Community > Forum > Technology & Science > Escape Velocity

Escape Velocity

Posted by: roygrif - Tue Oct 19, 2004 5:27 pm
Post new topic Reply to topic
 [ 19 posts ] 
Escape Velocity 
Author Message
Spaceflight Trainee
Spaceflight Trainee
avatar
Joined: Mon Jan 19, 2004 12:29 pm
Posts: 25
Location: Enschede, The Netherlands
Post    Posted on: Tue Oct 26, 2004 2:26 pm
OK, I'm glad we've got that little issue cleared up. :)


Back to top
Profile
Spaceflight Trainee
Spaceflight Trainee
avatar
Joined: Thu Jul 29, 2004 4:19 pm
Posts: 42
Location: Indianapolis
Post    Posted on: Tue Nov 02, 2004 4:51 pm
Ok, so if I am in a stationery craft, how far from the Earth would I have to be to not be dragged back to the Earth.

_________________
www.theshiversmusic.com


Back to top
Profile
Moon Mission Member
Moon Mission Member
User avatar
Joined: Tue Oct 05, 2004 5:38 pm
Posts: 1361
Location: Austin, Texas
Post    Posted on: Tue Nov 02, 2004 5:08 pm
In theory you need to be infinitely far away to be stationary and never fall back. Basically that is the definition of escape velocity. In real life you only need to be far enough away to fall under the influence of some other planet's gravity. Then you fall there instead of back to Earth. And in real REAL life it is way more complicated, because you are never stationary. All planets orbit the sun, the sun orbits the galaxy, the galaxy moves in space relative to other galaxies, and so on.


Back to top
Profile WWW
Spaceflight Trainee
Spaceflight Trainee
avatar
Joined: Mon Jan 19, 2004 12:29 pm
Posts: 25
Location: Enschede, The Netherlands
Post    Posted on: Tue Nov 02, 2004 9:48 pm
campbelp2002 is right. There is no such distance, but there are other objects you can get influenced by. Let's get "realistic" and see what distance you need to be for an escape velocity of 10^-3 m/s (that's 1 mm/second, not much right?)

M (Mass of earth ) = 5.9736×10^24 kg
G (Gravity constant) = 6.67 × 10^-11 N m^2 kg^-2

Earlier, I demonstrated the relationship between the escape velocity V and the distance to the centre of the earth r (see earlier in this thread). That relation is:

1) V = sqrt( 2 * M * G / r)

We need the oposite relationship. Rewriting gives:

2) V = sqrt( 2 * M * G / r) <=> V² = 2 * M * G / r <=> r = 2 * M * G / V²

Filling in M and G, and using 10^-3 for V, this results in

r = 7.97 *10^20 meter, or roughly 135 million times the (average) distance between the sun and Pluto.

Please note that this does by no means set you free from the gravitational pull of the sun, let alone from the galaxy...
I'll leave the question how fast you'd have to go from earth in the direction of the sun to get free of the influence of earth and be captured by the sun as an exersise to the reader ;-)

BTW: I just discovered http://en.wikipedia.org/wiki/Escape_velocity :-) It seems my mathematical derivation was correct 8)


Back to top
Profile
Display posts from previous:  Sort by  
Post new topic Reply to topic  [ 19 posts ] 
 

Who is online 

Users browsing this forum: No registered users and 10 guests


© 2014 The International Space Fellowship, developed by Gabitasoft Interactive. All Rights Reserved.  Privacy Policy | Terms of Use