(II) The masses of the Earth and Moon are 5.98 x 10²⁴

Question

(II) The masses of the Earth and Moon are 5.98 x 10²⁴ kg and 7.35 x 10²² kg, respectively, and their centers are separated by 3.84 x 10⁸ m. What can you say about the motion of the Earth–Moon system about the Sun, and of the Earth and Moon separately about the Sun?

Accepted Answer

Hello, fellow physicists today, we're gonna solve the following practice problem together. So first off, let us read the problem and highlight all the key pieces of information that we need to use in order to solve this problem. What can you say about the orbital dynamics of the planet delta and its moon epsilon around a nearby star zeta? And how do delta and epsilon orbit separately around zeta? If the system consists of a planet and a moon separated by 4.12 multiplied by 10 to the power of 8 m and having masses of 6.88 multiplied by 10 to the power 24 kg and 8.25 multiplied by 10 to the power of 22 kg. Respectfully given that RD equals 6.42 multiplied by 10 to the power of 6 m. So that's our end goal. Our end goal is we ultimately trying to figure out what we can say about the orbital dynamics of this planet delta and its moon epsilon around a nearby star called zeta. And that's our final answer we're ultimately trying to solve for. We're also given some multiple choice answers let's read them off to see what our final answer might be. A is planet Delta exhibits a wobble due to the center of mass being located inside its volume. And moon epsilon performs a combination of smaller rapid orbits around planet delta and slower larger orbits around the star. Zeta B is both planet delta and moon epsilon maintain stable separate orbits around star zeta, unaffected by each other's gravitational pull. C is planet delta orbits the star zeta in a stable path. While moon epsilon is captured in a tight orbit around planet delta causing minimal influence on its trajectory around the star. Zeta. And D is the system center mass dictates a joint orbit around the star zeta with moon epsilon experiencing more pronounced oscillations due to its smaller mass and the gravitational effects from planet delta. Awesome. So now our first step is, is we need to, firstly, we need to figure out what the center of mass of this planet moon epsilon system is. So for you to do that, we need to recall and use the center of mass equation for which states that XCM, which is the center of mass, which in this case, we're trying to figure out the center of mass for the planet delta moon epsilon system. And this is equal to MP multiplied by XP plus lowercase M subscript capital, M multiplied by X subscript capital M Awesome. And this is all divided by MP plus M subscript capital. M OK. And let's call this equation one. So here like once again, XCM is the center of mass of the planet delta system. So delta planet delta system relative to the star zeta. And MP is the mass of the planet delta and M subscript capital M is the mass of the moon epsilon. And XP is the center of mass of the planet delta. And X capital M or X subscript capital M is the center of mass of the moon epsilon. So now we need to plug in our known variables into this equation and to solve for the numerical value of X CM. So when we do that, let us recall that MP is equal to 6.8 eight multiplied by 10 to the power of 24 kilograms. And we need to note that once again that MP is the mass of the planet delta. And now we need to recall that XP is equal to zero. And XP once again XP is the center of mass of planet delta. So we need to use zero plus an M subscript capital M is equal to 8.25 multiplied by 10 to the power of 22 kg. And this once again is the mass of the moon epsilon. And then we need to multiply this by our XM value. And XM is the center of mass of the moon epsilon, which is given to us as 4.12 multiplied by 10 to the power of 8 m. And this is all divided by 6.88 multiplied by 10 to the power of 24 kilograms plus 8.25 multiplied by 10 to the power of 22 kilograms. So when we plug this into our calculator, we will find that XCM is equal to 4.8 eight, multiplied by 10 to the power of six meters. Awesome. So the reason why we sol for the center of mass first is because we need to use the center of mass to help us create a statement to figure out what we can say about the orbital dynamics of the planet delta and its moon epsilon as it orbits a nearby star named Zeta. So now we need to note that the center mass for the planet delta moon epsilon system is 4.88 multiplied by 10 to the power of 6 m from the center of planet delta. So now we need to talk about the motion of the planet delta moon epsilon system. So we need to note that the orbit of the delta epsilon system will trace around the star zeta with respect to the center mass location of the delta epsilon system. And we need to note that both delta and epsilon will orbit about this path in approximately circular orbits. We also need to now discuss the motion of delta about the star. So the center of mass of the delta epsilon system is 4.88 multiplied by 10 to the power 6 m from the center of delta. And the radius of delta, which is given to us in the prom itself as 6.42 multiplied by 10 to the power 6 m. Therefore, the center of mass location of the delta epsilon system is located inside the volume of delta. Therefore, delta will be observed to wobble about the center mass of the delta epsilon system. So now we need to discuss the motion of epsilon about the star. So the motion of epsilon around the star zeta will be the sum of two motions. First, the motion of epsilon about the delta epsilon center of mass. And secondly, the motion of the delta epsilon center of mass about zeta. Therefore, to an external observer, epsilon's motion will appear to be a smaller radius, a small radius, higher frequency circular orbit combined with a larger radius and lower frequency circular motion. Therefore, without a doubt, we can say that multiple choice answer letter A has to be our final answer. So a states that planet delta exhibits a wobble due to the center mass being located inside its volume and moon epsilon will perform a combination of smaller rapid orbits around planet delta and slower larger orbits around star zeta. And that's it we've solved for this problem. Thank you so much for watching. Hopefully, that helped and I can't wait to see in the next video. Bye.

(II) The masses of the Earth and Moon are 5.98 x 10²⁴ kg and 7.35 x 10²² kg, respectively, and their centers are separated by 3.84 x 10⁸ m. What can you say about the motion of the Earth–Moon system about the Sun, and of the Earth and Moon separately about the Sun?

Key Concepts

Gravitational Force

Orbital Motion

Center of Mass

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(II) The masses of the Earth and Moon are 5.98 x 1024 kg and 7.35 x 1022 kg, respectively, and their centers are separated by 3.84 x 108 m. What can you say about the motion of the Earth–Moon system about the Sun, and of the Earth and Moon separately about the Sun?

Key Concepts

Gravitational Force

Orbital Motion

Center of Mass

Watch next

(II) The masses of the Earth and Moon are 5.98 x 10²⁴ kg and 7.35 x 10²² kg, respectively, and their centers are separated by 3.84 x 10⁸ m. What can you say about the motion of the Earth–Moon system about the Sun, and of the Earth and Moon separately about the Sun?