Our Milky Way galaxy is 100,000 ly in diameter. A spaceship cr...

Question

Our Milky Way galaxy is 100,000 ly in diameter. A spaceship crossing the galaxy measures the galaxy’s diameter to be a mere 1.0 ly. How long is the crossing time as measured in the galaxy’s reference frame?

Accepted Answer

Hello, fellow physicists today, we're gonna solve the following practice pro together. So first off, let us read the problem and highlight all the key pieces of information that we need to use. In order to solve this problem. A galaxy has a radius of 30,000 light years, a spacecraft traveling through the galaxy perceives the radius of the galaxy to be 125 light years determine the time needed as measured in the galaxy frame for the spacecraft to cross the entire galaxy. So that's our end goals. We're trying to figure out how much time is required as measured from the galaxy's frame of reference for the spacecraft to cross the entire galaxy. That's the end goal. That's our end goal. That's ultimately what we're trying to solve for. So that's our final answer. Awesome. We're also given some multiple choice answers and they're all in the same units of years. So let's read them off to see what our final answer might be. A is 15 multiplied by 10 to the power of two B is 12 multiplied by 10 to the power of three C is 60 multiplied by 10 to the power of three and D is 75 multiplied by 10 to the power of three. So first off, let us note that the reference frame s within the galaxy, which is the diameter of this galaxy or the length of this galaxy will be L equals two multiplied by 30 1000 light years, which is equal to 60,000 light years. And I'll just put it in parentheses, but this is also our diameter, which I'm just gonna denote as D. So don't get confused. This is the diameter of our galaxy. So this length will be denoted as the proper length L and this value will remain constant since it is measured from rest. OK. Since the galaxy obviously isn't moving, the only thing that's moving in this particular problem is the spacecraft, the galaxy isn't moving but the spacecraft is however, in the reference frame of the a spacecraft, which we're gonna denote as S prime. So let's make a little note here in blue. So we don't get confused. So S is the galaxy and S prime and where is our spacecraft? OK. So we don't get confused. So, so our reference frame of a spacecraft which is denoted as S prime is moving towards the galaxy with a velocity of V. So the galaxy appears to have a much shorter length or a shorter diameter which is measured as, and we can write this equation as L prime is equal to two multiplied by 125 light years. Thus, there is a difference in perception based on the relative motion between the spacecraft and the galaxy. So we can apply and use the following equation to sulfur the velocity V. So L prime. So this is the spacecraft's frame of reference for the length is equal to L which is our proper length of the galaxy. And it's multiplied by the square root of one minus V squared divided by C squared. So now we need to take this equation and rearrange it to isolate and solve for V. So let's start to do that. So L prime divided by L is equal to the square root of one minus V squared divided by C squared, which we can simplify and write that L prime divided by L all to the power of two is equal to one minus V squared divided by C squared, which we can simplify again and write V squared divided by C squared is equal to one minus L prime divided by L all to the power of two which you can simplify and write that V squared is equal to C squared multiplied by one minus L prime divided by L all to the power of two, which we can simplify once again to get that V is equal to C multiplied by the square root of one minus L prime divided by L all to the power of two. So now that we finally got it down to our final form. We can now plug in our known variables and solve for V. So let's do that. So V is equal to C, which if I haven't mentioned it before, C is the speed of light multiplied by the square root of one minus two, multiplied by 125 light years, all divided by two multiplied by 30,000 light years. And this is all to the power of two. So when we plug that into a calculator, we will get that V is equal to 0.9999 nine. So five nines 1319 C as this is the number you'll see in your calculator. Hopefully if you typed everything in correctly. So in the galaxy's frame of reference, s the spacecraft travels 60,000 light years at a speed of 0.999991319 C. Thus, the time required will be, and that's called as delta T is equal to L divided by V which is equal to 60,000 light years, all divided by 0.9999 91319, multiplied by one light year, divided by years, which is approximately equal to 60,000 point 52084. Therefore, the time needed as measured from the galaxy's frame of reference for the spacecraft to cross the entire galaxy or the galaxy's diameter value is 60,000 years or in the notation which is like a form of scientific notation. Like all of our multiple choice answers are in, it'd be 60 multiplied by 10 to the power of three years. And this is our final answer. Hooray, we did it. So looking at our multiple choice answers, the correct answer has to be the letter C 60 multiplied by 10 to the power of three years. Thank you so much for watching. Hopefully, that helped and I can't wait to see you in the next video. Bye.

Our Milky Way galaxy is 100,000 ly in diameter. A spaceship crossing the galaxy measures the galaxy’s diameter to be a mere 1.0 ly. How long is the crossing time as measured in the galaxy’s reference frame?

Key Concepts

Length Contraction

Time Dilation

Reference Frames

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