If I'm standing at the equator, jump, and land 1 second later, the Earth does NOT move 1000mph (or .28 miles per second) relative to me, since my velocity while jumping is also 1000mph.

However, the Earth is moving in a circle (albeit a very large one), while I, while jumping, am moving in a straight line.

How much do I move relative to my starting point because of this? I realize it will be a miniscule amount, and not noticeable in practise, but I'd be interested in the theoretical answer.


5 Answers 5


$\newcommand{\xhat}[0]{\hat{x}}$ $\newcommand{\yhat}[0]{\hat{y}}$ $\newcommand{\zhat}[0]{\hat{z}}$ $\renewcommand{\c}{\cos \theta}$ $\newcommand{\s}{\sin \theta}$ $\newcommand{\fce}{F_\textrm{cent}}$ $\newcommand{\fco}{F_\textrm{cor}}$ $\newcommand{\vfce}{\vec{F}_\textrm{cent}}$ $\newcommand{\vfco}{\vec{F}_\textrm{cor}}$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.

  • $\begingroup$ People usually jump perpendicular to the surface, not directly in the direction from the Earth's center. This means you jump to the North and land to the South, thus without any change in the North-South position. $\endgroup$
    – Anixx
    Oct 12, 2013 at 16:41
  • $\begingroup$ I was using a spherical earth approximation. In that approximation the two directions are the same. $\endgroup$ Oct 13, 2013 at 20:37

While you jump, just like earth, you continue to move in a circle around sun. This is simply because you and earth are both continuing to undergo a gravitational acceleration towards sun.

However, while you jump, due to your and earth's difference in positions, earth and you will experience a miniscule difference in gravitational acceleration towards sun, resulting in so-called tidal forces.

The solar tidal acceleration at the Earth's surface along the Sun-Earth axis is about $0.52$ $10^{−6} m/s^2$. You would have to specify where on the equator you are jumping relative to the sun-earth axis, but as a rough estimate, the effect during a 1 second jump will not be larger than 0.5 micrometer. In reality, it will be much less, as a significant part of the effect would be in changing the height of your jump, not in changing your landing position. Also, you have to include tidal forces due to the moon (which are of the same order of magnitude), and effects due to the rotation of the earth around it's own axis. The tidal effects due to the Milky Way can safely be ignored.

Edit: The effect due to earth's rotation around it's own axis can be estimated as follows. At the equator, the circumference of the Earth is 40,000 kilometers, and the day is 86,400 seconds long, so the speed of earth's surface at the equator is roughly 460 m/s. When during the jump you spend your time on average roughly one meter above earth's surface, your velocity lags 460 m/s times 1/6.4x10^6 (the denominator corresponding to earth's radius in meters) which equates to about 70 micrometer per second. So, when jumping exactly vertically, after a second you land roughly 70 micrometer west of where you started.

  • $\begingroup$ I was referring to the Earth's rotation about its own axis when I said "moving in a circle", not the Earth's revolution around the Sun. $\endgroup$
    – user854
    Jan 4, 2013 at 4:40
  • $\begingroup$ Included the rotation effect in an edit. $\endgroup$
    – Johannes
    Jan 4, 2013 at 13:20
  • $\begingroup$ Why wouldn't my velocity remain constant at 460m/s, albeit in a different direction (straight line vs circular around the Earth's center)? $\endgroup$
    – user854
    Jan 4, 2013 at 15:18
  • $\begingroup$ Your velocity lags in the sense that at radius 6.4x10^6 + 1 m away from earth's center, the rotational speed corresponding to a radius of 6.4x10^6 m is not enough to keep up with earth's rotation. In other words, you keep your speed, and therefore you lack angular speed. $\endgroup$
    – Johannes
    Jan 4, 2013 at 15:28

EDIT: As suggested in the comments, I solved this problem after writing it symbolically. See https://physics.stackexchange.com/questions/227391 for details.


For small jumps, the answer is 122micrometers*s^3, where s is the time of the jump in seconds. I used numerical methods in Mathematica. Can someone improve on and/or verify this solution?

Consider this diagram:

enter image description here

I then ran the following:

(* Earth's radius in meters *) 
r = 40000000/2/Pi 

(* seconds in day *) 
d = 86400 

(* acceleration at Earth surface *) 
a = -10 

(* initial velocity, using 50m/s as an example *) 
vinit = 50 


The initial position is (0,r), as per the diagram above. 

The initial x velocity is 2*Pi/r/d, the speed at which the equator is rotating. 

The initial y velocity is vinit, by definition. 

The total gravitational force is inversely proportional to the square 
of the distance from the Earth's center, and equals "a" on the Earth's 
surface. Thus, in terms of x and y, the total gravitation force is: 


We further break this into x and y components. 

All this yields the below 


diffsolve[v0_] := NDSolve[{ 
 x[0] == 0, 
 y[0] == r, 
 x'[0] == 2*Pi*r/d, 
 y'[0] == v0, 
 x''[t] == a*(r^2/(x[t]^2+y[t]^2))*(x[t]/Sqrt[x[t]^2+y[t]^2]), 
 y''[t] == a*(r^2/(x[t]^2+y[t]^2))*(y[t]/Sqrt[x[t]^2+y[t]^2]) 
}, {x[t],y[t]}, {t,0,vinit*3}][[1]] 

(* we then solve for x and y based on the initial velocity *) 
x1[t_] = x[t] /. diffsolve[vinit] 
y1[t_] = y[t] /. diffsolve[vinit] 

(* the angle theta at time t *) 
ang[t_] = Pi/2-ArcTan[x1[t],y1[t]] 
(* the angle my original starting point makes at time t *) 
eangel[t_] = 2*Pi*t/d 

(* my distance from the surface of the Earth at time t *) 
rad[t_] = Sqrt[x1[t]^2+y1[t]^2]-r 


Time it take me to land. If the earth were flat, and using the numbers 
above, I would land in 10s. However, the actual landing time is 
10.034s. The extra 34ms are important 


landingtime = t /. FindRoot[rad[t]==0,{t,vinit/20,vinit/5}] 

(* The angular distance between me and my starting point by the time I 
land, multipled by the earth's radius to give me total distance *) 


(* Values for various vinit: 

vinit=5    122 micrometers (1 second jump) 
vinit=50   12 cm (~10 second jump) 
vinit=500  122m (~100 second jump) 
vinit=1000 985m (~200 second jump) 
vinit=3000 29km (~10m jump) 
vinit=4000 73km (~13m20s jump) 

For values over 4000, the cubic function breaks down rapidly; once 
orbital velocity is obtained, I would never land at all 


EDIT: ignoring the x component of gravity (treating x''[t] as 0) appears to introduce a significant amount of inaccuracy.

  • 1
    $\begingroup$ I would highly recommend against using numbers in formulas like these, just you $\omega$ or something for angular frequence, etc. The way you wrote in now, makes is pretty unreadible. $\endgroup$
    – Bernhard
    Jan 6, 2013 at 18:28
  • $\begingroup$ No idea if this is right, as Bernhard says the maths is pretty unreadable, but +1 for the banter $\endgroup$ Oct 11, 2013 at 8:11
  • $\begingroup$ "Page not found" when I follow the link in the EDIT at the top of this answer. $\endgroup$
    – Mörre
    Jun 2, 2021 at 16:44

Ugh, assuming constant radial gravity $g$ I need to solve the equations of motion in polar coordinates $r$, $\theta$ as

$$ \ddot{r} = r \dot{\theta}^2 - g \\ \ddot{\theta} =- \frac{2 \dot{r} \dot{\theta}}{r} $$

which I do not know how to do. When I find out I will add to this answer. This system varies the direction of gravity and not it's magnitude for an approximate solution that should be fairly accurate.

related notes

There is trivial solution with $\theta=\dot\theta=\ddot\theta=0$ and $\ddot{r}=-g$, but this does not match the initial conditions of $$r(0)=R \\ \dot{r}(0)=v_{jump} \\ \theta(0)=0 \\ \dot\theta(0) = \Omega$$ where $R$ is the radius of the Earth and $\Omega$ it's rate of rotation and $v_{jump}$ is the take off speed.

The ODE system is $g-r^2 \omega^2 + \ddot{r}=0$ and $r \dot{\omega} + 2 \dot{r} \omega = 0$ with $\omega = \dot{\theta}$.

The solution to the second equation is $$ \omega = \frac{\Omega R^2}{r^2} \\ \dot{\omega} = -\left( \frac{2 \Omega R^2}{r^3}\right) \dot{r} $$

and so the first equation becomes $$ \frac{{\rm d} \dot{r}}{{\rm d} t} = \frac{\Omega^2 R^4}{r^3} - g $$

which is solved by direct integration $\int \dot{r}\,{\rm d}\dot{r} = \int \left( \frac{\Omega^2 R^4}{r^3} - g \right)\,{\rm d} r $ as

$$ \frac{\dot{r}^2}{2} = - \left( \frac{\Omega^2 R^4}{2 r^2} g r\right) + \left( \frac{\Omega^2 R^2}{2} + R g + \frac{v_{jump}^2}{2} \right) $$

Now for an approximation. Change variables to $y = r - R$ and $\dot{y}=\dot{r}$ to get

$$\boxed{ \dot{y}^2 = v_{jump}^2 + \Omega^2 R^2 - 2 g y - \frac{\Omega^2 R^4}{(y+R)^2} }$$ $$ \dot{y}^2 \approx v_{jump}^2 + y \left( 2 \Omega^2 R - 2 g \right) $$

$$ t = \int_0^y \frac{1}{\sqrt{v_{jump}^2 + y \left( 2 \Omega^2 R - 2 g \right)}}\,{\rm d} y $$

$$ y = v_{jump} t + \frac{1}{2} \left(\Omega^2 R-g \right) t^2 $$

which is Doh! nothing more than a projectile under constant gravity.

Let's do a 2nd order approximation of $\dot{y}^2$ above

$$ \dot{y}^2 \approx v_{jump}^2 + y \left( 2 \Omega^2 R - 2 g \right) - 3 \Omega^2 y^2 $$

with solution

$$ \boxed{ y(t) = \left( \frac{g}{3 \Omega^2} - \frac{R}{3} \right) \left( \cos(\sqrt{3} \Omega t)-1 \right) + \frac{v_{jump}}{\sqrt{3} \Omega} \sin(\sqrt{3} \Omega t) }$$

with time in the air:

$$ t = \frac{\pi}{\sqrt{3}\Omega} + \frac{ 2 \arctan\left( \frac{\Omega^2 R -g}{\sqrt{3} \Omega v_{jump}}\right) }{\sqrt{3}\Omega} $$

These equations match the numerical solution


My excel sheet with both the numeric and the above solution is at Public Dropbox. Caution you need to have macros enabled as they are used for the numeric results.

  • $\begingroup$ OK, so you're taking into account that the direction of gravity depends on position, but assuming that the magnitude of gravity remains constant? I believe this will give you a fairly accurate approximation. $\endgroup$
    – user854
    Oct 12, 2013 at 3:50
  • $\begingroup$ Yes, that is my reasoning. I am hoping we can collaboratively come up with a solution. Any ideas on how to proceed? $\endgroup$ Oct 12, 2013 at 13:35
  • $\begingroup$ I'm going to plug this into Mathematica, but I think radial gravity means the result will be an ellipse, which has no clean polar formula. $\endgroup$
    – user854
    Oct 12, 2013 at 14:27
  • $\begingroup$ I have solve the angular equation (it makes sense from a angular momentum point of view) and used to to reduce the radial equation. $\endgroup$ Oct 12, 2013 at 14:47
  • $\begingroup$ Ok, look at my solution now. $\endgroup$ Oct 12, 2013 at 16:15

In the rotating frame of the earth's surface, there are two fictitious forces acting on you, a centrifugal force and a Coriolis force. These are both quite small in absolute terms, and the centrifugal force is also not noticeable because it simply feels equivalent to a slight change in the over-all direction and magnitude of the gravitational field. The Coriolis force has a horizontal component and will therefore in theory cause you to come down in a slightly different place then the one from which you jumped up. In practice, this effect is much too small to observe in this situation due to all the other systematic and random errors.

There is identically zero effect due to the earth's motion around the sun, since the earth is free-falling and therefore constitutes an inertial frame in the general-relativistic sense.