(I) The asteroid Icarus, though only a few hundred meters acro...

Question

(I) The asteroid Icarus, though only a few hundred meters across, orbits the Sun like the planets. Its period is 410 d. What is its mean distance from the Sun?

Accepted Answer

Hey, everyone in this problem, we're told that astronomers spot a previously unseen planetoid with a diameter on the order of hundreds of meters. It's orbiting the sun in the same plane as the other planets in the solar system. Given that its period is 560 days. We are asked to determine its mean distance from the sun. And we're told to assume that the mean distance between the sun and the earth is 1.5 multiplied by 10 to the exponent eight kilometers. We're given four answer choices here all in meters. Option A 7.8 multiplied by 10 to the exponent 10. Option B 1.1 multiplied by 10 to the exponent 11. Option C 2.0 multiplied by 10 to the exponent 11. And option D 2.9 multiplied by 10 to the exponent 11. OK. So in this case, we're looking at some planetoid that's orbiting the sun. OK. Now we know that earth also orbits sun, right? So we can do is we can think about Kepler's third law because we have these two objects that are both orbiting the same point, the same point. OK. So Kepler's third law recall tells us that the ratio of the squares of the periods of any two planets revolving above the sun is equal to the ratio of the cubes of this their semi major axis. And so if we write that down in terms of the variables, we can say that T one divided by T two. OK. So the ratio of their periods squared, it's gonna be equal to S one divided by S two. Cute. OK. So the ratio of the cubes of their semi major axis. And so again, T is representing the period S is representing that um semi major axis. And then we're just using one and two for the 1st and 2nd plans. Now, if we wanna go ahead and write this in terms of the problem that we have, we can choose some different notations. So let's use subscript P to indicate our planetoid subscript E to indicate the earth. So we can write this as TP divided by TE squared is equal to sp divided by se Cute. Now, what we're gonna try to do is we're gonna try to figure out what this ratio of this semi major axis is. OK? If we can figure out that ratio, OK. The mean distance between the sun when we're talking about the planetoid versus the earth is gonna have that same ratio. OK? So we're gonna be able to use that to find this mean distance that we're looking. So in order to find this ratio, we're gonna have sp divided by se and we have to take the cube root on both sides. So we can write this as TP divided by te all to the exponent two thirds. Now we know the period of both the planetoid and of the earth. So we can write this out. Now, the period of the planetoid was given in days. So let's stick with days. OK? So what we have is 560 days divided by the period of earth. So 365 days, the number of days in a full year and this is raised to the exponent two thirds. And when we work this out, we're gonna get that this is approximately equal to 1.33024. So again, if we take a look at what this number actually represents, what it's saying is that the semi major axis of our planetoid is gonna be 1.33 times bigger than earth's OK. Now again, if that's the case, then the mean distance between the sun and the planetoid is gonna be that same amount, 1.33 times bigger than that between the sun and the earth, right? So what we can do is we can take that mean distance between the sun and the earth, 1.5 multiplied by 10 to the exponent eight kilometers. We were given that the problem multiply it by this 1.33024. And we get that the mean distance between the planetoid and the sun is approximately equal to 1.99 536 multiplied by 10 to the exponent eight coom. Now, I have to be a little bit careful. The answer choices we were given were in meters. So we need to convert to go from kilometers to meters. We multiply by 1000. So this is gonna be approximately equal to 1.99536 multiplied by 10 to the exponent 11 m. And that is the final answer we were looking for. That is gonna be that mean distance from the plane to, to the sun. If we go up and compare it to our answer choices, we are gonna round to two significant digits and we can see that the correct answer is gonna be option C 2.0 multiplied by 10 to the exponent 11 m. Thanks everyone for watching. I hope this video helped see you in the next one.

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(I) The asteroid Icarus, though only a few hundred meters across, orbits the Sun like the planets. Its period is 410 d. What is its mean distance from the Sun?

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