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Chapter 8: Problem 52

One method of detecting extrasolar planets involves looking for indirect evidence of a planet in the form of wobbling of its star about the star-planet system's center of mass. Assuming that the Solar System consisted mainly of the Sun and Jupiter, how much would the Sun wobble? That is, what back-and- forth distance would it move due to its rotation about the center of mass of the Sun-Jupiter system? How far from the center of the Sun is that center of mass?

Short Answer

Expert verified
Answer: The Sun wobbles approximately 7.784 * 10^11 meters back-and-forth due to Jupiter's presence in the solar system.

Step by step solution

01

Given information

Mass of the Sun (M₁) = 1.989 * 10^30 kg Mass of Jupiter (M₂) = 1.898 * 10^27 kg Distance between Sun and Jupiter (R) = 7.784 * 10^11 meters
02

Center of Mass Formula

The center of mass formula for a two-body system can be written as: Center of Mass (Rcm) = (M₁R₁ + M₂R₂)/(M₁ + M₂) where Rcm is the distance between the center of mass and the reference point, R₁ and R₂ are the distances of the two bodies from the reference point.
03

Reference Point and Distances

In this scenario, the reference point will be the center of the Sun: R₁ = 0 (since the distance of Sun from itself is 0) R₂ = R
04

Calculate the Center of Mass

Now we can plug the values into the equation: Rcm = (M₁R₁ + M₂R₂) / (M₁ + M₂) Rcm = (1.989 * 10^30 * 0 + 1.898 * 10^27 * 7.784 * 10^11) / (1.989 * 10^30 + 1.898 * 10^27) = 741330.64 meters So, the center of mass of the Sun-Jupiter system is 741,330.64 meters away from the center of the Sun.
05

Calculate the Back-and-Forth Distance of the Sun

Since we now have the distance between the Sun and the center of mass of the Sun-Jupiter system, we can calculate how much the Sun wobbles. We can do this by finding the difference between the center of the Sun and the center of mass: Wobble Distance = R - Rcm = 7.784 * 10^11 - 741330.64 = 7.784 * 10^11 meters (approximately) Due to Jupiter's presence, the Sun wobbles approximately 7.784 * 10^11 meters back-and-forth.

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Most popular questions from this chapter

A student with a mass of \(40.0 \mathrm{~kg}\) can throw a \(5.00-\mathrm{kg}\) ball with a relative speed of \(10.0 \mathrm{~m} / \mathrm{s}\). The student is standing at rest on a cart of mass \(10.0 \mathrm{~kg}\) that can move without friction. If the student throws the ball horizontally, what will the velocity of the ball with respect to the ground be?

Find the \(x\) - and \(y\) -coordinates of the center of mass of the flat triangular plate of height \(H=17.3 \mathrm{~cm}\) and base \(B=10.0 \mathrm{~cm}\) shown in the figure.

A baseball player uses a bat with mass \(m_{\text {bat }}\) to hit a ball with mass \(m_{\text {ball }}\). Right before he hits the ball, the bat's initial velocity is \(35.0 \mathrm{~m} / \mathrm{s}\), and the ball's initial velocity is \(-30.0 \mathrm{~m} / \mathrm{s}\) (the positive direction is along the positive \(x\) -axis). The bat and ball undergo a one-dimensional elastic collision. Find the speed of the ball after the collision. Assume that \(m_{\text {bat }}\) is much greater than \(m_{\text {ball }}\), so the center of mass of the two objects is essentially at the bat.

A system consists of two particles. Particle 1 with mass \(2.0 \mathrm{~kg}\) is located at \((2.0 \mathrm{~m}, 6.0 \mathrm{~m})\) and has a velocity of \((4.0 \mathrm{~m} / \mathrm{s}, 2.0 \mathrm{~m} / \mathrm{s}) .\) Particle 2 with mass \(3.0 \mathrm{~kg}\) is located at \((4.0 \mathrm{~m}, 1.0 \mathrm{~m})\) and has a velocity of \((0,4.0 \mathrm{~m} / \mathrm{s})\) a) Determine the position and the velocity of the center of mass of the system. b) Sketch the position and velocity vectors for the individual particles and for the center of mass.

A man standing on frictionless ice throws a boomerang, which returns to him. Choose the correct statement::: a) Since the momentum of the man-boomerang system is conserved, the man will come to rest holding the boomerang at the same location from which he threw it. b) It is impossible for the man to throw a boomerang in this situation. c) It is possible for the man to throw a boomerang, but because he is standing on frictionless ice when he throws it, the boomerang cannot return. d) The total momentum of the man-boomerang system is not conserved, so the man will be sliding backward holding the boomerang after he catches it.
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