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Escape velocity from Earth's surface is 11.2 Kilometres/second

How long would one need to maintain this escape velocity to actually escape Earth's gravitational pull? Must this 11.2 km/s velocity need to be maintained all the way up to geosynchronous orbit?

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4 Answers 4

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No, the escape velocity doesn't need to be maintained for any length of time. Escape velocity is the minimum speed you need to have at the Earth's surface to be able to escape the gravitational pull, without using a rocket or other continuous propulsion.

In other words, ignoring all sources of gravity other than the Earth, if you launch a projectile straight up at escape velocity from Earth's surface, it will slow down as it rises but will never quite stop. Any slower, and the projectile will stop at some point and fall back to Earth.

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Forgive me for being dense. If the escape velocity is maintained for only one second, surely the body will achieve an altitude of 11.2 kilometres in that period? At the end of that period, surely the gravity and air resistance will take over? –  Everyone Jun 8 '12 at 21:22
Assume there's no air resistance - that's part of the definition of escape velocity. Given that, if escape velocity is maintained for one second, then yes, the body will achieve an altitude of 11.2 km in that time. Then, if its propulsion cuts out, it will be slowed down by gravity, but not enough to ever stop it. –  David Z Jun 8 '12 at 21:26
To be more general, the escape velocity at any point is the instantaneous velocity needed then and there to get away. Note that escape velocity is really escape energy and is thus a scalar: any direction will do as long as it is a non-intersecting route. –  dmckee Jun 8 '12 at 23:39
This answer is interesting. You launch the ball straight up. About the slowing of the ball as it rises, will it be more important here than if I launch the ball horizontally ? –  Nicolas Barbulesco Jun 7 at 9:43
@Everyone — The escape velocity is from the Earth. Yes, after 1 s, the ball will have a speed < 11.2 km/s. But it will be far from the Earth. So it will be less subject to Earth's gravity. This is what I understand. And, as David said, there is no air resistance. We have decided so, because in physics we don't like air resistance. ^^ –  Nicolas Barbulesco Jun 7 at 9:50
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No - it's the energy you need to have to leave Earth completely, essentially how fast you need to fire a bullet for it to leave.

The idea of an escape velocity only comes about because current launch vehicles burn all their fuel very soon after launch and need to achieve a high speed to have enough kinetic energy to escape. If it was possible to build a rocket where you could have the engine running 'all the way up' you could reach space at any slower vertical speed you wanted.

So if you had a ladder you could climb into space. Although if you want to be in orbit at the top you do have to have enough tangential (sideways) velocity to maintain that orbit.

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Related: Launch loop

Assuming you are useing a rocket style launch vehicle, everyone has so far.

To reach orbit, the rocket must impart to the payload a delta-v of about 9.3–10 km/s.

This figure is mainly (~7.8 km/s) for horizontal acceleration needed to reach orbital speed, but allows for atmospheric drag (approximately 300 m/s with the ballistic coefficient of a 20 m long dense fuelled vehicle), gravity losses (depending on burn time and details of the trajectory and launch vehicle), and gaining altitude.

First stage - External Tank

Engines 3 SSMEs located on Orbiter

Thrust 5.45220 MN total, sea level liftoff (1,225,704 lbf)

Specific impulse 455 s

Burn time 480 s

Fuel LOX/LH2

Second stage - Orbiter

Engines 2 OME

Thrust 53.4 kN combined total vacuum thrust (12,000 lbf)

Specific impulse 316 s

Burn time 1,250 s

Fuel MMH / N 2 O 4.

Referances: Launch Vehicle , Orbital Spaceflight , Burn Time

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Velocity equation of a body thrown straight up from Earth surface:

$v^2(h)=(v_o^2 – 2\bf{g}R)+2gR\frac{R}{h}$

$v_o$ - initial velocity, $R$ - Earth radius, $h$ - distance from Earth center, $ \bf{g}$ - gravitational acceleration at $R$

If initial velocity is escape velocity $v_o=v_e=\sqrt{2\bf{g}R}$ then

$v^2(h)= v_e^2\frac{R}{h}$


$v(h)= v_e\sqrt{\frac{R}{h}}$

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