Physics, 05.05.2020 07:55, levilugar

Astronomical observations of our Milky Way galaxy indicate that it has a mass of about 8.0×1011 solar masses. A star orbiting on the galaxy's periphery is about 6.0×104 light years from its center. (a) What should the orbital period of that star be? (b) If its period is 6.0×107 y instead, what is the mass of the galaxy? Such calculations are used to imply the existence of "dark matter" in the universe and have indicated, for example, the existence of very massive black holes at the centers of some galaxies.

Answers: 2

Show answers

Answers

answered: 182075

A) The Orbital period of the star in milky way galaxy is : 2.88 * 10⁸ y

B) The mass of the galaxy when the period is 6.0 * 10⁷ y is = 1.53 * 10¹³ solar mass

Given data :

Mass of milky way galaxy = 8.0 * 10¹¹ solar masses

Radius of star orbiting on the galaxy periphery = 6.0 * 10⁴ light years

A) Determine the orbital period ( T ) of the star

The orbital period of a star in a milky way galaxy is calculated using the relation below

T² = $(\frac{4\pi ^{2} }{GM} )R^{3}$ ---- ( 1 )

First step : convert the mass and radius to Kg and M

Where : Mass of milky way galaxy ( M ) = 8.0*10¹¹ solar masses = 1.59 * 10⁴² kg, Radius ( R ) = 6.0 * 10⁴ light years = 6.07 * 10²⁰ m, G = 6.673 * 10⁻¹¹ N.m²/kg²

Insert the values into equation ( 1 )

∴ T = 2.88 * 10⁸ years .

B) Determine the mass of the galaxy

From equation ( 1 )

M = $\frac{4\pi ^{2} }{G} * \frac{R^{3} }{T^{2} }$ ---- ( 2 )

where : G = 6.673 * 10⁻¹¹ N.m²/kg², R = 5.7 * 10²⁰ m, T = 1.896 * 10¹⁵ s ( 6.0 * 10⁷ y )

Insert values the values into equation ( 2 )

M = 1.53 * 10¹³ solar mass

Hence we can conclude that the The Orbital period of the star in milky way galaxy is : 2.88 * 10⁸ y, The mass of the galaxy when the period is 6.0 * 10⁷ y is = 1.53 * 10¹³ solar mass.

learn more about milky way galaxy : link

answered: claudia3776

a) $T = 26.147\times 10^{7}\,y$ , b) $1.520\times 10^{13}$ solar masses

Explanation:

Let suppose that galaxy can be treated as a puntual mass.

a) The acceleration experimented by the star is:

$a_{r} = G\cdot \frac{M}{r^{2}}$

$a_{r} = \left(6.674\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}} \right)\cdot (8.0\times 10^{11})\cdot\frac{\left(1.989\times 10^{30}\,kg\right)}{\left[(6.0\times 10^{4}\,ly)\cdot \left(9.461\times 10^{15}\,\frac{m}{ly} \right)\right]^{2}}$

$a_{r} = 3.296\times 10^{-10}\,\frac{m}{s^{2}}$

The angular speed of the star is:

$\omega = \sqrt{\frac{a_{r}}{r} }$

$\omega = \sqrt{\frac{3.296\times 10^{-10}\,\frac{m}{s^{2}} }{(6.0\times 10^{4})\cdot (9.461\times 10^{15}\,\frac{m}{ly} )} }$

$\omega = 7.62\times 10^{-16}\,\frac{rad}{s}$

The period is:

$T = \frac{2\pi}{\omega}$

$T = \frac{2\pi}{7.62\times 10^{-16}\,\frac{rad}{s} }$

$T = 8.246\times 10^{15}\,s$

$T = 26.147\times 10^{7}\,y$

b) The period is:

$T = 6\times 10^{7}\,y$

$T = 1.892\times 10^{15}\,s$

The angular speed is:

$\omega = \frac{2\pi}{1.892\times 10^{15}\,s}$

$\omega = 3.321\times 10^{-15}\,\frac{rad}{s}$

The acceleration experimented by the star is:

$a_{r} = \left(3.321\times 10^{-15}\,\frac{rad}{s} \right)^{2}\cdot (6\times 10^{4}\,ly)\cdot \left(9.461\times 10^{15}\,\frac{m}{ly} \right)$

$a_{r} = 6.261\times 10^{-9}\,\frac{m}{s^{2}}$

The mass of the galaxy is:

$M = \frac{a_{r}\cdot r^{2}}{G}$

$M = \frac{\left(6.261\times 10^{-9}\,\frac{m}{s^{2}} \right)\cdot \left[(6.0\times 10^{4}\ly)\cdot (9.461\times 10^{15}\,\frac{m}{ly} )\right]^{2}}{6.674\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}} }$

$M = 3.023\times 10^{43}\,kg$

Which is equal to $1.520\times 10^{13}$ solar masses.

answered: Guest

when two balls of the different masses are dropped from equal height, they reach the ground at the same time .this is because they experience the same acceleration.

the acceleration is independent of mass.

thus when they are dropped they will always maintain the same velocity and travel the same distance (neglect air resistance of course) at the same time.

answered: Guest

Drink a pepsi cherry iced fruit shoot apple juice milkshake mocktail

Other questions on the subject: Physics

Physics, 21.06.2019 18:00, lizarroyo

Acommercial passenger jet can travel up to 970 km/hr at an altitude of 30,000 feet. at this speed, how long will it take the jet to travel 1,865 kilometers? a. 2.3 hrs b. 1.9 hrs c. 1.7 hrs d. 1.5 hrs

Answers: 1

continue

Physics, 22.06.2019 11:30, ghernadez

This punnett square shows the cross between two pants. one parent has round seeds (rr). and the other parent has wrinkled seeds (rr) which best describes their offspring as shown in this cross?

Answers: 1

continue

Physics, 22.06.2019 14:00, tonytashaqua

Me pl give an example of a collision in real life. use the law of conservation of energy to describe the transfer of momentum. be sure and discuss the momentum before and after the collision occurs. you will need at least 3 sentences to thoroughly answer this question.

Answers: 3

continue

Physics, 22.06.2019 16:20, jmackenzie7

How does a circuit breaker protect a refrigerator? a. when the current is too high, a metal strip in the fuse melts and opens the circuit. b. when the resistance is too high , a re-settable which opens a circuit c. when the current is too high , a re-settable switch opens the circuit d. when the resistance is too high a metal strip in the fuse melts and opens the circuit

Answers: 2

continue

Do you know the correct answer?

Astronomical observations of our Milky Way galaxy indicate that it has a mass of about 8.0×1011 sola...