(III) The comet Hale–Bopp has an orbital period of 2400 years....

Question

(III) The comet Hale–Bopp has an orbital period of 2400 years.

(c) What is the ratio of the speed at the closest point to the speed at the farthest point?

Accepted Answer

Hey, everyone in this problem, a celestial body located beyond Neptune's orbit takes 1200 years to complete one revolution around the sun. Given that its closest distance from the sun is three A U and its furthest distance from the sun is 223 A U. We are asked to determine the ratio of its speed at the perihelion, the closest distance from the sun to this speed at the aphelion. OK. So the furthest distance from the sun and we're told that one A U is equal to the mean distance between earth and the sun. We have four answer choices in this problem. Option A three divided by 223. Option B one divided by one, option C 113 divided by 110 and option D 223 divided by three. So let's start by writing out some of this information we're given because we're told that this celestial body takes 1200 years to complete one revolution around the sun. And so its period we'll call TC. The period of this celestial body is equal to 1200 years. Bye now, we have these distances, the closest distance and the furthest distance from the sun. Ok. So we're gonna say that RP that radius at the perihelion, the closest distance is gonna be three A U and the furthest distance are A and the distance at the affiliate is 223 A U. Now, let's go ahead and draw what we have going on as well. So we have our elliptic orbit, we have the sun, it's gonna be towards one side. We have our perihelion on the left and that's gonna be closest to the sun and we have our helium on the right. That's the furthest distance from the sun. OK. So we think about the distance directly from that perihelion to the sun. That's that distance RP. OK. And we can draw a bit of a triangle here and imagine in this area really close to RP, our celestial body is moving with some speed VP. OK. And we can do the same on the other side. OK. So we have this distance between the sun and the ophelia. That's gonna be that distance R A. And we can draw a similar triangle where this celestial body will be moving with a velocity we'll call VA or speed VA. Now, in this case, hey, we want to think about the ratio of the speeds and what we're actually gonna use is Kepler's Second law. OK. So recall it, Kepler's second law tells us that each planet moves such as an imaginary line line drawn from the sun to the planet sweeps out equal areas in equal periods of time. OK. So what that means is if we look at the diagram, we've drawn the area that this blue line encompasses and the area that the green lines encompass those are gonna be equal if we look at an equal amount of time, an equal period of time for both of them. Hm. So what we can do is approximate these regions by a triangle. OK. So this is reasonable if we consider a small enough delta T, we've obviously drawn them a little bit bigger so that we can see them more clearly and understand them. But we're, we can think about having a really small delta T that would allow us to approximate these by triangles. And if we did the base of the triangle. And so if we think about the base of the triangle here and we think about this being really, really small delta T, the base of the triangle is just gonna be equal to the distance traveled during that time period. Delta T. OK. Well, how do we calculate a distance? I recall that a distance is gonna be equal to the speed multiplied by the time. OK. So that base, that distance is gonna be equal to the multiplied by T. Now the speed V we have either VP or VA and then there are time T we're gonna have delta T and now we can see how we get these speeds introduced into this equation. Remember we want this ratio of speed. So this allows us to introduce those speeds into our equations. OK. So now let's get to Kepler's Second law. OK. And it tells us like we mentioned that the area, OK. For this para heelan triangle is gonna be equal to the area of this athel triangle that we drew. How do we calculate area? Area is one half multiplied by the base multiplied by the height because we're dealing with a triangle. OK. So the same for both, we just have a different base, a different height. Now, for the pelion, OK, our base is gonna be the speed VP multiplied by delta T and the height HP of our triangle is gonna be that distance RP. We're gonna have one half multiplied by VP, delta T multiplied by RP equal to one half VA delta T multiplied by R. Now, on both sides, we have this one half so we can divide everything by one half. That's gonna go away. We have a delta T on both sides, we can divide by delta T on both sides. And what we are left with is V pr P is equal to VA R A. Can you remember again, we are looking for the ratio of the speed at the perion, the speed at the feeling? Hey, so we're gonna divide both sides by va divide both sides by RP. And we get that VP divided by va that ratio of speeds that we're looking for is just equal to R A divided by RP. We know the value of R A, we know the value of RP. We can substitute those in, they are both in the same unit of A U which will divide out so we can use those units and we are gonna have 223 A U divided by three A U which gives us 223 divided by three. OK. So using Kepler's second law, approximating these areas using some information we have that relates the speed to the distance. We were able to find that the ratio of the speed at the para heel to the speed at the ethel is gonna be equal to approximately 223 divided by three which corresponds with answer choice. D Thanks everyone for watching. I hope this video helped see you in the next one.

Struggling with Physics?

(III) The comet Hale–Bopp has an orbital period of 2400 years.
(c) What is the ratio of the speed at the closest point to the speed at the farthest point?

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