Okay, let's try to tackle this problem step by step. So, the question is about a supernova in our Galaxy and the movement of the ejecta part. The key here is understanding reference frames and how distances and times are perceived differently in different frames. Hmm, special relativity might come into play here because we're dealing with high speeds, but the problem mentions that the ejecta and the Galaxy are inertial frames, so maybe we can use some basic principles without getting into full relativistic calculations. Let me think.

First, the problem states that in the ejecta's reference frame, the center of the Galaxy is moving towards it at 60,000 km/s. Then, we need to find the distance the ejecta travels in the Galaxy's reference frame during 50 seconds as measured in the ejecta's frame. Wait, but the time here is in the ejecta's frame. So, there's a time dilation effect if the speeds are significant compared to the speed of light, but 60,000 km/s is 6e4 km/s. Let me convert that to meters per second to check: 6e4 km/s is 6e7 m/s. The speed of light is about 3e8 m/s, so 6e7 is 0.2c. Hmm, 20% the speed of light. That's not negligible, so maybe we need to consider time dilation. Wait, but the problem says both are inertial frames. So, perhaps we need to use the Lorentz transformation here.

Wait, but the question is asking for the distance traveled in the Galaxy's frame when 50 seconds pass in the ejecta's frame. So, first, in the ejecta's frame, the time is 50 seconds. But in the Galaxy's frame, the time would be different due to time dilation. Because the ejecta is moving at 0.2c relative to the Galaxy. So, the time experienced in the Galaxy's frame (t') would be t = t0 * gamma, where gamma is the Lorentz factor. Gamma is 1 / sqrt(1 - v²/c²). Let's calculate gamma here. v is 0.2c, so v²/c² is 0.04. So, 1 - 0.04 = 0.96. sqrt(0.96) is approximately 0.9798. So, gamma is 1 / 0.9798 ≈ 1.0206. So, t' = 50 * 1.0206 ≈ 51.03 seconds. But wait, no, wait. Wait, in the ejecta's frame, the time is t0 = 50 s. The Galaxy's frame sees the ejecta moving at 0.2c, so the time dilation would make the Galaxy's frame time longer. So, t' = t0 * gamma. So, yes, around 51.03 seconds. But maybe we need to do the exact calculation. Let me compute gamma more accurately. Let's compute 1 - (0.2c)^2 / c^2 = 1 - 0.04 = 0.96. So sqrt(0.96) is sqrt(96/100) = sqrt(24/25) = (sqrt(24))/5 ≈ 4.89898/5 ≈ 0.9798. So gamma is 1/0.9798 ≈ 1.02062. So, t' = 50 * 1.02062 ≈ 51.031 seconds. But maybe the problem expects us to use the Galilean transformation instead of Lorentz? Because 0.2c is not that small, but maybe in the context of the problem, since it's a supernova ejecta moving at a significant fraction of the speed of light, we need to use relativity. Wait, but the question says both are inertial frames, so the correct way is to use Lorentz transformations. Hmm. Alternatively, maybe the problem is simplified and we can ignore time dilation, but that seems unlikely. Let's check the answer options. The options are in the range of about 3 million km. Let's see: if we use Galilean transformations, which neglect relativity, then the distance would be speed multiplied by time in the Galaxy's frame. But wait, if the ejecta is moving at 0.2c towards the center, and in the Galaxy's frame, the ejecta is moving at 6e4 km/s, then the time in the Galaxy's frame would be t = t0 * gamma. So, the distance would be v * t. Wait, but v is the speed of the ejecta in the Galaxy's frame. Wait, the ejecta is moving at 6e4 km/s towards the center, and the time in the Galaxy's frame is t' = 50 * gamma. So, the distance would be v * t' = 6e4 km/s * 50 * gamma. Let's compute that. 6e4 * 50 = 3e6 km. Then multiply by gamma ≈ 1.0206, so 3e6 * 1.0206 ≈ 3,061,800 km. The closest option is C, 3,060,000 km. But let's check the exact calculation. Gamma is 1/sqrt(1 - 0.04) = 1/sqrt(0.96) ≈ 1.0206207. So, 3e6 * 1.0206207 ≈ 3,061,862 km. That's approximately 3,061,862 km, which rounds to 3,060,000 km if we consider significant figures. But let's see if there's another approach.

Alternatively, maybe the problem expects us to use the fact that in the ejecta's frame, the Galaxy is moving towards it at 6e4 km/s. Then, the time in the ejecta's frame is 50 seconds. But the distance in the ejecta's frame would be the speed multiplied by time, so 6e4 km/s * 50 s = 3e6 km. But that's the distance in the ejecta's frame. But the question asks for the distance in the Galaxy's frame. So, we need to calculate the distance in the Galaxy's frame. So, perhaps using length contraction? Wait, length contraction is in the direction of motion. But in this case, the ejecta is moving towards the center, so in the Galaxy's frame, the ejecta is moving at 6e4 km/s. Wait, but the length contraction formula is L = L0 * sqrt(1 - v²/c²), where L0 is the proper length (in the rest frame of the object). But here, the ejecta is moving relative to the Galaxy, so the distance in the Galaxy's frame would be contracted. Wait, but the ejecta's frame is moving at 6e4 km/s relative to the Galaxy. So, in the Galaxy's frame, the ejecta's length (the distance from the ejecta to the center) is contracted. Wait, but the ejecta is moving towards the center, so the distance from the ejecta to the center in the Galaxy's frame would be less than the distance in the ejecta's frame. Wait, but the ejecta's frame is moving at 6e4 km/s towards the center relative to the Galaxy. So, in the Galaxy's frame, the ejecta is approaching the center at 6e4 km/s. The time elapsed is 50 seconds in the ejecta's frame. But how to relate the times between the frames? Alternatively, perhaps we need to use the Lorentz transformation equations. Let's try that.

Let me denote the Galaxy's frame as S and the ejecta's frame as S'. In S, the ejecta is moving at velocity v = 6e4 km/s = 6e7 m/s. The speed of light is c = 3e8 m/s, so v = 0.2c. The time in S' is t' = 50 s. We need to find the distance traveled in S (the Galaxy's frame). The distance in S would be x = v * t, but we need to relate t in S to t' in S'. Using Lorentz transformations:

t' = gamma (t - (v x)/c²)
x' = gamma (x - v t)

But in the ejecta's frame (S'), the center of the Galaxy is moving towards it at v = 0.2c. The ejecta starts at some point, and after 50 seconds in S', the center has moved closer. The distance in S' would be x' = v * t' = 6e4 km/s * 50 s = 3e6 km. So, in S', the center is 3e6 km closer. Now, we need to find the corresponding time t in S. Using the time transformation equation:

t' = gamma (t - (v x)/c²)

But in S', the position of the center is x' = 3e6 km. Wait, but in S', the ejecta is at rest (x' = 0) and the center is moving towards it. Wait, maybe I need to set up the coordinates properly. Let's define the events. Let's say at t' = 0 in S', the ejecta is at x' = 0, and the center of the Galaxy is at x' = -3e6 km (since it's moving towards the ejecta). Then, at t' = 50 s, the center would be at x' = -3e6 km + 0.2c * 50 s. Wait, 0.2c is 6e7 m/s, 50 s is 50 s. So, 6e7 m/s * 50 s = 3e9 m = 3,000,000 km. So, x' = -3e6 km + 3e6 km = 0 km. Wait, that can't be right. Wait, no. Wait, the speed is 0.2c, so in 50 seconds, the center moves 0.2c * 50 s = 6e7 m/s * 50 s = 3e9 m = 3,000,000 km. So, if the center was initially at x' = -3e6 km (relative to ejecta's position at x' = 0), after 50 s, it's at x' = -3e6 km + 3e6 km = 0. So, the center is coinciding with the ejecta's position in S'? That seems like a coincidence, but perhaps it's because the ejecta's frame is moving towards the center at the same speed as the distance in their frame. But regardless, we need to find the corresponding t in S. Let's use the Lorentz transformation. The Lorentz factor gamma is 1/sqrt(1 - v²/c²) = 1/sqrt(1 - 0.04) = 1/sqrt(0.96) ≈ 1.02062. Now, the events are in S': at t' = 50 s, x' = 0 (since the center is at the ejecta's position). So, we can use the Lorentz transformation equations:

t = gamma (t' + (v x')/c²)
x = gamma (x' + v t')

But x' = 0, so:

t = gamma * t' = 1.02062 * 50 ≈ 51.031 s
x = gamma * v * t' = 1.02062 * 0.2c * 50 s

Wait, but v is 0.2c, so v = 0.2c. Let's compute x:

x = gamma * 0.2c * 50 s = 1.02062 * 0.2 * 3e8 m/s * 50 s

First, 0.2 * 3e8 = 6e7 m/s. Then, 6e7 m/s * 50 s = 3e9 m = 3,000,000 km. Then, multiply by gamma ≈ 1.02062: 3e6 km * 1.02062 ≈ 3,061,860 km. So, the distance in the Galaxy's frame is approximately 3,061,860 km. The closest option is C, 3,060,000 km. But the options are given as exact numbers. Let me check if there's an exact calculation.

Alternatively, perhaps we can use the velocity addition formula. Wait, but the problem is about distance over time, so maybe that's not necessary. Alternatively, perhaps the problem expects us to use the fact that in the ejecta's frame, the center is approaching at 6e4 km/s, and in the Galaxy's frame, the ejecta is moving at 6e4 km/s. So, the time in the Galaxy's frame is t = t' * gamma. Then, the distance the ejecta travels is v * t = 6e4 km/s * 50 * gamma. Gamma is 1/sqrt(1 - (0.2)^2) = 1/sqrt(0.96) ≈ 1.02062. So, 6e4 * 50 = 3e6. 3e6 * 1.02062 ≈ 3,061,860 km. So again, approximately 3,060,000 km. The answer options are A:3e6, B:2.94e6, C:3.06e6, D:2.88e6. So, 3.06e6 is option C. But let's check if there's another approach.

Wait, maybe the problem expects us to use the length contraction formula. In the ejecta's frame, the distance to the center is 3e6 km. But in the Galaxy's frame, the ejecta is moving at 0.2c, so the distance would be contracted. Length contraction is L = L0 * sqrt(1 - v²/c²). So, L = L0 * sqrt(1 - 0.04) = L0 * sqrt(0.96) ≈ L0 * 0.9798. So, if L0 is 3e6 km, then L = 3e6 * 0.9798 ≈ 2,939,400 km. But that's the distance in the Galaxy's frame. Wait, but the problem is asking for the distance the ejecta travels in the Galaxy's frame when 50 seconds pass in the ejecta's frame. So, maybe this is not the right approach. Wait, length contraction applies to the object's rest frame. In this case, the ejecta is moving, so the distance between the ejecta and the center in the Galaxy's frame is shorter than in the ejecta's frame. But the ejecta's frame is moving relative to the Galaxy. So, the distance in the Galaxy's frame would be Lorentz contracted version of the ejecta's frame distance. So, if in the ejecta's frame, the distance is 3e6 km, then in the Galaxy's frame, it's 3e6 * sqrt(1 - v²/c²) = 3e6 * sqrt(0.96) ≈ 2,939,400 km. But that's the distance between the ejecta and the center in the Galaxy's frame. However, the problem is asking for the distance the ejecta travels in the Galaxy's frame during 50 seconds in the ejecta's frame. Wait, maybe that's a different scenario. Wait, the ejecta is moving at 0.2c relative to the Galaxy. So, in the Galaxy's frame, the ejecta is moving at 0.2c. So, over a time t, the distance it travels is 0.2c * t. But the time in the Galaxy's frame is not the same as the ejecta's frame. So, we need to relate the time in the Galaxy's frame to the time in the ejecta's frame. Using time dilation, t = gamma * t'. So, t = 1.02062 * 50 ≈ 51.031 s. Then, distance = 0.2c * t ≈ 0.2 * 3e8 m/s * 51.031 s ≈ 0.2 * 3e8 * 51.031 ≈ 3.06186e9 m ≈ 3,061,860 km. Which matches the previous calculation. So, the answer should be approximately 3,061,860 km, which is closest to option C: 3,060,000 km. But let's check the exact calculation. Gamma is exactly 1/sqrt(1 - 0.04) = 1/sqrt(0.96) = sqrt(25/24) = 5/(2*sqrt(6)) ≈ 1.02062. So, 50 * 5/(2*sqrt(6)) = 250/(2*sqrt(6)) ≈ 125/sqrt(6) ≈ 125/2.4495 ≈ 51.03. Then, 0.2c * t = 0.2 * 3e5 km/s * t. Wait, 3e5 km/s is 300,000 km/s. Wait, 0.2c is 0.2 * 3e5 km/s = 6e4 km/s. So, 6e4 km/s * 51.03 s ≈ 6e4 * 51.03 ≈ 3.0618e6 km. So, exactly 3,061,800 km, which is 3,061,800 km. But the options are in thousands, so 3,060,000 km is option C. Therefore, the correct answer should be C. But let me check if there's an alternative approach where we use the velocity in the Galaxy's frame and the time in the ejecta's frame without considering time dilation. If someone ignores relativity and just uses t = 50 s and v = 6e4 km/s, then distance = 6e4 * 50 = 3e6 km, which is option A. But this ignores the fact that time in the Galaxy's frame is longer due to time dilation. So, the correct answer should account for time dilation. Therefore, the answer is C.
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