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ewcommand{\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.