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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?

$4.56 \times 10^{31} k g$3.07$y r$

Physics 101 Mechanics

Chapter 5

Dynamics of Uniform Circular Motion

Newton's Laws of Motion

Applying Newton's Laws

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03:28

Newton's Laws of Motion are three physical laws that, laid the foundation for classical mechanics. They describe the relationship between a body and the forces acting upon it, and its motion in response to those forces. These three laws have been expressed in several ways, over nearly three centuries, and can be summarised as follows: In his 1687 "Philosophiæ Naturalis Principia Mathematica" ("Mathematical Principles of Natural Philosophy"), Isaac Newton set out three laws of motion. The first law defines the force F, the second law defines the mass m, and the third law defines the acceleration a. The first law states that if the net force acting upon a body is zero, its velocity will not change; the second law states that the acceleration of a body is proportional to the net force acting upon it, and the third law states that for every action there is an equal and opposite reaction.

03:43

In physics, dynamics is the branch of physics concerned with the study of forces and their effect on matter, commonly in the context of motion. In everyday usage, "dynamics" usually refers to a set of laws that describe the motion of bodies under the action of a system of forces. The motion of a body is described by its position and its velocity as the time value varies. The science of dynamics can be subdivided into, Dynamics of a rigid body, which deals with the motion of a rigid body in the frame of reference where it is considered to be a rigid body. Dynamics of a continuum, which deals with the motion of a continuous system, in the frame of reference where the system is considered to be a continuum.

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here we have to find the mass of a star that two separate planets are orbiting around. Given that we know both their speeds and one of their periods, the first part of the problem asks us to find the star's mass. Um, and the first thing that we want to do is look at all our givens and make sure that they're in quantities that we can use. Um, everything we read me notice is that our losses were given in kilometers per second. Would you wanna rewrite and meters per second? That's the units that we always work with. Physics. That's just gonna be 43.3 times 10 cubed meters per second on the same thing here. We're just gonna multiply by 10 cubed to get everything in terms of meters. Now, our period for first planet is actually given in years. Um, but you congratulate and just double check. There's 3.154 times under the seven seconds in a year, so we will have end up just multiplying those quantities to get our period in seconds when we end up looking at it. All right. So, as in all planets where we have our little problems where we have something orbiting another object. We know that our gravitational force is responsible for a centripetal force. Gravitational attraction is what causes the object to move around the other object in a circular motion that means that n v squared over r must be equal to and one which is gonna be the mass for satellite capital M, which I'm going to use to be the mass of our planet times the gravitational constant all over r squared. Now from here we want to do is sulfur capital and the mass of our planet. So we can rewrite this to be m equals r squared times V squared over RG was just simplifies of course toe r v squared over G. Now this problem we're given the velocity, but we're not given the radius. But since we're given the period we consult for that. We know that t is to pi over omega, which means that it is equal to two pi r over b does weaken solve to get our in terms of quantities that we know tv over to pie. So now we have the m is equal to t v cubed over to Pi G in order to solve for em all of us to do is plug in. We know our velocity So we get 43 3 times 10 cubed meters per second on DDE. That is all cubes well multiplied by our period 7.60 times, 3.15 times 10 to the seventh. This could be in seconds and then all over two times pi times the gravitational constant which is 6.74 times 10 to the negative 11 Newtons meters squared over kilograms squared. We do all that math, Albert, looking into a calculator we get that answer is 4.56 times 10 to the 31 kilograms and we do in fact find that our units around to be kilograms so that all works out and make sense. All right, so from there we can move on to the second part of the problem here. We need to find the orbital period of our second planet in years so we can use the equation that we had earlier that our mass is equal to t v cubed over to pi G. Since we just sulfur our capital m so we can rearrange the sulfur tea which we find is equal to two pi g times m over be cubed and we know RV because they are givens We know our teeth We're looking for a tea and we know our m So when we plug in all our quantities that we found earlier to pi times not gonna write g out again, that's quite a long quantity. But you could find it earlier on our M that we just found to be equal to 4.56 times 10 to the 31 all over a velocity which we know is, uh, still a check at the beginning of the problem for 58.6 times 10 to the tempting cubed meters per second and that'll Cuba's well, and we find that the period of our second um planet is 9.96 times 10 to the seven seconds the problem ask for this quantity in years. So we're gonna divide by the number of seconds in a year which ever done earlier. It's to put 15 times tended to seven, and that turned out to be a period of 3.7 years, which is our final solution for party

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