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.

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Answer from: 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

Answer from: 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.

Answer from: Quest

he kicked it 10 feet straight up into the air? i think this works.

Answer from: Quest

answer: i think he can see forward ten seconds and move forward ten seconds but giorno can undo his actions with ger.

explanation:

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

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