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Akimi4 [234]
2 years ago
13

Astronomers have observed a small, massive object at the center of our Milky Way Galaxy. A ring of material orbits this massive

object; the ring has a diameter of about 13 light-years and an orbital speed of about 180 km/s .A) Determine the mass of the massive object at the center of the Milky Way galaxy.Take the distance of one light year to be 9.461×1015m. I was able to get this it is 4.26×1037kg.B) Express your answer in solar masses instead of kilograms, where one solar mass is equal to the mass of the sun, which is 1.99×1030. I got this to it is 2.14×107 solar massesC)Observations of stars, as well as theories of the structure of stars, suggest that it is impossible for a single star to have a mass of more than about 50 solar masses. Can this massive object be a single, ordinary star? I got this it is no.
Physics
1 answer:
Mkey [24]2 years ago
3 0
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Answer:

r=1.14m

Explanation:

\theta is the angle between the velocity and the magnetic field. So, the magnetic force on the proton is:

F_m=qvBsen\theta\\F_m=qvBsen(90^\circ)\\F_m=qvB

A charged particle describes a semicircle in a uniform magnetic field. Therefore, applying Newton's second law to uniform circular motion:

F_m=F_c\\qvB=F_c(1)

F_c is the centripetal force and is defined as:

F_c=m\frac{v^2}{r}

Here v is the proton's speed and r is the radius of the circular motion. Replacing this in (1) and solving for r:

qvB=\frac{mv^2}{r}\\r=\frac{mv^2}{qvB}\\r=\frac{mv}{qB}

Recall that 1 J is equal to 6.242*10^{12}MeV, so:

4.9MeV*\frac{1J}{6.242*10^{12}MeV}=7.85*10^{-13}J

We can calculate v from the kinetic energy of the proton:

K=\frac{mv^2}{2}\\\\v=\sqrt{\frac{2K}{m}}\\v=\sqrt{\frac{2(7.85*10^{-13}J)}{1.67*10^{-27}kg}}\\v=3.06*10^{7}\frac{m}{s}

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