- #1
catalyst55
- 24
- 0
Why do people always use the misleading term 'escape velocity'?
If I had a rocket whose engines provided a thrust of 1N more than its weight (thus yielding a miniscule net force up), i would be able to go to Mars with it, given the required amount of time, right? Thrust > Weight, so it should accelerate up...
The space shuttle, for example, never escapes the Earth's gravitational field -- nor can it ever (strictly) because of the horizontal asymptote in g vs. distance graph.
The shuttle merely reaches the required velocity to attain a stable orbit around the Earth. So, when in stable orbit, the shuttle is virtually accelerating towards to Earth at whatever g is at that distance from the centre of the Earth but does not fall down because its velocity is always tangential to its orbit (and perpendicular to its acceleration).
Am i correct? I'm having a little argument with a friend.
Cheers
If I had a rocket whose engines provided a thrust of 1N more than its weight (thus yielding a miniscule net force up), i would be able to go to Mars with it, given the required amount of time, right? Thrust > Weight, so it should accelerate up...
The space shuttle, for example, never escapes the Earth's gravitational field -- nor can it ever (strictly) because of the horizontal asymptote in g vs. distance graph.
The shuttle merely reaches the required velocity to attain a stable orbit around the Earth. So, when in stable orbit, the shuttle is virtually accelerating towards to Earth at whatever g is at that distance from the centre of the Earth but does not fall down because its velocity is always tangential to its orbit (and perpendicular to its acceleration).
Am i correct? I'm having a little argument with a friend.
Cheers