If gravity of the Earth is so strong and pulling the moon

Question

If the gravity of the earth is so great that it is pulling the moon, then why aren't we - humans - so strongly attracted to earth that we can't even lift ourselves up?

Answer 1

Gravity is proportional to the mass of both objects: in this case, a human is way less massive than the moon, so the Earth attracts much less a human than the moon.

Answer 2

The gravitational acceleration due to the Earth's gravitational field is given by:

$$ a = -\frac{GM}{r^2} \tag{1} $$

where $r$ is the distance to the centre of the Earth. Let's see what the gravitational acceleration of the Moon is. The Earth-Moon distance is $r = 384,400$ km, so the acceleration is:

$$ a \approx -0.0027\,\text{ms}^{-2} $$

That's about $0.00027g$, while the acceleration at the Earth's surface is $1g$. So the Earth does accelerate us humans on its surface far more strongly than it accelerates the Moon.

The force on the Moon is much greater because, as Newton's second law tells us, $F=ma$, i.e. you have to multiply the acceleration by the mass to get the force. But it's the acceleration that determines how the object is going to move.

The point is that the acceleration is independent of the mass. That was the point made by Galileo all those years ago. Objects in a gravitational field fall at the same rate regardless of their mass. This means we'd expect light to accelerate at the same rate as well, because it doesn't matter what the mass of the light is. Though in fact light is bent about twice as much as we'd expect due to relativistic effects. This is discussed in How can gravity affect light?

You might think it's odd that the Moon is accelerating so slowly, but because the Moon moves so slowly around the Earth (i.e. the angular velocity is slow) it doesn't need to be accelerated very much to keep it in orbit. The centripetal acceleration in a circular orbit is given by:

$$ a = r\omega^2 \tag{2} $$

The Moon completes an orbit, i.e. a rotation of $2\pi$ in $27.322$ days, so its angular velocity is:

$$ \omega \approx 2.66 \times 10^{-6}\,\text{radians per second} $$

If we feed this angular velocity into equation (2) we get:

$$ a \approx -0.0027\,\text{ms}^{-2} $$

which of course is the same as the gravitational acceleration.