Mathemidiotics

You droppa my fountain pen, I breaka your face

I hope you all learned something about Newton and Kepler in school. In that case you won't be surprised if I tell you that if you drop something, not only it falls to the Earth, but the Earth also falls a bit to that object as well. Those two masses attract each other.
Will you notice it? No, you won't. As the mass of the Earth is close to $6 \cdot 10^{24}$ kg, anything you drop is vastly outnumbered by that 6,000,000,000,000,000,000,000,000 kg.
But will it actually happen? Yes, it will actually happen, because of, well, because of laws of physics. They work always, not only if they please you.

Let's however assume that you come to my home town, and drop my precious Mont Blanc Meisterstück fountain pen from our beautiful Dom tower, standing 112 meter tall. Let's make that 100 meters. You drop the pen, the pen will fall down, the Earth will fall towards the fountain pen. How much?

Well, we know some high school formulas, Newton and things.
How long will it take for the pen to cover that 100 meter distance, minus the Earth moving up?
The formula $S_t=V_0t+\frac{1}{2}at^2$ tells you the answer.
This tells you that the distance at a certain point is equal to the sum of the star velocity (which is zero) times the time it falls and a half times the gravitational acceleration times the square of the time it falls. The Earth's gravitational acceleration is $9.8\frac{m}{s^2}$, so we find out that it takes about 4.5 seconds to fall down. At that time it has a speed defined by $V_t=V_0+at$ which tells us without friction it would accelerate up to 160 kph. Fine.
But what is the gravitational acceleration of my 30 grams fountain pen? We use this formula: $a=\frac{G\cdot m_1}{r^2}$ where G is the gravitational constant (very small) and $m_1$ is the mass of the pen (in kilograms!). The number r is the distance of the falling object to the center of my pen. That's 100 meters in my case, as I'm 100 meters up.

Some calculus might lead to the conclusion that the Earth moves 20 picometer towards the pen. That is, the whole Earth, not the ground. The impact of the pen might well make a larger dent than 20 picometer.

If not, it would lead to some weird results. A thought experiment: stand in front of your car. You want to open the hood. Now let your body fall towards it. The bonnet will not - I repeat: not - open due you your mutual attraction. The dent in the bonnet will tell you that. Yet the car falls towards you, guaranteed.
If you want, try it out, I won't repair the damage. But you'll see there's something wrong with your assumptions. Anyhow.

There's even another consequence of this mutual attraction. The Earth attracts me a lot. I attract the Earth a little. But the Sun also attracts the Earth and me. It's at 150 million kilometers away and has a mass more than 330,000 times that of the Earth. You bet it attracts us, otherwise we would be in the cold dark interstellar space by now. But what is that attraction? The Earth pulls at me with such a force that I fall towards it with $9.8 \frac{m}{s^2}$. How ' bout the Sun? It's mass is about $2\cdot 10^{30} kg$, ($m_s$) and we are at a distance of $1.5\cdot 10^{11} m$ away. There's that formula again: $a=\frac{G\cdot m_s}{r^2}$.

Input of all those goodies gives you the answer that the Sun achieves a gravitational acceleration of $0.006 \frac{m}{s^2}$ with respect to the Earth. That is $\frac{1}{1633}$th part of 1G, the gravitational acceleration we experience here, on the surface of the earth.

But with my 72 kg (I know, weight is not mass, etcetera etcetera, 700 or so Newton it should be, but for heaven' s sake) that means when I'm on the daylight side the Sun pulls me away from the Earth with about 40 grams and at night they join in pulling me down with an extra of about 40 grams!
So at night I'm 40 grams heavier, at noon I'm 40 grams lighter. That's why you have a shootout at High Noon.

Am I making sense? No? Good.