<p>This is something that crops up a lot in Astrophysics and Astronomy, so here is the general approach.</p>

<p>Get your equation into SI units. Thus
$$M = \frac{5v^2R}{G} = 7.5\times10^{10} v^2 R,$$
with $M$ in kg, $v$ in m/s and $R$ in m.</p>

<p>Now if you wish to replace $v^2$ by $(v/1000)^2$ (which is velocity expressed in km/s), you would simultaneously multiply the numerical constant at the front by $1000^2$. Similarly, to replace $R$ by $R/3.09\times 10^{22}$ (which is $R$ expressed in Mpc), then you have to multiply the numerical constant by $3.09\times10^{22}$ to leave everything unchanged. This then leaves your result still expressed in kg. To express it in solar masses, simply divide by a solar mass of $2\times 10^{30}$ kg.</p>

<p>Thus
$$\frac{M}{M_{\odot}} = \frac{7.5\times 10^{10}\times 10^6 \times 3.09\times10^{22}}{2.0\times 10^{30}} \left(\frac{v}{1 {\rm km/s}}\right)^2 \left(\frac{R}{1 {\rm Mpc}}\right)$$
$$\frac{M}{M_{\odot}} =  1.16\times10^{9} \left(\frac{v}{1 {\rm km/s}}\right)^2 \left(\frac{R}{1 {\rm Mpc}}\right)$$</p>

<p>Finally, check your result makes sense. Galaxy clusters are of order a Mpc in size, have velocity dispersions of hundreds of km/s and consists of hundreds of galaxies with total mass of $\sim 10^{14} M_{\odot}$ or so. This seems to check out - usually if a mistake has been made you will be many, many orders of magnitude out.</p>

CC BY-SA 3.0
2017-01-14T22:33:31.820Z
I know the comments aren't here to say thank but I just wanted to say that I really appreciated your great explanation :)
Amaluena
CC BY-SA 3.0
2017-01-14T20:09:13.110Z
2017-01-14T20:09:13.110Z
ProfRob
<p>I have a question that I don't know how to solve, any help is appreciated.</p>

<p>Here it is:</p>

<p><em>Rewrite the equation for the virial mass of a cluster so that if velocities are entered in km/sec and dimensions in Mpc, the resulting mass is in solar mass</em></p>

<p>Well the equation is: $M=\frac{5(v_r)^2R}{G}$ </p>

<p>I think I have to use dimensional analysis but I really don't know how to get there</p>

<p>$$[M]=\mathrm{\frac{(\frac{km}{s})^2 km}{\frac{m^3}{kg \cdot s^2}}}$$</p>

<p>with one $\mathrm{mpc}$ being $3.09 \times 10^{19} \ \mathrm{km}$</p>

<p>Thanks in advance :)</p>

CC BY-SA 3.0
2017-01-14T17:50:44.730Z
I'm voting to close this question as off-topic because it's a homework question.
AtmosphericPrisonEscape
CC BY-SA 3.0
2017-01-14T17:56:59.583Z
Homework questions are not necessarily off topic. The questioner should clarify if it is a homework question. Make clear what they have done so far, and make sure that they ask a specific question about their attempt to solve. "I really don't know how to get there" If the questioner can't do this, this can be closed as "unclear", or "too broad" not "off topic".
James K
CC BY-SA 3.0
2017-01-14T17:57:38.193Z
Even if that would be the case we would be allowed to discuss the homework with our classmates, as well as generally help eachother, why can't I discuss such a question with fellow stack exchange members? As I said any help is appreciated, so not just a solution, but also tips or just how to start
Amaluena
CC BY-SA 3.0
2017-01-14T22:35:33.963Z
[Cross-posted from Physics.](http://physics.stackexchange.com/q/305099/56299) Please don't cross-post.
HDE 226868
CC BY-SA 3.0
2017-01-14T22:39:19.067Z
I'm sorry, but after posting it here I thought it might have a greater audience in physics since it had to do with units and stuff, that's why
Amaluena
<mass>
Rewriting the equation for the virial mass of a cluster
Amaluena
# Rewriting the equation for the virial mass of a cluster

(Asked by: username_0 on 2017-01-14)

I have a question that I don't know how to solve, any help is appreciated.

Here it is:

*Rewrite the equation for the virial mass of a cluster so that if velocities are entered in km/sec and dimensions in Mpc, the resulting mass is in solar mass*

Well the equation is: $M=\frac{5(v\_r)^2R}{G}$

I think I have to use dimensional analysis but I really don't know how to get there

$$[M]=\mathrm{\frac{(\frac{km}{s})^2 km}{\frac{m^3}{kg \cdot s^2}}}$$

with one $\mathrm{mpc}$ being $3.09 \times 10^{19} \ \mathrm{km}$

Thanks in advance :)

**Tags:** mass

> **Comments:**
> **username_1**: I'm voting to close this question as off-topic because it's a homework question.  
> **username_2**: Homework questions are not necessarily off topic. The questioner should clarify if it is a homework question. Make clear what they have done so far, and make sure that they ask a specific question about their attempt to solve. "I really don't know how to get there" If the questioner can't do this, this can be closed as "unclear", or "too broad" not "off topic".  
> **username_0**: Even if that would be the case we would be allowed to discuss the homework with our classmates, as well as generally help eachother, why can't I discuss such a question with fellow stack exchange members? As I said any help is appreciated, so not just a solution, but also tips or just how to start  
> **username_3**: [Cross-posted from Physics.](http://physics.stackexchange.com/q/305099/56299) Please don't cross-post.  
> **username_0**: I'm sorry, but after posting it here I thought it might have a greater audience in physics since it had to do with units and stuff, that's why  

---

## 1 answers

### Answer by username_4 on 2017-01-14. Score: 3 | Accepted answer

This is something that crops up a lot in Astrophysics and Astronomy, so here is the general approach.

Get your equation into SI units. Thus
$$M = \frac{5v^2R}{G} = 7.5\times10^{10} v^2 R,$$
with $M$ in kg, $v$ in m/s and $R$ in m.

Now if you wish to replace $v^2$ by $(v/1000)^2$ (which is velocity expressed in km/s), you would simultaneously multiply the numerical constant at the front by $1000^2$. Similarly, to replace $R$ by $R/3.09\times 10^{22}$ (which is $R$ expressed in Mpc), then you have to multiply the numerical constant by $3.09\times10^{22}$ to leave everything unchanged. This then leaves your result still expressed in kg. To express it in solar masses, simply divide by a solar mass of $2\times 10^{30}$ kg.

Thus
$$\frac{M}{M\_{\odot}} = \frac{7.5\times 10^{10}\times 10^6 \times 3.09\times10^{22}}{2.0\times 10^{30}} \left(\frac{v}{1 {\rm km/s}}\right)^2 \left(\frac{R}{1 {\rm Mpc}}\right)$$
$$\frac{M}{M\_{\odot}} = 1.16\times10^{9} \left(\frac{v}{1 {\rm km/s}}\right)^2 \left(\frac{R}{1 {\rm Mpc}}\right)$$

Finally, check your result makes sense. Galaxy clusters are of order a Mpc in size, have velocity dispersions of hundreds of km/s and consists of hundreds of galaxies with total mass of $\sim 10^{14} M\_{\odot}$ or so. This seems to check out - usually if a mistake has been made you will be many, many orders of magnitude out.

> **Comments:**
> **username_0**: I know the comments aren't here to say thank but I just wanted to say that I really appreciated your great explanation :)  

---

