude and are the same type (gravity). But they are on two
different objects (earth & keys) and point in opposite directions. F(k-e) +
F(e-k) = 0 where F(k-e) is the force of gravity the keys exerts on the earth
and F(k-e) is the force of gravity the earth exert on the keys (weight of
keys). 

Lets do some substitutions:
	   F(k-e) = -F(e-k)           ; Newton's 3rd Law
	  M(e)*a(e) = -M(k)*a(k)      ; Newton's 2nd Law
	M(e)*∆V(e)/∆t = -M(k)*∆V(k)/∆t; definition of acceleration
	 M(e)*∆V(e) = -M(k)*∆V(k)     ; time intervals are the same for each

Since we should assume that both bodies were initially at rest (v1=0), we
can consider that your (v1 + v2) is the same as (v2-v1); so we can use ∆v/2
for both objects.

So:	M(e)*2d(e)/∆t = -M(k)*2d(k)/∆t         ; substitute for distance
	    M(e)*d(e) = -M(k)*d(k)             ; cancel like terms (∆t same)
	         d(e) = -M(k)/M(e) * d(k)      ; rearrange terms

Now it looks like the distance the earth moves is (6.0*10^24/1.5*10^-1) times
the distance the keys fall. Or 4*10^-25 times the distance of the keys.
The minus sign in the equation says that they move in opposite directions.
That ratio is so small that even if the keys fell in from outer space, the
earth would move less than an atomic diameter.

If everyone in the world (except you) were to meet in a desert and jump off
a chair, do you think you would feel the earth move?

There are other ways to reach this same expression, but this is one that
uses your equation. See if you can find a shorter way to get the same answer.

Good luck,
Gene