(II) A science fiction writer imagines a planet that has three...

Question

(II) A science fiction writer imagines a planet that has three times the mass of the Earth orbiting a star that has twice the mass of our Sun. The writer wants the planet’s “year” to be 10% longer than Earth’s orbital period. At what mean distance should she place the planet from its star?

Accepted Answer

Hey, everyone in this problem, we are told that astronomers discovered a potentially habitable planet with a mass equal to five times that of earth. It revolves around a star of mass similar to that of our sun. If its orbital period is 5% more than earth, we are asked to determine the average distance it is from its star. We're told to assume that the mean distance between the sun and earth is 1.5 multiplied by 10 to the exponent eight kilometers. And we're given four answer choices A through D all in meters. Option A two multiplied by 10 to the exponent 10. Option B 1.5 multiplied by 10 to the exponent 11. Option C 1.6 multiplied by 10 to the exponent 11. And option D 1.7 multiplied by 10 to the exponent 11. OK. What, so let's start. And what we're gonna do is we're gonna recall that the period of orbit T squared, it can be written as four pi squared multiplied by R cute divided by capital G multiplied the mass. OK. So this is a form of orbiting. And in this case, we've said the mass of the sun or in this, it really is the mass of the star that we're orbiting around. OK. R is gonna be that average distance from the star G is our gravitational constant. OK. So this is a general formula that applies to these orbits around some star. So we think about this and we're given some information about earth here and then we're given some information about this other planet. OK. So we can write this equation out for earth and we can write this equation out for that other planet. So let's do that. We're gonna use subscript E to indicate the earth. When we do that, we get that te is gonna be equal to four high squared multiplied by REQ. So re is gonna be the average distance from the earth to the sun. OK? That's gonna be divided by capital G multiplied by the mass of the sun. OK? Notice that this is not the mass of the earth. This is the mass of the star that we're orbiting around, which in the case of earth is the sun. OK? So that's for the earth, we can do the same for our planet. We're gonna use subscript P. So TP is gonna be four high squared multiplied by RP cubed, all divided by capital G multiplied by the mass of this star that it's orbiting about. Now RP, this is the value we're looking for this average distance from this planet to the star. OK. Now, what we're told is that the orbital period is 5% more than earth for this plant. OK. Now, the period where call is given by t so that means that the period for the planet is gonna be 1.05 multiplied by the period for earth because it's 5% more. OK. So let's actually substitute this in to our equation that we wrote on the right hand side. If we go ahead and substitute that in what we have is at 1.05 multiplied by the period of the earth is gonna be equal to four pi squared multiplied by RP cute divided by G multiplied by the mass of the star. Now, we know what TE is, we have an equation for TE in terms of these similar things, Pi RGM. So let's go ahead and substitute in what we know for TE as well. So now we can write at 1.05 multiplied by four pi squared re cubed divided by G multiplied by the mass of the sun. It's gonna be equal to four pi squared multiplied by R pc divided by G multiplied by the mass of the star. Now, some of these terms are gonna cancel. OK. So we can divide both sides by four pi squared. That's gonna go away. We can multiply both sides by capital G that's gonna go away. And we recall that in the problem, we were told that the mass of this star is equivalent to the mass of the sun. And they're very similar. So the mass of the sun, the mass of the star is equivalent, which means we can divide that out as well. When we do that, we are left with RP cubed being equal to 1.05 multiplied by re cubed. We have this direct relationship just between these average distances. Now, based on what we were told. And so we didn't need all of the particular details about the mass of each planet. We were able to eliminate that just by knowing the relationship between the two. OK. So again, we're looking for RP, the last step will be to take the cube route. RP will be equal to the Q root of 1.05 multiplied by REQ. Now re we are told in the problem to assume that the mean distance between the sun and earth is 1.5 multiplied by 10 to the exponent eight kilometers. That is gonna be that re value. So RP is gonna be equal to the cube root of 1.05 multiplied by 1.5 multiplied by 10 to the exponent eight kilometers. All cute. And when we work this out on our calculator, we're gonna find that RP is approximately equal to 1.52 459 multiplied by 10 to the exponent eight kilometers. We have to be a little bit careful. The answer choices here were given in meters, not kilometers. So we're gonna have to convert this into meters by multiplying by 1000. So we have 1.52 459, multiplied by 10 to the exponent 11 kilometers, right. And that is the average distance from this planet to its star that we were looking for. If we go back to our answer choices, we're gonna have to round a little bit. So we're going round to two significant digits. And when we do that, we can see that the correct answer is going to be option B but 1.5 multiplied by 10 to the exponent 11 m. Thanks everyone for watching. I hope this video helped see you in the next one.

(II) A science fiction writer imagines a planet that has three times the mass of the Earth orbiting a star that has twice the mass of our Sun. The writer wants the planet’s “year” to be 10% longer than Earth’s orbital period. At what mean distance should she place the planet from its star?

Key Concepts

Kepler's Third Law of Planetary Motion

Gravitational Force

Orbital Period

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