Dark Jedi

The speed needed to escape earth's gravity well.

Is it a matter of force exerted, speed, or what? Meaning: assuming we can design an engine for such a trip, can't we leave earth's gravity with a thrust of 1.00001G, eventually?

Abacus

Imagine you build a really big cannon and point it straight up. Escape velocity would be the minimum speed that the projectile would need (neglecting air friction) as it leaves the end of the barrel so that it never falls back to the earth. The escape velocity from near the earth's surface is about 25000 mph.

elwoodblues

Actually, I think it's the speed needed to put yourself in orbit around the earth. Somewhat different.

Here is the definition:

So it's a matter of velocity, or speed, which is a function of force exerted, mass, and wind resistance.

Abacus

Be sure not to confuse an elliptical orbit with a parabolic or hyperbolic orbit. An elliptical orbit would be where the satellite continues to go around the earth. Only about 17000 mph is required to put a satellite in low-earth orbit.

Any object with a speed greater than or equal to the escape velocity (about 25000 mph near the earth) will follow a parabolic or hyperbolic orbit. That is, they will escape from the earth's gravity completely.

Shadowy Man

"Escape velocity" is very easily defined. It's basically the velocity needed to enter a parabolic orbit (i.e. have zero velocity at infinite distance from the Earth). It is easy to calculate, too. Just take the gravitational potential energy at the earth's surface and transfer it all into kinetic energy.

E_grav = GmM/R
KE = 1/2 m v^2

v_escape = sqrt (2GM/R)

where M is the mass of the Earth, R is the radius of the Earth and G is the universal gravitational constant.

Easy Be

Dark Jedi,
I think escape velocity is the initial velocity on earth's surface needed to send an object into hyperbolic orbit. However this is assuming no additional force will be exerted. If you were to create a rocket with a continuous thrust of 1.00001G you would be able to escape but, since G will continually decrease to 0 your velocity will approach 0 at an increasing rate. Assuming no foreign gravitational forces if the thruster ever quit you would be sucked back to earth at exactly G.

I'm not sure, but I think even if the rocket traveled at #G where #<(EV/G) if the thruster ever quit you would still be sucked back to earth.

Earth's gravity well extends forever it will just get increasingly easy to escape it the further away from earth you are. An object with zero velocity at any real distance conceptually could fall to earth. Once again assuming no other gravitational force besides the earth.

Dark Jedi

Thanks, folks. That answers my question.

The main failing of my question was one of reference. If a ship was to thrust at 1.0001g in reference to G at sea level, then it would accelerate away from the Earth at an increasing rate.

At the point that it could escape Earth, would that translate to a given velocity? Would that velocity be immutable, meaning no matter what your relative thrust, you won't escape until you acheive that relative velocity?

I am just trying to understang a concept of a slow, easy rise off earth and in to space.
Considerable amounts of energy would be expended, but is it possible to escape Earth at 200mph with an engine that can maintain this relative speed no matter what your attitude in relation to Earth's gravitational pull?

Shadowy Man

You don't want to confuse acceleration with velocity here. If you shot a rocket straight up with an initial velocity of 200 mph, it would not escape the Earth. It would reach some height, its velocity would become zero and it would fall back. If you shot a rocket straight up with velocity equal to escape velocity, it's final velocity wouldn't reach zero until it had traveled an infinite distance away from the Earth.

elwoodblues

To clarify what Shadowy Man just said, if you give a rocket an initial 'kick' that got it to 200 mph immediately, and then no other power, it'd do a graceful swan dive right back into the launch pad. But, if you give it an initial 'kick' so that it achieved a velocity of 25,000 mph (or more) it'd clear the earth and not return, not even to orbit. The gravity would slow it down on the way out, continuously, but less and less as it gets farther and farther away. So, if it's 'kicked' off the earth going, at first, 25,500 mph, it's final velocity should approach 500 mph.

These rockets aren't exerting any thrust after they've been kicked off the surface. They're very odd rockets. Think of them more as packages thrown by a slingshot.

Easy Be

Dark Jedi,
This has to do with acceleration and differential equations.
I'm fairly sure that acceleration and distance won't coincide with the same escape velocity. The equations with acceleration are trivial.

Just solve the equation using EV = 200 and you'll find the distance the rocket needs to travel before it can escape.