Calculating Third Cosmic Velocity: What Am I Missing?

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The discussion centers on calculating the third cosmic velocity needed for a body on Earth's surface to escape both Solar and Terrestrial gravity. The equation presented yields a value of 13 km/s, while most sources cite 16 km/s, leading to questions about potential errors in the calculation. Clarifications indicate that the distance from the Earth to the Sun (R_s) is correctly used, but the vector nature of the velocities may not have been fully accounted for. The conversation highlights that the angle between the Earth's orbital speed and the escape velocity affects the final result. Ultimately, the discrepancy in values stems from differing interpretations of how to combine these velocities.
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... is the minimal speed (relative to the Earth) which must be imparted on a body resting on the Earth's surface to escape the Solar (and Terrestrial) gravity. So, if one uses the conservation of energy (neglecting the Earth's rotation about its axis), one has

1/2 m (v_3 + v_o)^2 = G m (M_s / R_s + M_e / R_e), where

m - mass of the body
M_s/e, R_s/e - mass of the Sun (Earth) and distance from its center to the body
v_3 - third cosmic velocity
v_o - the speed at which the Earth orbits the Sun
G - gravitational constant

Now, this equation gives v3 = 13 km/s, while all the sources cite v3 = 16 km/s. Am I missing something?

Thanks
 
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Shouldn't your R_s be the distance of the Earth from the Sun?
 
Yes, R_s is the distance form the Sun to the Earth. Anyhow, all other refrences give completely different expression than mine (e.g. Irodov - problem circa 1.230). I think that my energy calculation is ok. Or am I missing something?
 
The sum v_o+v_3 should be a vector sum. Your answer is for the two velocities parallel.
There is a range of answers for v_3, depending on the angle between them.
 
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