What is the Mass of the Andromeda Galaxy?

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The discussion focuses on estimating the mass of the Andromeda Galaxy based on the orbital speed of a star located at its edge. The star's speed is approximately 200 km/s, and its distance from the galaxy's center is given as 5 x 10^9 AU. Participants clarify the formula used for calculating mass, emphasizing the need to equate gravitational and centripetal forces while noting that one mass can be omitted in the calculation. A common mistake identified was the incorrect interpretation of the distance variable, with clarification that "g" should not be included. The final estimated mass of the Andromeda Galaxy is approximately 4.47 x 10^41 kg.
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1. The problem statement, all variables and given/known

A star at the edge of the Andromeda galaxy appears to be orbiting the center of that galaxy at the speed of about 200km/s. The star is about g*10^9AU from the center of the galaxy. Calculate a rough estimate of the mass of the Andromeda galaxy. Earth orbital radius is 1.40*10^8

Homework Equations



Fg=gm1m2/r^2

The Attempt at a Solution



1 a.u. = 1.40*10^8 km = 1.40*10^11 m

For the mass of the entire galaxy, M, you will have to assume a spherical distribution of mass, with the star in question at the outside, at distance R. ( what does that mean by assume spherical)

In that case, the centripetal acceleration of the star is
V^2/R = G M/R2

Solve for M.

M = R V^2/G

I got this solution from a web page however I'm trying to understand how it works.

Isn't fg=Gm1m2/r^2

And then that will be equal to centripetal force
Why in the above solution did they omit the one of the mass??
 
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Coco12 said:
1. The problem statement, all variables and given/known

A star at the edge of the Andromeda galaxy appears to be orbiting the center of that galaxy at the speed of about 200km/s. The star is about g*10^9AU from the center of the galaxy. Calculate a rough estimate of the mass of the Andromeda galaxy. Earth orbital radius is 1.40*10^8

Homework Equations



Fg=gm1m2/r^2

The Attempt at a Solution



1 a.u. = 1.40*10^8 km = 1.40*10^11 m

For the mass of the entire galaxy, M, you will have to assume a spherical distribution of mass, with the star in question at the outside, at distance R. ( what does that mean by assume spherical)

In that case, the centripetal acceleration of the star is
V^2/R = G M/R2


That formula derives from equating centripetal and gravitational forces on m:
GmM/R^2 = mv^2/R.
m is divided out.

Don't write V for velocity. Use v.
V is for Volts.
 
So I did M=2000m/s^2* 7.45*10^20m/ 6.67*10^-11

And I still didn't get the right ans which is 4*10^41kg

Note: the 7.45*10^20 was obtained by taking the 5*10^9 AU multiplying it by 1.49*10^8km and then by 1000 to convert it to m
 
200 km/s is not 2000 m/s!

I don't know what "g x 10^9 AU" is. What is the "g" in that expression for R?
 
rude man said:
200 km/s is not 2000 m/s!

I don't know what "g x 10^9 AU" is. What is the "g" in that expression for R?

I know my mistake now. I converted it wrong . The g is supposed to be a 5
 
This person wrote the question wrong. There is no g, it's "The star is about 5 x 10^9 AU from the centre..." The answer is 4.47 x 10^41 kg :)

- convert the velocity to m/s
-use the formula m=(v^2*r)/G
-multiply the star's radius with the Earth's orbital radius then convert to m

then plug everything in and you're good to goooo!
 
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