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The mass of the Sun is 2 × 1030 kg, the mass of the Earth is 6 × 1024 kg, and their center-to-center distance is 1.5 × 1011 m. Suppose that at some instant the Sun's momentum is zero (it's at rest). Ignoring all effects but that of the Earth, what will the Sun's speed be after 3 days? (Very small changes in the velocity of a star can be detected using the "Doppler" effect, a change in the frequency of the starlight, which has made it possible to identify the presence of planets in orbit around a star.)

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Answer:

0.00461031264 m/s

Explanation:

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

M = Mass of the Earth =  6 × 10²⁴ kg

r = Distance between Earth and Sun = [tex]1.5\times 10^{11}\ m[/tex]

t = Time taken = 3 days

Acceleration is given by

[tex]a=\dfrac{GM}{r^2}\\\Rightarrow a=\dfrac{6.67\times 10^{-11}\times 6\times 10^{24}}{(1.5\times 10^{11})^2}\\\Rightarrow a=1.77867\times 10^{-8}\ m/s^2[/tex]

Velocity of the star

[tex]v=u+at\\\Rightarrow v=0+1.77867\times 10^{-8}\times 3\times 24\times 60\times 60\\\Rightarrow v=0.00461031264\ m/s[/tex]

The Sun's speed will be 0.00461031264 m/s

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