# Does the Earth's rotation affect an object in the stratosphere?

Does the Earth's rotation affect an object in the stratosphere?

Example: I attach an object to a weather balloon and let it go at latitude +40.7, longitude -73.9 (New York). It goes up until it hits the stratosphere (about 40 km up). Then the air is not dense enough for the balloon to go any higher so it sits there. Taking wind out of the equation, will the object change latitude and/or longitude due to the Earth's rotation?

• "Taking wind out of the equation..." I do not think this is realistic assumption. The motion of the balloon will certainly be strongly affected by the motion of the air during its ascent and thereafter. Jan 15, 2014 at 15:39
• A better model (if you want zero wind) is to assume the atmosphere rotates with the earth. Then the balloon will stay above one point of the surface, not changing latitude or longitude at all. We measure wind speed relative to the ground for this reason. If there is wind, assume the balloon moves relative to the ground with exactly the wind speed. Jan 15, 2014 at 22:52

Since we can remove the wind from equation, we conclude that the balloon's angular momentum will remain unchanged (the only forces acting on the balloon will be parallel to the line connecting the balloon to the center of Earth, thus exerting no torque).

We denote $m$ as the mass of the balloon, $r_1=6371\,km$ for the initial distance from the centre of Earth (the average radius of Earth), $r_2=6411\,km$ for the final distance from the centre of Earth and respectively $v_1$ and $v_2$ for the initial and final linear velocity.

From the conservation of angular momentum we have: $$L_1 = L_2$$ $$m(r_1\cdot v_1\cdot sin\alpha) = m(r_2\cdot v_2\cdot sin\alpha)$$ where $\alpha$ is the angle between the radius and velocity. Since the velocity is always perpendicular to the radius, $sin\alpha$ is equal to 1.

$$v_2 = \frac{r_1}{r_2}v_1$$

So, we see that the balloon will be flying around the centre of Earth slower, than it would be rotating, if it was placed on the ground. Let's calculate the angular velocity:

$$v=\omega r \implies \omega_2 r_2 = \frac{r_1}{r_2} \omega_1 r_1$$ $$\omega_2 = \frac{r_1^2}{r_2^{2}} \omega_1 \approx \frac{355^{\circ}}{24\,h}$$

Thus, while the Earth makes a full spin, the balloon only rotates for about $355^{\circ}$.

If we want to calculate the exact velocity the balloon is flying with we come back to $v_2 = \frac{r_1}{r_2}v_1$, but for this it's necessary to calculate the speed $v_1$, which is equal to $\frac{2\pi R}{24\,h}$ where R is the distance from New York to the axis the Earth rotates around. Why do we need it? Because New York doesn't rotate around the Earth's centre (which we used in our calculations as a point from where we calculated angular momenta), but around the Earth axis. Everything rotates around this axis, but things closer to equator have a greater velocity, because the circle they're making is longer.

$$R = r_1 \cdot cos\alpha$$ where $\alpha$ is the latitude.

If $\alpha = 40.7^{\circ}$, then

$$v_2 = \frac{r_1}{r_2}\cdot \frac{2\pi R}{24\,h} = \frac{r_1}{r_2}\cdot \frac{2\pi r_1 \cdot cos\alpha}{24\,h} = 1256\,\frac{km}{h}$$

• Thanks so much for you fast response. What do your think the speed of the object would be? Jan 15, 2014 at 11:24
• I've updated the answer with a more suitable way of finding the angular momentum of the balloon (which we can treat as a particle with mass m). Jan 15, 2014 at 19:06
• I don't think removing the wind (presumably differential speed wrt. to the group below) from the equation is the same as removing the atmosphere (apparently whoosing by at 1256 km/h). Jan 15, 2014 at 21:07
• I didn't remove the effect of atmosphere, as far as I'm concerned - air resistance doesn't create a torque and the same equations apply to the air particles - meaning, that they will be moving with the same velocity as the balloon, so they won't be moving relative to the balloon, thus excerting no horizontal force on it (which would be exerted by "wind"). Jan 15, 2014 at 21:50