🎉 Announcing Numerade's $26M Series A, led by IDG Capital!Read how Numerade will revolutionize STEM Learning California State Polytechnic University, Pomona ### Problem 48 Easy Difficulty # Find the position of the center of mass of the system of the sun and Jupiter. (Since Jupiter is more massive than the rest of the planets combined, this is essentially the position of the center of mass of the solar system.) Does the center of mass lie inside or out-side the sun? Use the data in Appendix F. ### Answer ##$7.42 \times 10^{8} \mathrm{m}\$

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Moment, Impulse, and Collisions

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##### Christina K.

Rutgers, The State University of New Jersey

##### Andy C.

University of Michigan - Ann Arbor

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University of Winnipeg

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### Video Transcript

{'transcript': "welcome to a new problem. This time we have the sun, Okay? And then we have Jupiter right there on. We want to find this the center ofthe moss. You know, this is this is what we're given. We know that the mass the moss off the sun, uh, is the same was the marks off. The sign is a simmers one point nine nine times us huge. Ten to the thirty kilograms. That's what we're given. Also, given that the mass off Jupiter happens to be one point nine times ten to the twenty seven kilograms is also Marcy. Okay, so we want to find this, uh uh, sent off muscle this system. You know, this is a system, you know. Want to find the scent of musk of these two thie orbital radius off the sun also happens to have ah come on a distance of zero meters per second, zero meters and then the the orbital radius of Jupiter eyes the same a seven point seven eight times, ten to the eleven meters per second. So that's the information we were given. Ah, simple algebraic problem where we want to find the center ofthe muss. Okay, we want to find the center off mus off this system and so use mu cento muss. He calls too. The moss off the sun times three orbital radius off the sun plus the mass ofthe Jube Jupiter Times the orbital radius of Jupiter all that divided by the must off the sun plus the, um, mass of Jupiter. And then we plug in the numbers, Remember, Since the orbital radius off the sun is zero, this whole thing will just be zero. And so we have not one point nine times ten to the twenty seven kilograms on, then seven point seven eight times ten to eleven meters all over one point nine nine time. Stand to the thirty kilograms us the muscle the sun Uh, plus one point nine times ten to the twenty seven kilograms. Okay, that becomes the mass of Jupiter. So when you simplify this problem, you get that the the center ofthe muss for this system. Me who happens to be seven point four two times ten to the eight meters. Hope you enjoy the problem. Feel free to send any questions or comments and looking forward to the next video and have a wonderful day. Okay, thanks. Bye."}

California State Polytechnic University, Pomona

#### Topics

Moment, Impulse, and Collisions

##### Christina K.

Rutgers, The State University of New Jersey

##### Andy C.

University of Michigan - Ann Arbor

##### Jared E.

University of Winnipeg

Lectures

Join Bootcamp