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So I need to estimate the mass of the andromeda galaxy using: $$ M = \frac{v^2r}{G}$$ where $v$ is the rotational velocity, $r$ the galaxy's radius and $G$ Netwon's gravitational constant. I'm told the following:

  1. Distance (m) = 2.403×$10^{22}$
  2. Radius (arcmin) = 12250.40
  3. Velocity (km/s) = 300

I went so far as to convert the radius: $$ \frac{12250.4}{60*57.3} = 3.563\ rad$$ then, calculated the radius: $$0.5*3.563*2.403\times10^{22} = 4.2757\times10^{22}$$ Then subbed into original formula to get: $$\frac{ 4.2757\times10^{22} \times 300^2}{G} = 1.7595\times10^{36} kg$$ It should be $2\times10^{42}kg$. where have I gone wrong?

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    $\begingroup$ Andromeda does not have a radius of 204 degrees. $\endgroup$ – Rob Jeffries Feb 10 at 17:07
  • $\begingroup$ What is it? I'm not sure how to calculate it $\endgroup$ – H98 Feb 10 at 17:08
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You have gone wrong because the radius of Andromeda is not 12250.40 arcminutes. It is difficult to be more helpful. Possibly the units are arcseconds?

Neither do I see why you have inserted a factor of 0.5 in your second equation. or why you have inserted $300^2$ into the third equation, when $v$ is given in km/s, so it should be $(3\times 10^5)^2$.

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  • $\begingroup$ I think OP inserted 0.5 in the second calculation, because it's for radius. Not necessary, of course. He's just misunderstood radius. $\endgroup$ – Jim421616 Feb 10 at 19:41
  • $\begingroup$ @Jim421616 He?? $\endgroup$ – Rob Jeffries Feb 10 at 19:49
  • $\begingroup$ He/she/it. There's no information in his/her/its profile. $\endgroup$ – Jim421616 Feb 10 at 19:58
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H98. There are a couple of issues we need to work through.

First, using your figures, I get a value of $5.78\times10^{37}$ kg, which is different to yours. I'm including here, the fact that the numbers are wrong. What value are you using for $G$?

Second,

a) I don't know where your information is coming from, but according to enter link description here, M31 has an angular radius of 178 arcminutes, not 12250.4 arcminutes.

b) One arcminute (written as $1'$) is $\frac{1}{60}$ of a degree, so $178'=2.97^o$.

c) To convert $2.97^o$ to metres, you'd use the formula $S=D\theta$

d) As @RobJeffries stated, the velocity is $300$ km/s, which is $3\times10^5\rm{m/s}$.

Have another go and let us know how you get on.

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