After hammering away at this problem, I arrived at the right answer according to my online homework program but it still isn't very clear.

Question

I have the correct answer for a but I'm still hazy on how I got there. For part b, I'm tol

 

A massive black hole is believed to exist at the center of our galaxy (and most other spiral galaxies). Since the 1990s, astronomers have been tracking the motions of several dozen stars in rapid motion around the center. Their motions give a clue to the size of this black hole.
(a) One of these stars is believed to be in an approximately circular orbit with a radius of about 1.50 ? 10^3 AU and a period of approximately 30 yr. Use these numbers to determine the mass of the black hole around which this star is orbiting.

 

(b) What is the speed of this star?

 

 

How does it compare with the speed of the Earth in its orbit?

 

 

How does it compare with the speed of light?

Answer 1

Just as a disclaimer, I'm answering your question by looking up formulas online with very little astronomical knowledge. I can't check whether the answers I get are right or even reasonable. So...if the answers aren't right, hopefully this will at least give you some ideas on where to go!

 

 

 

For the first part of this question, you probably used Kepler's Third Law. The equation is:

 

P² = (4π²a³) / (G(M₁ + M₂))

 

P is the period of the orbiting object (in seconds), a is the radius of the orbit (in meters), G is a constant, M₁ is the mass of the first object (in kilograms), and M₂ is the mass of the second object (in kilograms). In cases where M₁ is way bigger than M₂ (like when M₁ is the mass of a massive black hole), M₁ + M₂ is approximately equal to M₁. So we'll just call that sum M₁, the mass of the black hole.

 

 

 

We'll first convert years to seconds (depending on how you measure one year, this could vary):

 

P = 30 years × 31556926 seconds / 1 year = 946707780 seconds = 9.4670778×10⁸ seconds

 

 

 

Now, we'll convert Astronomical Units to meters:

 

a = 1.5×10³ AU × 149597870700 meters / 1 AU = 224396806050000 meters = 2.2439680605×10¹⁴ meters

 

 

 

Plug those numbers in, as well as G = 6.673×10^-11:

 

(9.4670778×10⁸)² = (4π²(2.2439680605×10¹⁴)³) / (6.673×10^-11(M₁))

59807137.6M₁ = 4.46076916×10⁴⁴

M₁ = 7.45858996 × 10³⁶ kilograms

 

 

 

For the second part of this question, you can use the formula for speed of an orbiting object:

 

v = √((GM₁)/a)

 

v is the speed (in meters per second), G is our constant from above, M₁ is the mass of the object being orbited (in kilograms), and a is the orbital radius (in meters).

 

Plug those numbers in:

 

v = √((6.673×10^-11)(7.45858996×10³⁶))/(2.2439680605×10¹⁴))

v = √(2.21799818×10¹²)

v = 1489294.52 m/s, or 1489.29452 km/s

 

 

 

The Earth's orbital speed is 30 km/s, so the star is much faster (almost 50 times faster).

 

 

 

The speed of light is 299792.458 km/s, so the star is much slower (about 200 times slower).