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soldier1979 [14.2K]
3 years ago
8

(a) The Sun orbits the Milky Way galaxy once each 2.60 x 108y , with a roughly circular orbit averaging 3.00 x 104 light years i

n radius. (A light year is the distance traveled by light in 1 y.) Calculate the centripetal acceleration of the Sun in its galactic orbit. Does your result support the contention that a nearly inertial frame of reference can be located at the Sun? (b) Calculate the average speed of the Sun in its galactic orbit. Does the answer surprise you?
Physics
1 answer:
PilotLPTM [1.2K]3 years ago
5 0

Answer:

Part a)

a_c = 1.67 \times 10^{-10} m/s^2

Part b)

v = 2.18 \times 10^5 m/s

Explanation:

Time period of sun is given as

T = 2.60 \times 10^8 years

T = 2.60 \times 10^8 (365 \times 24 \times 3600) s

T = 8.2 \times 10^{15} s

Now the radius of the orbit of sun is given as

R = 3.00 \times 10^4 Ly

R = 3.00 \times 10^4 (3\times 10^8)(365 \times 24 \times 3600)m

R = 2.84 \times 10^20 m

Part a)

centripetal acceleration is given as

a_c = \omega^2 R

a_c = \frac{4\pi^2}{T^2} R

a_c = \frac{4\pi^2}{(8.2\times 10^{15})^2}(2.84 \times 10^{20})

a_c = 1.67 \times 10^{-10} m/s^2

Part b)

orbital speed is given as

v = \frac{2\pi R}{T}

v = \frac{2\pi (2.84 \times 10^{20})}{8.2 \times 10^{15}}

v = 2.18 \times 10^5 m/s

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Thickness of the tennis ball, t = 2.0 mm = 2.0\times 10^{- 3}\ m

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We can calculate the tunneling probability as:

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P(t) = e^{-2.63\times 10^{32}}

Taking log on both the sides:

logP(t) = -2.63\times 10^{32} loge

P(t) = 10^{- 1.17\times 10^{32}}

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