# Escaping from the solar system

1. Mar 21, 2004

### Ed Quanta

I have to solve a problem regarding the minimum velocity required to escape from the solar system while taking into account the gravitational force of the earth as well as the sun. What I am supposed to do is calculate the velocity of an object after escaping from earth. Then I am supposed to go to a sun fixed frame and calculate the minimum velocity for escape from the sun. The answer that I am striving for is the following:

v^2= v(escape from earth)^2 +v1^2

v1=v(escape from sun) - v(orbit)

I am not sure how these equations above are derived. I can calculate the escape velocity for the sun and earth, but I am not sure how to relate these two separate frame.

2. Mar 27, 2004

### Janus

Staff Emeritus
calculate your escape v from Earth.
Calculate your escape velocity from the sun at Earth's distance.

Now the Earth is alrady moving with respect to the sun and the rocket was lauched from the Earth, so it carries this velocity with it. Thus thus you can subtract the Earth's orbital velocity from the escape v from the Sun, to get the amount of velocity you need to add to your ship in order to leave the solar system after escaping the Earth.

3. Mar 27, 2004

### mathman

Earth's orbital velocity won't help. It is pointing essentially perpendicular to the sun-earth line. To get out of the solar system, you need a radial (from the sun) high speed.

4. Mar 27, 2004

### Janus

Staff Emeritus
No, Escape velocity is escape velocity, The direction doesn't matter (as long as your trajectory doesn't intersect the body you are trying to escape.) If you attain escape velocity while traveling perpendicular to the radius vector, this means you are at the periapis of a parabolic orbit, one that will take you out of the Solar system.

5. Apr 15, 2004

### hellfire

I will give you a derivation of the formula, but I am not completely sure whether it is correct.

Take v as the required escape velocity, m the mass of the rocket or object which is going to escape, Me the mass of the earth, Ms the mass of the sun, re and rs the radius of earth and the distance from the sun to the earth respectively, v_es escape velocity from sun, v_e escape velocity from earth and v_o orbital velocity of earth.

As already mentioned, escape velocity is escape velocity, and it should not be dependent on trayectories and so on.

Thus, consider an rocket escaping from the earth and going to a point in the earth's orbit, which is far from the influence of the earth's gravitational field. Consider also velocities only relative to earth. In such a situation the required velocity to escape the sun is (v_es - v_o).

Aplying energy conservation to this situation in relation to the launch of the rocket:

½ m v^2 - (G Me m) / re - (G Ms m) / rs = ½ m (v_es - v_o)^2 - (G Ms m) / rs

½ m v^2 - (G Me m) / re = ½ m (v_es - v_o)^2

Since (2 G Me) / re = v_e^2, then

v^2 = v_e^2 + (v_es - v_o)^2

Regards.

6. Oct 6, 2010

### lifelearner

I just had to mention that this reply was really helpful in solving one of my problems as well. Thank You!