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