For simplicity, consider the milky way galaxy to be a ball of about 100 billion
ID: 2104586 • Letter: F
Question
For simplicity, consider the milky way galaxy to be a ball of about 100 billion stars, with average masses equal to the sun's. Our solar system resides outside the "ball" and is about 26.6 kilo light years from the center. Our solar system moves around the center of the Milky Way in a roughly circular orbit at a tangential speed of about 220 km/s (about 489,000 MPH). How many times faster than this speed would our solar system need to be moving to reach escape "velocity" and drift out of our galaxy? Given: 1 LY=9.461 X10^15 m, G=6.6743 x 10^-11 N-m^2/kg^2, Mass of sun = 1.989 x 10^30 kg.
Explanation / Answer
since i bagged on the other two who poorely attempted this problem.. ill show ALL how to do it...
V escape=sqrt(2GM/r)...
in this case, our r will be the distance from our solar system to the milkyway galaxy in meters... if its 26.6 kly away, convert that into ly away which is 26,600 ly away... convert that to meters by multiplying by the given conversion and you get 2.516626x10^20...
now our M is not the mass of the sun, but the mass of the sun times 100 billion...
now solve for V escape... it equals sqrt((2 x6.6743x10^-11x1.989x10^30x100 billion)/2.516626x10^20)
put it into your calc and it spits out 324807.3974 m/s.. convert to km/s and you get 324.8 km/s... this is the required orbital speed escape the mily way galaxy... to determine how many times faster, simply take 220 km/s and divide it by 324.8 km/s...
final answer gives 1.476, round it up to 1.48, and it is 1.48... so escape velocity is 1.48 times the oribital speed of our solar system