The atmosphere rotates along with the Earth for the same reason you do.
Force isn't needed to make something go. That's a basic law of physics - that a thing that's moving will just keep moving if there's no force on it.
Force is needed either to make something change its speed, or to make its motion point in a new direction. A force can do both or just one of these. Most forces do both, but a force that pushes in the exactly the same direction you're already going only changes your speed, and does not change your direction. A force that pushes at a right angle to the direction you're already going only changes your direction, and does not add any speed. A force at "10 o'clock", for example, will change both your speed and your direction.
As you stand still on Earth, you continue going the same speed, but your direction changes; between day and night you move opposite directions. So the forces on you must be at a right angle to your direction of motion. Indeed, they are. Your motion is from west to east along the surface of the Earth, and the force of gravity pulls you down towards the center of the Earth - the force and your motion are at right angles. Similarly for the atmosphere. It is moving along with the Earth, and moving at a constant speed. It does not need anything to push it along with the Earth. Since only its direction of motion is changing, it only needs a force at a right angle to its motion, the same as you, and the force that does the job is again gravity.
That's not the whole picture, because the amount that your direction of motion changes depends on how strong the right-angle force is. It turns out gravity is much too strong for how much our direction of motion changes as the Earth spins. There must be some other force on us and on the atmosphere canceling out most of the gravity. There is. For me it's the force of the chair on my butt. For the atmosphere, it's the air pressure.
So gravity doesn't "make the air rotate". The air is already going, and gravity simply changes its direction to pull it in a circle.
You may be wondering why the air doesn't just sit there and have the Earth spin underneath it. One answer to that is that from our point of view that would mean incredibly strong wind all the time. That wind would run into stuff and eventually get slowed down to zero (that's from our point of view - the air would "speed up" to our speed of rotation from a point of view out in space watching everything happen). Even the air high up would eventually rotate with the Earth because although it can't slam into mountains or buildings and get stopped from blowing, it can essentially "slam into" the air beneath it due to friction in the air. (This is a little redundant with dmckee's answer; I was half way done when he beat me to the punch)
$\newcommand{\xhat}[0]{\hat{x}}$ $\newcommand{\yhat}[0]{\hat{y}}$ $\newcommand{\zhat}[0]{\hat{z}}$ $\renewcommand{\c}{\cos \theta}$ $\newcommand{\s}{\sin \theta}$ $\newcommand{\fce}{F_\textrm{cent}}$ $\newcommand{\fco}{F_\textrm{cor}}$ $\newcommand{\vfce}{\vec{F}_\textrm{cent}}$ $\newcommand{\vfco}{\vec{F}_\textrm{cor}}$I saw a question I wanted to answer which was a duplicate of this one.
So I guess I will answer this question.
Instead of restricting to the special case of jumping at the equator, I will consider the general case of jumping at a latitude $\theta$.
First I will give an intuitive argument for which direction you will move.
Then I will give a leading order estimate for how far you will move.
First I will present the intuitive argument.
The first thing to remember is that the earth rotates from west to east.
Now let's say you are standing in london.
You turn on your mods and jump very high.
You will notice two things as you go very into orbit.
First, your angular velocity about the earth's axis of rotation will have decreased because of conservation of angular momentum.
Second, you will be moving south, because you are in orbit around the earth, and it would be weird if you stayed at the same latitude; you are supposed to be orbiting the center of the earth.
So as you look down, you would see the earth spinning beneath your fit, and london would head east and you would be somewhere in the americas.
Also you would be heading south, so you might land in central or south america.
From this we can see that you will land southwest of where you started.
(If you were in the southern hemisphere you would land to the northwest.)
Now let's try to estimate the magnitude of the effect.
We will consider only the earth's gravity and rotation; we will not consider effects from other planets, moons or stars, since these should be weaker effects.
We will also assume a spherical earth.
I am not sure how much this assumption affects the answer.
We will start by picking a coordinate system.
The origin of the coordinate system will be the place that you jump from.
We will choose the $\xhat$ axis to be pointing east, the $\yhat$ axis to be pointing north, and the $\zhat$ axis to be pointing up.
Some quantities which will be important to this problem are: $\theta$, the latitude you are jumping from; $R_e$, the radius of the earth; $\rho_0 = R_e \cos \theta$, the your initial distance from the axis of rotation of the earth; $\omega = 2 \pi / \textrm{day}$, the angular velocity of the earth; $H$, how high you can jump; $m$, your mass; and $g$, gravitational acceleration.
There are three force acting on you when you are in the air.
One force is gravity, and the other two are (fictitious) inertial forces.
One is centrifugal force, and the other is the coriolis force.
Let's start by considering gravity.
Gravity points down to the center of the earth, so the force of gravity is $\vec{F}_g=mg\zhat$.
Next is the centrifugal force $\fce$.
This has magnitude $m \omega^2 \rho$, where $\rho$ is the distance from the axis of rotation.
It will be sufficient to approximate this distance to be the constant $\rho_0$ even though you will be jumping.
Thus we will make the approximation that the magnitude of the centrifugal force is $\fce=m \omega^2 \rho_0$.
The direction of the centrifugal force is away from the axis of rotation.
Thus $\vfce = \fce (\zhat \c-\yhat \s).$
The third force is the coriolis force.
This force is given by $\vfco = -2m \vec{\omega} \times \vec{v}$, where $\vec{v}$ is your velocity.
The magnitude of this force is then $\fco = 2m \omega v \c$.
On your way up, the direction of the force will be west, as we expect.
Thus $\vfco = -2m \omega v_z \c \xhat$.
Having analyzed the forces, we are now ready to calculate the your motion as you jump.
We assume you are jumping straight up a height $H$.
Your initial speed must be $v_0 =\sqrt{2gH}$, and so the time you will spend in the air is $\Delta t = t_f - t_i = 2 \sqrt{\frac{2H}{g}}$.
Your distance from the center of the earth as a function of time $r(t) = R_e + v_0 t - \frac{1}{2} g t^2$ where we have taken $t_i = 0$.
Already we can calculate the displacement due to the centrifugal force.
To lowest order, we can neglect the $z$ component of the centripetal force.
Then we get a acceleration due to the centripetal force which is directed along the $y$ axis.
The $y$ component is $ \omega^2 \rho_0 \s = \omega^2 R_e \c \s = \frac{1}{2} \omega^2 R_e \sin(2 \theta)$.
The $y$ component of the displacement due to this acceleration is $\frac{1}{2} \frac{1}{2} \omega^2 R_e \sin(2 \theta) * (\Delta t) ^2 = \frac{1}{4} \omega^2 R_e \sin(2 \theta) * 8\frac{H}{g} = 2 \sin(2\theta) \frac{\omega^2 R_e}{g} H$.
Thus the fraction of $H$ that you move toward the equator is roughly the fraction of centrifugal acceleration to gravity.
Next we can calculate the displacement due to the coriolis force.
We saw the coriolis force gives an acceleration with $x$ component equal to $-2\omega v_z \c $.
Integrating once, we find $v_x = -2 \c \omega z$.
Plugging in our formula for $z$, we find $v_x = -2 \c \omega (v_0 t - \frac{1}{2} g t^2)$.
Integrating this once with respect to time, we find $$\Delta x = - \c \omega (v_0 (\Delta t)^2 - \frac{1}{3} g (\Delta t)^3) \\
= - \c \omega (\sqrt{2gH} 8 \frac{H}{g} - \frac{1}{3} g 8 \frac{H}{g} 2 \sqrt{\frac{2H}{g}} ) \\
= - \frac{8}{3} \sqrt{2} \c \sqrt{\frac{\omega^2 R_e}{g}} \sqrt{\frac{H}{R_e}} H.$$
Here we see that the fraction of the height you jump that you move west is roughly the product of the square roots (i.e.
geometric mean) of two fractions: the ratio of centrifugal force to gravity, and the ratio of the amount you are able to jump to the radius of earth.
Now let's calculate distances for the case where the height $H$ you jump is one meter, and the latitude is $45^ \circ$. In this case the centrifugal force will move you $6.91 \textrm{ mm}$ south and the coriolis force will move you $62.1 \textrm{ $\mu$m}$ west. In the spherical earth approximation, I think these values should be good to about 1%, the largest error being the uncertainty in the acceleration due to gravity.
Best Answer
Actually, you do land a little west of where you jumped. However, that distance is so miniscule for a human jumping that you don't notice it.
The reason something projected straight up doesn't fall back onto the same spot is that its radius from the center of rotation increases as it gets higher. Initially, the projectile and the surface of the earth are moving at the same speed horizontally. As the projectile goes higher, it moves further from the center of rotation and would have to move faster laterally to keep above the same spot on the ground below it. It doesn't, so appears to move west to a observer fixed to the ground. Note that this effect is proportional to sin(90° - latitude). If the projectile was launched straight up at either pole, it would fall back to the same spot with the same orientation.
For a human jumping, the radius change is so small and the flight time so small that the effect is so tiny as to be swamped by many other sources of errors.