Physics, 05.07.2019 00:30 AgarioEdit

The large magellanic cloud is a small galaxy that orbits the milky way. it is currently orbiting the milky way at a distance of roughly 16 light-years from the galactic center at a velocity of about 300 km/s.
1) use these values in the orbital velocity law to get an estimate of the milky way's mass within 16 light-years from the center. (the value you obtain is a fairly rough estimate because the orbit of the large magellanic cloud is not circular.)
express your answer using one significant figure.

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Answer from: joey333

$2(10)^{42}kg$

Explanation:

Approaching the orbit of the Large Magellanic Cloud around the Milky Way to a circular orbit, we can use the equation of velocity in the case of uniform circular motion:

$V=\sqrt{G\frac{M}{r}}$ (1)

Where:

$V=300km/s=3(10)^{5}m/s$ is the velocity of the Large Magellanic Cloud's orbit, which is assumed as constant.

is the Gravitational Constant and its value is $6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}$

is the mass of the Milky Way

$r=160000ly=1.51376(10)^{21}m$ is the radius of the orbit, which is the distance from the center of the Milky Way to the Large Magellanic Cloud.

Now, if we want to know the estimated mass of the Milky Way, we have to find from (1):

$M=\frac{V^{2} r}{G}$ (2)

Substituting the known values:

$M=\frac{(3(10)^{5}m/s)^{2}(1.51376(10)^{21}m)}{6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}}$ (3)

$M=\frac{1.362384(10)^{32}\frac{m^{3}}{s^{2}}}{6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}}$

Finally:

$M=2.0416(10)^{42}kg\approx 2(10)^{42}kg$ >>>This is the estimated mass of the Milky Way

Answer from: Quest

The answer would be a

Answer from: Quest

m₁v₁ + m₂v₂ = (m₁ + m₂)vmomentum is conserved.the left-hand side of the equation represents the momentum in the system before the collision.the right-hand side represents the momentum in the system after the collisionm₁ and v₁ represents the mass and the velocity of the physics blockm₂ and v₂ represents the mass and velocity of the ball of puttyv represents the final velocity of the combined mass.the question states the two masses stick together. they combined mass has its own velocity with the body mass the sum of the two objects' massestheir combined momentum is (m₁ + m₂)vto find the magitude of (m₁ + m₂)v, we need to add the vectors on the left-hand side of the equation: m₁v₁ + m₂v₂.note the diagram. i have positioned the vectors in accordance with their directions.the magnitude of the resultant can be calculated using the pythagorean theorem since this is a right triangle situation (north and west are perpendicular)|(m₁ + m₂)v| = √[ (m₁v₁)² + (m₂v₂)² ]|(m₁ + m₂)v| = √[ (1.5kg·8.0m/s)² + (0.40kg·20m/s)² ]|(m₁ + m₂)v| = 14.4 kg m/sfor two significant figures (assuming that 20m/s is two), this rounds to 14 kg m/s

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