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Two planets P1 and P2 orbit around a star S in circular orbitswith speeds v1 = 40.6 km/s, and v2 = 59.4 km/s respectively. (a) Ifthe period of the first planet P1 i...

Question

Two planets P1 and P2 orbit around a star S in circular orbitswith speeds v1 = 40.6 km/s, and v2 = 59.4 km/s respectively. (a) Ifthe period of the first planet P1 is 750 years what is the mass, inkg, of the star it orbits around? kg (b) Determine the orbitalperiod, in years, of P2. yr

Two planets P1 and P2 orbit around a star S in circular orbits with speeds v1 = 40.6 km/s, and v2 = 59.4 km/s respectively. (a) If the period of the first planet P1 is 750 years what is the mass, in kg, of the star it orbits around? kg (b) Determine the orbital period, in years, of P2. yr



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Two newly discovered planets follow circular orbits around a star in a distant part of the galaxy. The orbital speeds of the planets are determined to be 43.3 km/s and 58.6 km/s. The slower planet’s orbital period is 7.60 years. (a) What is the mass of the star? (b) What is the orbital period of the faster planet, in years?

Problem. We are given to newly discovered planet Follow circular orbit about a star in a distant part of the galaxy. The orbital dispute of the planets are deter. Mined 43.3 kilometers per 2nd, 43.3 kilometer per 2nd and 58.6 kilometer per 2nd, 58.6 kilometers per second. The slower planets orbit. Orbital period is 7.60 years, 7.60 years. What is the mass of the star? What is the orbital period of the faster planet in years? So first we know that formula it must will be equal to Speed Cube in two time period divided by buy into gravitational constant G capital D. So here time period is given 7.60 years. So first we will convert it into second. So we know that 17.60 years and in one year there are 3.156 and 2 10 to the power seven seconds. So the time period will be equal to 2.40 into 10 to the power eight seconds. This is the time period now speed for this planet is given to us 43.3 kilometer per second. So this will be called to 10 to the power three meters per second. So now we will put the value of V and time period in this formula, so the masks will be massive. Planet will be called to you. Speak Q just 43.3 into 10 to the power three whole skill into time period with just 2.40 in 2 10 to the power eight divided by by in two G. So we know that here to five will be there because, like this. So the value of G 6.67 to 10 to the power minus 11. Now we will solve this. The mass of planet Slower planet will be called to 4.65 into 10 to the power 31. Okay, Katie, now for part, will be able to find that time period. So the time period is a call to, but to buy G into em. The railway, his Speed Cube. So here we will put the value of G, which is 6.667 10 to the power minus 11. And Mars Mass is 4.65 to 10 to the Power 31 and his fetus 58.6 kilometre bus again. So this will be called to 58.6 into 10 to the power three Q. So the time period will be called to. It isn't second, which is 69.69 into 10 to the power seven seconds. So I converted in two years. So this will be called to 9.69 and 10 to the power seven divided by we 0.157 and 2 10 to the power 3.156 and 2 10 3 Power 76 So this is a quarto around 3.7 years.

In question 19 were asked to determine how much mass in our galaxy lies within the sun's orbit about the galactic center. To solve this problem, we're going to need to use this formula given to us in the text. A cube is equal toe m one plus m two times p squared. This expression is another way of writing. Kepler's third law a is thesis Emmy major axis of Thea orbiting body which, for our problem, is the sun. It can also be thought of as being the sun's distance from the galactic center, which is given to us as 26,000 light years. P is thieve orbital period for the sun. The period of one orbit about the galactic center is equal toe 2.25 times 10 to the eight years Now. In order to perform our calculations, we're going to need to convert that 26,000 light years in tow. Astronomical units we're going thio need thio multiply by a couple conversion factors here. First, we're gonna need to multiply by how maney meters there are in one light year. That's 9.46 times 10 to the 15 m and then we're going to need to multiply by one over the number of meters There are in an astronomical unit when astronomical unit is the Earth sun distance, which is equivalent to 1.5 times 10 to the 11 m. We can now cancel out the light years and the meters and get our sons distance from the galactic center in astronomical units. It turns out to be 1.64 times 10 to the nine astronomical units. Now M one and M two are the masses of the two objects involved in our problem, which will be the son and the galaxy. Now the sun's mass in comparison to the Milky Way's mass is very, very small. So in reality we don't even need to consider the sun's mass. When solving this problem, we could just simply say, am one. The mass of our galaxy is approximately equal toe m one plus m two. With solar mass being negligible, we can get on our way to solving for EMS of one substituting in our values for A and P. Here we have 1.64 times 10 to the nine astronomical units. Cubed is equal thio ems of one times two point 25 times 10 to the eight years squared to solve for M one, we need only do a bit of division weaken. Take 1.64 times 10 to the nine astronomical units cubed and divide by 2.25 times 10 to the eight years squared, and our answer will actually come out in solar masses. The mass of the galaxy inward of the Sun's orbit, is equal to 8.7 times 10 to the 10 solar masses.

So for part a um we can see that the orbital speed of the planet V. Is going to be equal to the square root of the gravitational constant, multiplied by the mass divided by our This is simply using Newton's second law and relating it to the centripetal acceleration of a planetary body in orbit. We can we know rather all of these values in terms of other known values. So The gravitational constant is of course 6.6, 7 Times 10 to the -11. Newton's meters squared per kilogram squared. And the mass here is going to be a that's not truly 85% of the mass of the Sun. So we'll say .85 multiplied by the mass of the sun. The mass of the sun being 1.9, 9 Times 10 to the 30th kilograms, extend the square root and this will be divided by an orbital orbital distance essentially. Um That is 11% of the orbital radius um of Earth, and in this case we have 11% sore .11 times the orbital radius of Earth. So essentially the distance between the center of the sun and the center of the earth. This is approximately 1.5 Times 10 to the 11th kilometers. So, given that we can see that, then the velocity for this planet row cancri 8.3 times 10 to the fourth meters per second. Of course this is rounded to two significant figures. We can use them for part B. Kepler's third law. So Kepler's third law relates the orbital period and the radius of the orbit. So we can say that then According to Kepler's 3rd Law, T Is going to be equal to two pi R are raised to the three halves power. This would be divided by the square root of the gravitational constant times the mass. So again, this would be two pi. This would be multiplied by again 0.11 Times 1.5 times 10 to the 11th kilometers. This will be raised to the 3/2 power, all divided by the square root. Uh 6.6, 7 times 10 to the negative 11th to be newton's meters squared per kilogram squared. And this will be multiplied by that mass. Again .85 Times 1.9, 9 times 10 to the 30th kilograms. Extend the square root, extend the fraction and we see that then the orbital period t rounded to again, two significant figures 1.3 Times 10 to the 6th seconds. That is the end of the solution. Thank you for watching.

A star with the name I won't attempt to pronounce is 57 light years away from us. And it has a mass of 570.885 times that of our sun. There's a planet that is 0.11 times the distance that we are away from the sun and we wanted then no, in part A. What is the orbital velocity of this planet around that star? So as we've been using a lot in the section, we know that for any object in circular orbits, its orbital velocity is just the square root of G times. The mass usually would say of the planet. I guess we'll say of the star. Here's to the mass of the object being orbited, divided by the or real distance are so they've given us these in terms of values that we know so well. Just substitute those in so G and then the mass of this planet is 0.85 times than that of the sun, and the distance is 0.11 times the distance between the sun to the earth. So we look up those values in the appendix and we should get the velocity is a 2000 700 meters per second. So that's part a. And then in part B. They want to know what the orbital period is. So the period is just the distance. Travel in one cycle, the circumference two pi r divided by the philosophy. So to pie and again for our 0.11 times the distance of the sun to the earth and then divided by that velocity. We just found 8282700 meters per seconds. And then we should find that that is about 1.25 times 10 to the sixth seconds. And I don't have a great sense of what that is, so we do a bit of converting and we find that it's about 14.5 days.


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