Answer:
613373.65233 m/s
Explanation:
M = Mass of Sun = [tex]1.989\times 10^{30}\ kg[/tex]
m = Mass of Earth
v = Velocity of Earth
r = Distance between Earth and Sun = [tex]147.12\times 10^{9}\ m[/tex]
[tex]r_e[/tex] = Radius of Earth = [tex]6.371\times 10^6\ m[/tex]
[tex]r_s[/tex] = Radius of Sun = [tex]695.51\times 10^6\ m[/tex]
In this system it is assumed that the potential and kinetic energies are conserved
[tex]\dfrac{1}{2}Mv_2-\dfrac{GMm}{r_e+r_s}=0-\dfrac{GMm}{r}\\\Rightarrow v=\sqrt{2GM(\dfrac{1}{r_e+r_s}-\dfrac{1}{r})}\\\Rightarrow v=\sqrt{2\times 6.67\times 10^{-11}\times 1.989\times 10^{30}(\dfrac{1}{6.371\times 10^6+695.51\times 10^6}-\dfrac{1}{147.12\times 10^{9}})}\\\Rightarrow v=613373.65233\ m/s[/tex]
The velocity of Earth would be 613373.65233 m/s
