To be more specific on the problem, a 50kg astronaut, an 80kg astronaut and a 20kg rock are tied together by a light rope during a space walk. I am asked to find the center of mass.

Now, I've chosen my reference point as the 50kg astronaut and I've used the formula $(m_1r_1+m_2r_2+m_3r_3)/(m_1+m_2+m_3)$ to find the position of the center of mass ($r$) of the system. I am uncertain if this is the correct method used.

If someone could just inform me if i am on the right track. Thanks.


1 Answer 1


You are on the right track. Mathematically, the center of mass $\vec{c}$ is defined as

$$ \vec{c} \sum_i m_i = \sum_i m_i \vec{r}_i $$

where $m_i$ are the individual masses and $\vec{r}_i$ the individual position vectors.

  • $\begingroup$ Or more clearly $$ \vec{c} = \frac{\sum_i m_i \vec{r}_i}{\sum_i m_i} $$ $\endgroup$
    – Floris
    Mar 11, 2015 at 19:16
  • $\begingroup$ The reason I stated it as I did because it describes the fact that the net mass at $\vec{c}$ equals all the masses at their own $\vec{r}$ $\endgroup$ Mar 11, 2015 at 21:08
  • $\begingroup$ I think that both are good. Your definition leads to my calculation. $\endgroup$
    – Floris
    Mar 11, 2015 at 21:58

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