Escape Velocity from Solar System

[e(ho0n3]

Here is a celestial mechanics problem I can't seem to solve:To escape the solar system, an interstellar spacecraft must overcome the gravitational attraction of both the Earth and Sun. Ignore the effects of the other bodies in the solar system. Show that the escape velocity is

v = \sqrt{v_E^2 + (v_S - v_0)^2}

where v_E is the escape velocity from the Earth; v_S = \sqrt{2GM_S/r_{SE}} is the escape velocity from the gravitational field of the Sun at the orbit of the Earth but far from the Earth's influence (M_S is the mass of the Sun and r_{SE} is the Sun-Earth distance); and v_0 is the Earth's orbital velocity about the Sun.

The thing I don't understand here is how the orbital velocity of the Earth plays a role. Suppose I took the Earth out of the picture and the spacecraft is orbitting the sun with orbital velocity equal to that of Earth's. Isn't the escape velocity from the sun just v_S. What do I need v_0 for?

 

[cookiemonster]

v_0 gives you some extra energy at the beginning of the trip, so naturally it'll take less to get away.

cookiemonster

 

[e(ho0n3]

v_0 gives you some extra energy at the beginning of the trip, so naturally it'll take less to get away.

cookiemonster

If you happen to escape in the same direction as the orbital velocity vector, then I guess it would help.

I think this all depends on where on Earth I am taking of from. Or does it make no difference? I'm still confused.

 

[e(ho0n3]

From the equation for v, it seems that

\vec{v} = (v_E, v_S - v_0) = \vec{v}_E + \vec{v}_S - \vec{v}_0

where \vec{v}_E = (v_E, 0), \vec{v}_S = (0, v_S), and \vec{v}_0 = (0, v_0). This is interesting since the escape velocity vector from the Sun has the same direction as the Earth's orbital vel. vector and both of these vectors are perpendicular to Earth's escape vel. vector. I guess when you make these assumptions, it is easy to show what the escape velocity is. I really don't know how one would come to these sorts of conclusions though.