Astronomers have observed an otherwise normal star, called S2, closely orbiting an extremely massive but small object at the center of the Milky Way Galaxy called SgrA. S2 moves in an elliptical orbit around SgrA with a period of 15.2 yr and an eccentricity e = 0.87. In 2002, S2 reached its closest approach to SgrA, a distance of only 123 AU (1AU = 1.50×10^11m is the mean Earth - Sun distance).Determine the mass MM of SgrA, the massive compact object (believed to be a supermassive black hole) at the center of our Galaxy. State MM in kgkg and in terms of the mass of our Sun.

Express your answer using two significant figures. MSgrA/MSun =?

We can determine the mass \( M \) of SgrA (Sagittarius A\*) using Kepler's Third Law and Newton's Law of Gravitation. The relationship between the orbital period and the mass of the central object for an elliptical orbit is:

\[
T^2 = \frac{4 \pi^2 a^3}{G M}
\]

where:
- \( T \) is the orbital period,
- \( a \) is the semi-major axis,
- \( G \) is the gravitational constant (\( G = 6.67430 \times 10^{-11} \, \text{m}^3 \, \text{kg}^{-1} \, \text{s}^{-2} \)),
- \( M \) is the mass of the central object.

### Step 1: Find the Semi-major Axis \( a \)
For an elliptical orbit, the closest approach (periapsis) is related to the semi-major axis \( a \) and the eccentricity \( e \) by:

\[
r_{\text{periapsis}} = a(1 - e)
\]

Given:
- \( r_{\text{periapsis}} = 123 \, \text{AU} = 123 \times 1.50 \times 10^{11} \, \text{m} \),
- \( e = 0.87 \).

We solve for \( a \):

\[
a = \frac{r_{\text{periapsis}}}{1 - e}
\]

### Step 2: Use Kepler's Law to find the Mass \( M \)
Now that we have \( a \), we can use Kepler’s Third Law to solve for \( M \). The period \( T \) is given as 15.2 years, which we convert into seconds:

\[
T = 15.2 \, \text{years} = 15.2 \times 365.25 \times 24 \times 60 \times 60 \, \text{s}
\]

Substitute \( T \) and \( a \) into the equation:

\[
M = \frac{4 \pi^2 a^3}{G T^2}
\]

Finally, we will express \( M \) in terms of solar masses by dividing by the mass of the Sun:

\[
\frac{M}{M_{\odot}} = \frac{M}{1.989 \times 10^{30} \, \text{kg}}
\]

Let me calculate this for you.The mass of Sagittarius A* (SgrA) is approximately:

\[
M = 7.35 \times 10^{36} \, \text{kg}
\]

In terms of solar masses, this is:

\[
\frac{M}{M_{\odot}} \approx 3.69 \times 10^6 \, M_{\odot}
\]

So, the mass of SgrA is about 3.69 million times the mass of the Sun.

Would you like further details, or do you have any other questions?

Here are 5 related questions you might explore:
1. How does Kepler's Third Law apply to planetary orbits?
2. What observational evidence supports the existence of black holes like SgrA?
3. How does the eccentricity of an orbit affect the dynamics of a star?
4. What are the methods used to measure the mass of supermassive black holes?
5. What is the significance of the periapsis in an elliptical orbit?

**Tip:** Gravitational interactions in highly eccentric orbits can reveal much about the mass distribution of a system.

Astronomers have observed an otherwise normal star, called S2, closely orbiting an extremely massive but small object at the center of the Milky Way Galaxy called SgrA. S2 moves in an elliptical orbit around SgrA with a period of 15.2 yr and an eccentricity e = 0.87. In 2002, S2 reached its closest approach to SgrA, a distance of only 123 AU (1AU = 1.50×10^11m is the mean Earth - Sun distance). Determine the mass MM of SgrA, the massive compact object (believed to be a supermassive black hole) at the center of our Galaxy. State MM in kg and in terms of the mass of our Sun. Express your answer using two significant figures. MSgrA/MSun =?

Learn how to determine the mass of Sagittarius A*, the supermassive black hole at the center of the Milky Way, by applying Kepler's Third Law and Newton's Law of Gravitation. This problem involves elliptical orbit analysis, gravitational mechanics, and astronomical units. Suitable for undergraduate physics students.

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