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
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.
In that case, the centripetal acceleration of the star is
V^2/R = G M/R2
Solve for M.
M = R V^2/G
You did not type the distance R correctly. Is the "g" supposed to be a 9?
G is Newton's universal constant of gravity, 6.67*10^-11 N*m^2/kg^2
To estimate the mass of the Andromeda galaxy, we can use the concept of orbital velocity, as well as Newton's law of universal gravitation.
The orbital velocity of a star in a galaxy is related to the mass of the galaxy and the distance between the star and the center of the galaxy. We can use the equation:
v = √(GM/r)
Where:
v is the orbital velocity of the star,
G is the gravitational constant (approximately 6.67430 x 10^-11 m^3 kg^-1 s^-2),
M is the mass of the Andromeda galaxy, and
r is the distance of the star from the center of the galaxy.
Converting the distance from AU to meters:
g * 10^9 AU = g * 10^9 * 1.496 x 10^11 meters
Using Earth's orbital radius as a reference point:
r = 1.40 x 10^8 meters
Rearranging the equation to solve for M:
M = v^2 * r / G
Substituting the given values:
M = (200 km/s)^2 * (g * 10^9 * 1.496 x 10^11 meters) / (6.67430 x 10^-11 m^3 kg^-1 s^-2)
Now, we simply need to perform the calculations:
M = (200000 m/s)^2 * (g * 10^9 * 1.496 x 10^11 meters) / (6.67430 x 10^-11 m^3 kg^-1 s^-2)
M = (4 x 10^10 m^2/s^2) * (g * 1.496 x 10^20 meters) / (6.67430 x 10^-11 m^3 kg^-1 s^-2)
M = (4 * g * 1.496 x 10^30 kg) / 6.67430
M ≈ (2.986 x 10^30 kg/g)
Therefore, a rough estimate of the mass of the Andromeda galaxy is approximately 2.986 x 10^30 times the mass of Earth (g). Please note that this is a simplified estimation and may not be entirely accurate, as it assumes a constant velocity and neglects other factors.