Astronomy Problem Help?

May 2021 4 81 Report

Astronomy Problem Help?

Suppose you observe a star orbiting the galactic center at a speed of 1400 km/s in a circular orbit with a radius of 25 light-days.

What would your estimate be for the mass of the object that the star is orbiting? Need to find it in M_Sun

If anyone can help me that would be great. I have no idea how to solve this, so any help would be very helpful. Thanks.

Update:

Ok I got 9.2x10^13 as my answer, but it's wrong. I plugged it into that formula and got that answer. Am I doing something else wrong? Or maybe I'm plugging it in wrong in the calculator. Can you or anyone else confirm it? Thanks.

Science & Mathematics / Astronomy & Space

Big Daddy

Verified answer

http://spiff.rit.edu/classes/phys301/lectures/blac...

M = v^2 R / G

M = (1400 km/s)^2 * 25 light-days / G

1 lightday = 2.59*10^13 m

1 solar mass = 1.989*10^30kg

Plug and chug:

You have: (1400km/s)^2 * 25 light-days / G

You want: sunmass

* 9562089

/ 1.0457966e-07

So nearly 10^7 solar masses.

Kurt

Each star in the disk is on a very nearly circular orbit, anchored by all the mass enclosed within its orbit, whether it's luminous or not. Thus, the amount of mass within a star's orbit can be determined from Kepler's Third Law:

M + M* = a^3 / P^2,

where M = mass inside star's orbit (in solar masses)

M* = mass of the star (in solar masses)

a = radius of the star's orbit (in AU)

P = orbital period of star (in years)

A few clarifying words:

In the above equation, M is the total mass in a sphere of radius a, centered on the galactic center. (The mass outside the sphere doesn't have any net effect on the star's orbit).

Since the mass M includes the mass of the supermassive black hole at the galactic center, M is guaranteed to be much much greater than M*, the mass of a single star.

For the Sun's orbit:

a = 8000 parsecs = 1.65 billion A.U.

P = 220 million years

THEREFORE (get out your calculators if you want to check these numbers), the mass inside the Sun's orbit is M = a^3 / P^2 = 90 billion Msun

So,

convert your light-days into AU and your speed into an orbital period, and plug the numbers into the equation above.