Earth's orbit around the Sun is nearly circular. The period is 1 yr \(=365.25\) d. (a) In an elapsed time of 1 d what is Earth's angular displacement? (b) What is the change in Earth's velocity, \(\Delta \overrightarrow{\mathbf{v}} ?\) (c) What is Earth's average acceleration during 1 d? (d) Compare your answer for (c) to the magnitude of Earth's instantaneous radial acceleration. Explain.

To find the angular displacement (\(\Delta \theta\)) in one day, we'll first find the angular speed (\(\omega\)) of Earth around the Sun. Since Earth completes one full revolution (360 degrees or \(2\pi\) radians) in one year, we can compute the angular speed as: \(\omega = \frac{2\pi \text{ radians}}{365.25 \text{ days}}.\) Now, we can calculate the angular displacement in one day by multiplying the angular speed by the elapsed time: \(\Delta \theta = \omega \times 1 \text{ day} = \frac{2\pi \text{ radians}}{365.25 \text{ days}} \times 1 \text{ day}.\)

Earth's orbit is nearly circular, so we can use the formula of the radial acceleration to calculate the change in Earth's velocity (\(\Delta \overrightarrow{\mathbf{v}}\)): \(\Delta \overrightarrow{\mathbf{v}} = a_r \times t.\) Here, \(a_r\) is the radial acceleration, and \(t\) is the time elapsed (1 day). To find \(a_r\), recall that for a circular motion, \(a_r = \frac{v^2}{r}\). In our case, \(v\) is Earth's orbital speed, and \(r\) is the distance from the Sun to the Earth. The orbital speed can be calculated by dividing the total path length (the circumference of the Earth's orbit) by the time it takes to complete one revolution (the period): \(v = \frac{2\pi r}{\text{period}}.\) Now, we can plug this expression into the \(a_r\) formula and find the change in Earth's velocity: \(\Delta \overrightarrow{\mathbf{v}} = a_r \times t = \frac{v^2}{r} \times t = \frac{(2\pi r / \text{period})^2}{r} \times t.\)

The average acceleration (\(\overline{a}\)) during 1 day can be calculated as the change in velocity divided by the time elapsed: \(\overline{a} = \frac{\Delta \overrightarrow{\mathbf{v}}}{t}.\) Using the expression we found for the change in Earth's velocity in Step 2, we have: \(\overline{a} = \frac{\frac{(2\pi r / \text{period})^2}{r} \times t}{t}.\)

We will now compare the previously calculated average acceleration to the magnitude of Earth's instantaneous radial acceleration (\(a_r\)). Recall that: \(a_r = \frac{v^2}{r}.\) By looking at this expression and the expression for the average acceleration, we can see that they are equal: \(\overline{a} = \frac{(2\pi r / \text{period})^2}{r} = a_r = \frac{v^2}{r}.\) This means that the average acceleration during 1 day is equal to the magnitude of Earth's instantaneous radial acceleration. This result is expected since Earth's orbit is nearly circular, which means that its velocity is almost constant, resulting in the average acceleration being the same as the radial acceleration at any given instant.