Let's break down the problem step by step: --- ## **Step 1: Understanding the Geometry** The squared distance on the surface of a sphere (in spherical coordinates) is: \[ d\Omega^2 = d\theta^2 + \sin^2\theta\, d\phi^2 \] We're given two points on the **equatorial plane** (\(\theta = \pi/2\)): - Point A: \(\phi = \pi/2\) - Point B: \(\phi = \pi\) --- ## **Step 2: Finding the Angular Distance** At \(\theta = \pi/2\), \(\sin\theta = 1\), so: \[ d\Omega^2 = d\theta^2 + d\phi^2 \] But both points have the same \(\theta\), so \(d\theta = 0\), and: \[ d\Omega^2 = d\phi^2 \] \[ d\Omega = |d\phi| \] The difference in \(\phi\) is: \[ |\phi_2 - \phi_1| = |\pi - \pi/2| = \pi/2 \] So the angular distance \(\Delta\Omega = \pi/2\). --- ## **Step 3: Actual Distance on the Sphere** The actual arc length on a sphere of radius \(R\) is: \[ d = R \cdot \text{(angular distance)} \] So, \[ d = R \cdot \frac{\pi}{2} \] --- ## **Step 4: Radius as a Function of Time** Given: \[ R(t) = \sqrt{t/t_0} \] --- ## **Step 5: Distance as a Function of Time** \[ d(t) = R(t) \cdot \frac{\pi}{2} = \frac{\pi}{2} \sqrt{\frac{t}{t_0}} \] At \(t = t_0\): \[ d(t_0) = \frac{\pi}{2} \sqrt{\frac{t_0}{t_0}} = \frac{\pi}{2} \] --- ## **Step 6: When is the Distance Double?** We want \(d(t) = 2 d(t_0)\): \[ \frac{\pi}{2} \sqrt{\frac{t}{t_0}} = 2 \cdot \frac{\pi}{2} \] \[ \sqrt{\frac{t}{t_0}} = 2 \] \[ \frac{t}{t_0} = 4 \] \[ t = 4 t_0 \] --- ## **Final Answers** ### (A) **Distance between the two points at time \(t\):** \[ d(t) = \frac{\pi}{2} \sqrt{\frac{t}{t_0}} \] ### (B) **Time when the distance is double what it was at \(t_0\):** \[ t = 4 t_0 \] --- **Summary Table:** | Quantity | Expression | |-------------------------------|----------------------------------| | Distance at time \(t\) | \(\frac{\pi}{2} \sqrt{\frac{t}{t_0}}\) | | Time for double the distance | \(t = 4 t_0\) |