What is the Mass of the Andromeda Galaxy?

In summary, 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. Assuming a spherical distribution of mass with the star at the outside, at distance R, the centripetal acceleration of the star is V^2/R = G M/R2. M = R V^2/G.
  • #1
Coco12
272
0
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??
 
Physics news on Phys.org
  • #2
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.
 
  • #3
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
 
  • #4
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?
 
  • #5
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
 
  • #6
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!
 

1. What are Kepler's laws of planetary motion?

Kepler's laws of planetary motion are a set of three laws formulated by German astronomer Johannes Kepler in the early 17th century. These laws describe the motion of planets around the sun, and are considered to be some of the most important laws in the field of astronomy.

2. What is Kepler's first law?

Kepler's first law, also known as the law of elliptical orbits, states that all planets move in elliptical orbits around the sun, with the sun at one of the two foci of the ellipse. This means that the distance between a planet and the sun varies throughout its orbit.

3. What is Kepler's second law?

Kepler's second law, also known as the law of equal areas, states that a line connecting a planet to the sun will sweep out equal areas in equal time intervals. This means that a planet will move faster when it is closer to the sun and slower when it is farther away.

4. What is Kepler's third law?

Kepler's third law, also known as the law of harmonies, states that the square of a planet's orbital period is proportional to the cube of its semi-major axis. This means that the farther a planet is from the sun, the longer it takes to complete one orbit around the sun.

5. What is the significance of Kepler's laws?

Kepler's laws were groundbreaking in their time as they provided a mathematical explanation for the motion of planets, which was previously thought to be unpredictable. These laws also paved the way for further advancements in the field of astronomy and helped to shape our understanding of the solar system and the laws of physics.

Similar threads

  • Introductory Physics Homework Help
Replies
6
Views
986
  • Introductory Physics Homework Help
Replies
5
Views
909
  • Introductory Physics Homework Help
Replies
26
Views
835
Replies
6
Views
934
  • Introductory Physics Homework Help
Replies
7
Views
2K
  • Introductory Physics Homework Help
Replies
4
Views
554
  • Introductory Physics Homework Help
Replies
7
Views
1K
  • Introductory Physics Homework Help
Replies
8
Views
934
  • Introductory Physics Homework Help
Replies
17
Views
942
  • Introductory Physics Homework Help
Replies
16
Views
1K
Back
Top