## Acceleration due to gravity in a galaxy?

How strong is acceleration due to gravity ?
For example at the periphery at a galaxy ?
 PhysOrg.com science news on PhysOrg.com >> Ants and carnivorous plants conspire for mutualistic feeding>> Forecast for Titan: Wild weather could be ahead>> Researchers stitch defects into the world's thinnest semiconductor

 Quote by DaveC426913 http://en.wikipedia.org/wiki/Galaxy_rotation_curve
How strong is it approximately?

## Acceleration due to gravity in a galaxy?

To determine the acceleration due to gravity at the periphery of a galaxy, you need to know three numbers. The first is the distance from the galaxy centre where your acceleration is to be calculated (the radius). The second is the total mass of the galaxy contained with that radius. The third is the gravitational constant, G.

Lastly, you will need to learn about Newton's law of gravity (the wikipedia article is suffucient). With the above three numbers, some knowledge of how gravity works and a little bit of calculating, you should be able to work the numbers out yourself.

Recognitions:
Gold Member
 Quote by Bjarne How strong is acceleration due to gravity ? For example at the periphery at a galaxy ?
Put this into google "G*(mass of sun)/(1 AU)^2"
without the quotes
The google calculator will tell you the accel due to sun's mass at the Earth's usual distance of 1 AU.

The distance from center to apparent edge is sometimes estimated
about 15,000 parsecs. The mass of our galaxy within that radius has been put at 700 billion solar masses. Try this in the google search window, without the quotes:

"G*(700*10^9 mass of sun)/(15 kpc)^2"

I think you will get around 0.4 nanometer per second per second.
 Thank's PS.. 1.) I thought the equation was more complicated (Einstein’s Field Equation). But I assume GM/r^2 can be used as a rough estimation (?). 2.) I thought there are “only" 200 stars” in a typical galaxy (not 700 ?) 3.) 15,000 parsecs = 4,62e20 Meter (50.000 LY ) 700e9*2e30*6,67E-11/4,62e20^2= 4,37e-10 M^2