Astronomers using a 2.0-m-diameter telescope observe a distant su...

Question

Astronomers using a 2.0-m-diameter telescope observe a distant supernova—an exploding star. The telescope's detector records 9.1 x 10¯¹¹ J of light energy during the first 10 s. It's known that this type of supernova has a visible-light power output of 5.0 x 10³⁷ W for the first 10 s of the explosion. How distant is the supernova? Give your answer in light years, where one light year is the distance light travels in one year. The speed of light is 3.0 x 10⁸ m/s .

Accepted Answer

Hey, everyone. So this problem is dealing with light energy. Let's see what they're asking us. A telescope with a diameter of 1.8 m is used to view a star that is far away during a 32nd duration. The telescope's detector records a light energy of 8.25 times 10 to the negative 11 jewels using the temperature and size data from the star. It is known that the star radiates at a constant four times 10 to the 26 laws. During that 32nd duration were asked to determine the star's distance from the earth reporting the result in light years. You tell us then the speed of light is given as three times 10 to be eight m/s. And they remind us that one light year is the distance that light travels in one year. Okay. So our potential answer choices here are a 1150 light years, B 574 light years C 1200 light years or D 855 light years. So to tackle this problem, we need to recall two things about light and energy. So the intensity is given by power over area. And power is given by energy over delta T we have two different scenarios here where we have the energy and the intensity of the star at the point where the star exists. And then we have the energy at the telescope that the telescope is receiving. However, the intensity of that star isn't changing. What's changing is the energy that's transferred. So we can say that I S I, the intensity of the star is equal to I sub T, the intensity of the telescope. So we are going to solve for those two intensities equal to each other to find our distance. So we'll have E us over delta T times A S is equal to E sub T over delta T A sub T. So delta T is canceled because we're just working in that 32nd time frame are the energy from our telescope was given to us in the problem As 8.2, 5 times 10 to the -11 jewels, the energy form the star wasn't directly given to us, but the power from the star was. So we can using this power equation recall that the energy of the star is going to be equal to the power of the star multiplied by delta T and then the area. So the area of this distance from the star to the telescope will view that as a sphere where the radius of the sphere is the distance from the star to, to where the telescope is. So we can recall that the area of a sphere is given by four pi R squared. And then the telescope lens is a circle. So we can recall that the area of the circle is given by pi R squared, using all of that will plug back into this system of equations that we have started that we've identified. So we'll have P S power of the star times delta T over or pie R S squared equal to B T over pi R T squared. And what we are looking to solve for, we need to isolate this R S radius of the sphere distance from the star to the earth. So when we rearrange this equation that looks like R C S is equal to the square root of power given to us in the problem that was four times 10 to the 26th watts Times Delta T, 30 seconds times are telescope radius squared. So that our radius of our telescope, We were told the diameter is 1.8 m. It's the radius is half of that. What is right? 1.8 m divided by two. And then that radius term is squared all over or times Our energy from our telescope just given as 8.2, 5 Times 10 to the - Jones. We plug that into our calculator and we get 5.4, 3 Times 10 to the 18 m. So we are almost there, but not quite yet because that answer is in meters and they asked for the answer in light years. So luckily, they gave us the information that we need to solve this. So we'll write 5.43 times 10 to the 18 m. And we're told that the speed of light is three Times 10 to the 8th m/s. We can recall that there are 3600 seconds in an hour, 24 hours in a day And 365 days a year. And when we moved by all of that out, cancel our units. We are left with light years. And so that equals Our sub S. A distance of light years. We will look at our potential answer choices, that alliance with answer choice. B so B is the correct answer for this problem. That's all we have for this one. We'll see you in the next video.

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