(Please use the example problem solving solution as reference, it will be below the problem titled 1.6, no other references needed besides the internet)

Hello! Can you provide response to ONLY part b) and c) of the problem by using physics concepts or formulas? Please also provide all math/formula/conceptual steps so that I can check my own work. This question involves Physics Conversions, Center of Mass, Conservation of Energy, Potential Energy, Kinetic energy, Work-energy Theorem, etc.

Practice Problem: Searching for exoplanets It might look like the Sun stays still as the planets orbit it-even neglecting the Sun's motion around the Milky Way Galaxy-but it actually wobbles. (a) Following the Example Problem 1.6 (attached below), find the center of mass (we call this barycenter in astronomy) of the Sun-Jupiter system, neglecting all other objects in the Solar System, which is a reasonable approximation. Is the barycenter within the Sun? (b) From the vantage point of Jupiter, the Sun looks to be stationary as we orbit it. From the reference frame of the center of mass (the barycenter), the Sun and Jupiter orbit around the barycenter. In the center-of-mass reference frame, find the "orbital" velocity of the Sun. (c) When we view stars from far away, we usually can't see the planets, but we can see the star wobble around the barycenter. The most sophisticated telescopes and spectrographs have detected this wobbling of stars (called "radial velocities" ) as low as about 70 cm/s. Think about that. We can detect a star 70 light-years away that is wobbling at an ant's speed! In this case, the period is about 7 days, so we can watch it very closely. Would our instruments be able to detect the wobble of the Sun induced by Jupiter if we viewed it from outside the Solar System? Optional: could we detect the wobble of the Sun by the Earth?Example Problem 1.6 (for reference): Center of Mass of the Earth-Moon System Using data from text appendix, determine how far the center of mass of the Earth-moon system is from the center of Earth. Compare this distance to the radius of Earth, and comment on the result. Ignore the other objects in the solar system. Strategy We get the masses and separation distance of the Earth and moon, impose a coordinate system, and use Equation 9.29 with just / = 2 objects. We use a subscript "e" to refer to Earth, and subscript "m" to refer to the moon. Solution Define the origin of the coordinate system as the center of Earth. Then, with just two objects, Equation 9.29 becomes R = " mere + mmm me + mm From Appendix D, me = 5.97 x 10"* kg mm = 7.36 x 10"2 kg I'm = 3.82 x 10% m. We defined the center of Earth as the origin, so re = 0 m. Inserting these into the equation for R gives R = (5.97 x 1024 kg)(0 m)+(7.36 x 1022 kg) (3.82 x 108 m) 5.97 x 1024 kg+7.36 x 1022 kg = 4.64 x 100 m. Significance The radius of Earth is 6.37 x 10 m, so the center of mass of the Earth-moon system is (6.37 - 4.64) x 105 m = 1.73 x 100 m = 1730 km (roughly 1080 miles) below the surface of Earth. The location of the center of mass is shown (not to scale). RCM