anonymous
  • anonymous
The mass of the earth is 5.98 * 10^24 kg and the mass of the moon is 7.36 * 10^22 kg. The center to center distance of the earth and moon is about 3.84 * 10^8 m on average. The earth has a radius of 6.37 * 10^6 m. How far from the surface of the earth is the center of mass of the earth -moon system? Please explain.
Physics
schrodinger
  • schrodinger
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anonymous
  • anonymous
We find the center of mass of a system in the same way one might balance a pencil on their finger (this point, in-fact, is the center of mass of the pencil). Let's realize that can be represented mathematically as a summation of torques. If we sum the torques about the center of mass, their sum will be zero. The expression for center of mass is represented differently, but I'm going to walk you through how it derived. |dw:1334190351328:dw|Imagine the system shown is sitting on your desk and the rod between the two masses (earth and moon) is rigid. The force (F) is the normal force exerted on the rod by our pivot point (our finger in the case of the pencil). Realize that \[F = (m_m + m_e) \cdot g\]This force is placed some arbitrary distance (x) from earth. If we sum the torques about this point, we get the following\[\sum \tau = 0 \rightarrow -m_e \cdot g \cdot x + m_m \cdot g \cdot (R - x) = 0\]where R is the distance between the center of the earth and the center of the moon. Let's solve this equation for x, which represents the location of the center of mass. \[x(-m_e \cdot g - m_m \cdot g) + m_m \cdot g \cdot R = 0\]\[x = {m_m \cdot g \cdot R \over m_e \cdot g + m_m \cdot g}\]g will cancel out\[x = {m_m \cdot R \over m_e + m_m }\] Compare this result to the given definition of center of mass\[R = {1 \over M} \sum r_i m_i\]where M is the total mass of the system, r_i is the location of mass m_i. In this particular example, we need to subtract the radius of the earth from x, to get the location of the center of mass from the surface of the earth.

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