Question

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

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

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

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.

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

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.

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.