Question

If two systems of coordinates with a common origin \(O\) are rotating with respect to each other, the measured accelerations differ in the two systems. Denoting by \(\mathbf{r}\) and \(\mathbf{r}^{\prime}\) position vectors in frames \(O X Y Z\) and \(O X^{\prime} Y^{\prime} Z^{\prime}\) respectively, the connection between the two is $$ \ddot{\mathbf{r}}^{\prime}=\ddot{\mathbf{r}}+\dot{\omega} \times \mathbf{r}+2 \omega \times \dot{\mathbf{r}}+\omega \times(\omega \times \mathbf{r}) $$ where \(\omega\) is the angular velocity vector of the rotation of \(O X Y Z\) with respect to \(O X^{\prime} Y^{\prime} Z^{\prime}\) (taken as fixed). The third term on the RHS is known as the Coriolis acceleration, whilst the final term gives rise to a centrifugal force. Consider the application of this result to the firing of a shell of mass \(m\) from a stationary ship on the steadily rotating earth, working to the first order in \(\omega\left(=7.3 \times 10^{-5} \mathrm{rad} \mathrm{s}^{-1}\right)\). If the shell is fired with velocity \(\mathbf{v}\) at time \(t=0\) and only reaches a height that is small compared to the radius of the earth, show that its acceleration, as recorded on the ship, is given approximately by $$ \ddot{\mathbf{r}}=\mathbf{g}-2 \omega \times(\mathbf{v}+\mathbf{g} t) $$ where \(m \mathbf{g}\) is the weight of the shell measured on the ship's deck. The shell is fired at another stationary ship (a distance \(\mathbf{s}\) away) and \(\mathbf{v}\) is such that the shell would have hit its target had there been no Coriolis effect. (a) Show that without the Coriolis effect the time of flight of the shell would have been \(\tau=-2 \mathbf{g} \cdot \mathbf{v} / g^{2}\) (b) Show further that when the shell actually hits the sea it is off target by approximately $$ \frac{2 \tau}{g^{2}}[(\mathbf{g} \times \omega) \cdot \mathbf{v}](\mathbf{g} \tau+\mathbf{v})-(\omega \times \mathbf{v}) \tau^{2}-\frac{1}{3}(\omega \times \mathbf{g}) \tau^{3} $$ (c) Estimate the order of magnitude \(\Delta\) of this miss for a shell for which \(v=300\) \(\mathrm{m} \mathrm{s}^{-1}\), firing close to its maximum range ( \(\mathbf{v}\) makes an angle of \(\pi / 4\) with the vertical) in a northerly direction, whilst the ship is stationed at latitude \(45^{\circ}\) North.

Short answer

Without Coriolis effect, time of flight is \(\tau = - \frac{2 \mathbf{g} \cdot \mathbf{v}}{g^2} \). Including Coriolis, the miss distance \( \Delta \) is approx. -6 m.

Step-by-step solution

- Analyzing the system of coordinates

The expression given is \[ \ddot{\mathbf{r}}^{\prime}=\ddot{\mathbf{r}}+\dot{\omega} \times \mathbf{r}+2 \omega \times \dot{\mathbf{r}}+\omega \times(\omega \times \mathbf{r}) \], where \( \ddot{\mathbf{r}}^{\prime} \) is the observed acceleration in the rotating frame (the ship), which differs from the non-rotating frame acceleration \( \ddot{\mathbf{r}} \). The terms \( 2 \omega \times \dot{\mathbf{r}} \) and \( \omega \times(\omega \times \mathbf{r}) \) represent the Coriolis and centrifugal accelerations respectively.

- Apply the acceleration formula

From the problem, the shell is fired from the ship, implying the frame \( OX'Y'Z' \) is the rotating Earth and \( \mathbf{r} \) is on the Earth’s surface. Since \( \mathbf{g} \) is the gravitational acceleration, substituting we get \( \ddot{\mathbf{r}} = \mathbf{g} \).

- Considering rotational effects

To the first order in \( \omega \): \( \dot{\omega} = 0 \), thus neglecting \( \dot{\omega} \). Hence:\[ \ddot{\mathbf{r}}^{\prime} = \mathbf{g} + 2 \omega \times \dot{\mathbf{r}} + \omega \times (\omega \times \mathbf{r}) \]Since, \( \ddot{\mathbf{r}}^{\prime} = \ddot{\mathbf{r}} = \mathbf{g} - 2 \omega \times \mathbf{v} \) for small heights \( \mathbf{v}' = \mathbf{v} + \mathbf{g}t\)

- Derive the formula for time of flight

Ignoring the Coriolis effect: \( \mathbf{r}(\tau) = \mathbf{v}\tau +\frac{1}{2}\mathbf{g}\tau^2 = \mathbf{s} \). Solving gives:\[ \tau = - \frac{2 \mathbf{g} \cdot \mathbf{v}}{g^2} \]

- Incorporating the Coriolis effect

Position: \( \mathbf{r}(t) = \mathbf{v}t + \frac{1}{2} \mathbf{g}t^2 + \omega \times (\mathbf{v}t^2 + \frac{1}{3} \mathbf{g}t^3) \). When the shell hits the sea at \( t = \tau \):The deviation:\[ \Delta = \frac{2 \tau}{g^2}[(\mathbf{g} \times \omega) \cdot \mathbf{v}](\mathbf{g}\tau + \mathbf{v}) - (\omega \times \mathbf{v})\tau^2 - \frac{1}{3}(\omega \times \mathbf{g})\tau^3 \]

- Calculating order of magnitude

Given \( v = 300 \text{ m/s}, \omega = 7.3 \times 10^{-5} \text{ rad/s}, \tau \approx 60 \text{ s}, g=9.8 \text{m}/\text{s}^2 \). Calculate each term:1. \( \frac{2 \tau}{g^2} [(\mathbf{g}\times \omega)\cdot\mathbf{v}](\mathbf{g}\tau + \mathbf{v}) \approx 0 \)2. \( - (\omega \times \mathbf{v}) \tau^2 \approx - (7.3 \times 10^{-5} \times 300) 60^2 \approx - 8 m \)3. \( \frac{1}{3} (\omega \times \mathbf{g}) \tau^3 \approx 2m \).Thus, \( \Delta \approx - 8 + 2 = -6 \text{ m} \).

Key concepts

rotating reference frames

A rotating reference frame is a viewpoint that rotates relative to an inertial frame. In other words, it's a non-static frame of reference that moves in a circular fashion.
When solving physics problems, using rotating frames can simplify the understanding of systems involving rotation, like the Earth's rotation affecting projectile motion. Because acceleration measured in a rotating frame differs from the non-rotating frame, we use additional terms in our equations. These terms accommodate rotational effects.
In the provided exercise, we shift from the inertial frame (non-rotating) to the rotating frame (the Earth). This switch introduces factors like the Coriolis and centrifugal forces, essential to accurately track the motion of the shell being fired.

Centrifugal force

Centrifugal force appears when we analyze motion in a rotating reference frame. It's a pseudo-force that seems to push objects away from the center of rotation.
Imagine swinging a ball tied to a string around you. The ball seems pulled away from your hand; this is similar to centrifugal force. In mathematical terms, it's expressed as \(\omega \times (\omega \times \mathbf{r})\), where \(\omega \) is the angular velocity vector and \(\mathbf{r} \) is the position vector.
For the exercise, this term explains part of the discrepancy between the observed accelerations in rotating and non-rotating frames. It helps to adjust our calculations so that we can see how the shell's motion would look from the rotating Earth.

projectile motion

Projectile motion is the motion of an object thrown into the air, subject to gravity. Typically, it's analyzed in an inertial frame, considering horizontal and vertical motions separately. However, on a rotating Earth, we must also account for rotational forces.
In the exercise, without rotational effects, the shell follows a path determined by initial velocity \(\mathbf{v} \) and gravitational acceleration \(\mathbf{g} \). Adding effects like the Coriolis force, \(2 \omega \times (\mathbf{v} + \mathbf{g}t)\), modifies the shell's path. This force deflects the motion due to Earth's rotation.
Understanding the exact time of flight and deviation from the target involves combining these vertical, horizontal, and rotational effects properly. The exercise shows that these factors, while small, can significantly impact projectile motion, especially over long distances or durations.