(a) the sun orbits the milky way galaxy once every 2.60 ⨯ 108 years, with a roughly circular orbit averaging 3.00 ⨯ 104 light years in radius. (a light year is the distance traveled by light in one year.) calculate the centripetal acceleration of the sun in its galactic orbit. m/s2 (b) calculate the average speed of the sun in its galactic orbit. 217476.485 m/s

A) In order to find the centripetal acceleration, we need to apply the formula:

a = ω² · r
where
ω = angular velocity = (2π / T)

Let's transform the units of measurement:
T = 2.6×10⁸yr = 8.2×10¹⁵s
r = 3.0×10⁴ lyr = 2.8×10²⁰m

Therefore:
a = (2π / T)² · r
   = (2π / 8.2×10¹⁵)² · 2.8×10²⁰
   = 1.64 m/s²

B) The average speed can be calculated with the formula:
v = ω · r
   = (2π / T) · r
   = (2π / 8.2×10¹⁵) · 2.8×10²⁰
   = 2.15×10⁵ m/s

johanrusli johanrusli · Answer 2 · 2019-12-18T09:27:34+01:00

The magnitude of the net of force is about 37 Newton

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Further explanation

Centripetal Acceleration can be formulated as follows:

[tex]\large {\boxed {a = \frac{ v^2 } { R } }[/tex]

a = Centripetal Acceleration ( m/s² )

v = Tangential Speed of Particle ( m/s )

R = Radius of Circular Motion ( m )

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Centripetal Force can be formulated as follows:

[tex]\large {\boxed {F = m \frac{ v^2 } { R } }[/tex]

F = Centripetal Force ( m/s² )

m = mass of Particle ( kg )

v = Tangential Speed of Particle ( m/s )

R = Radius of Circular Motion ( m )

Let us now tackle the problem !

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Given:

period of the circular motion = T = 2.60 × 10⁸ = 8.20 × 10¹⁵ seconds

radius of the orbit = R = 3.00 × 10⁴ light years = 2.84 × 10²⁰ m

Unknown:

(a) centripetal acceleration = a = ?

(b) average speed = v = ?

Solution:

Question (a):

[tex]a = \omega^2 R[/tex]

[tex]a = (\frac{2\pi}{T})^2 R[/tex]

[tex]a = (\frac{2\pi}{8.20 \times 10^{15}})^2 \times 2.84 \times 10^{20}[/tex]

[tex]a \approx 1.67 \times 10^{-10} \texttt{ m/s}^2[/tex]

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Question (b):

[tex]v = \omega R[/tex]

[tex]v = (\frac{2\pi}{T}) R[/tex]

[tex]v = (\frac{2\pi}{8.20 \times 10^{15}}) \times 2.84 \times 10^{20}[/tex]

[tex]v \approx 2.18 \times 10^{5} \texttt{ m/s}[/tex]

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Learn more

Impacts of Gravity : https://brainly.com/question/5330244
Effect of Earth’s Gravity on Objects : https://brainly.com/question/8844454
The Acceleration Due To Gravity : https://brainly.com/question/4189441

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Answer details

Grade: High School

Subject: Physics

Chapter: Circular Motion