# Escape Velocity from Solar System

1. Jun 29, 2004

### 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?

2. Jun 29, 2004

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

3. Jun 29, 2004

### e(ho0n3

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.

4. Jun 29, 2004

### 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.