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Sorry if this question is very childish for experts here. This was the question I had in school and years later it has resurfaced. I did some google search but getting the same equations of gravitation.

So I have learned that to escape the gravitational field of Earth and go in outer space the rocket has to be accelerated greater than the gravitational acceleration of Earth i.e. 9.8 m/s2.

I am not able to understand as much as I think if I just send a rocket in an upward direction and keep it propelling with a constant velocity even as low as 1 m/s will it not escape gravity and go in outer space. Why does the acceleration constraint there?

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    $\begingroup$ And how do you keep that constant speed? $\endgroup$ – FGSUZ Jan 1 at 14:37
  • $\begingroup$ And how do you get to that 1 m/s initially? $\endgroup$ – Bill N Jan 1 at 14:41
  • $\begingroup$ @BillN FGSUZ assuming somehow with some technology that is achieved...let us accelerate it initially to reach 1m/s and keep it constant with some un-invented technology... won't it escape? I think it will escape. $\endgroup$ – user218564 Jan 1 at 14:51
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You CAN go at a constant speed.

In fact, if you apply a force upwards that would make a rocket accelerate at $9.8 m/s^2$ in empty space, when not under the influence of gravity, it WOULD go upwards at a constant speed!

The point is that since the earth's gravity pulls back on objects with that acceleration of $9.8 m/s^2$ (close to the earth's surface at least) you must give that acceleration upwards in order to counteract the earth's gravitational pull and move upwards at a constant velocity, and must give more than that acceleration to actually accelerate a rocket upwards.

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  • $\begingroup$ OK...if I understand you....you mean to say that the constraint on acceleration is actually a constraint on Force...meaning the greater acceleration means a greater force vector in an upward direction? So if I have a force in upward direction counteracting gravity, even if the velocity upwards is constant, it will go in space? $\endgroup$ – user218564 Jan 1 at 14:56
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    $\begingroup$ If you have a constant force upwards, not only will it go into space, but it will eventually accelerate into space as the distance from the earth grows and gravity weakens. So yes, you can go into space at a constant velocity, and a constant force upwards greater than the gravitational force the earth applies on the rocket will definitely get it into space. $\endgroup$ – Joshua Ronis Jan 1 at 14:57
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    $\begingroup$ That's right, Gravity is not constant, it changed with distance (getting weaker). So with enough force to counter 9.8 when you get high enough you will accelerate away from Earth. To be truly constant you need the rocket to wind down. +1 $\endgroup$ – ggcg Jan 1 at 15:00
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I think you are getting confused about the concepts of velocity, acceleration, and force. If you can somehow maintain a 1 m/s velocity component away from the surface of the earth you can get to "outer space." (There are circular components of velocity that real spaceships have to allow for, too. Just trying to address the radial, or to-and-from motion.)

The problem is that there is a gravitational force, $\vec{F_g}$, always pulling your ship back toward earth. To initially get to 1 m/s you must provide a thrust force in the direction opposite $\vec{F_g}$ and greater than it so that the net force is away from earth. Once you get to 1 m/s (or any other speed you want to consider), you still must provide an upward force or gravity will pull the ship back unless your speed is what is called escape velocity. exactly how to calculate that is beyond the scope of your question, but for earth starting near the surface it's approximately 11 km/s. If you turn your engines off before reaching that speed, you'll not escape earth, but you might orbit.

If you keep providing thrust and maintain an outward velocity of any value, you can escape. The problem with maintaining thrust is carrying enough fuel to do that. Fuel has mass, and you have to burn fuel to carry fuel with you for burning later. Eventually the fuel is gone and the thrust is gone.

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  • $\begingroup$ Sorry to bother you but, I just checked the wiki page for escape velocity and saw the derivation of the equation which I understood but I am not able to visualize this in terms of velocity...that question remains in my head...with the above answer by Jousha I am able to visualize that a force higher than Fg will be needed...I mean an 'escape force' leading later to constant velocity makes sense to me...but an 'escape velocity' is not making sense...why can't we just get to the point of 1m/s and no further and keep moving that ways? $\endgroup$ – user218564 Jan 1 at 15:15
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    $\begingroup$ @XIkkyuX If you throw a ball upwards with a velocity of 1m/s, what happens? The downward acceleration of gravity stops it and then makes it fall down again. If you throw it faster, it will go higher before it starts to fall. If you throw it at the escape velocity (or even faster than that) it will never stop and begin to fall back. But if you continuously apply a force to the ball, you can make it continue upwards at a constant 1m/s for as long as you keep applying the force. $\endgroup$ – alephzero Jan 1 at 15:40
  • $\begingroup$ @alephzero OK you mean to say that the escape velocity means that the initial velocity that an object should be thrown so that it escapes the gravitational field, and that is the practical implementation we need to send the rocket outside, but in a theoretical scenario 1m/s velocity upwards will do? $\endgroup$ – user218564 Jan 1 at 15:45
  • $\begingroup$ @XIkkyuX to respond to your comment to allephzero: If an upward velocity is maintained, that will "send the rocket outside." But that requires something to provide an upward force. $\endgroup$ – Bill N Jan 1 at 15:56
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    $\begingroup$ @XIkkyuX Just to emphasize, the escape velocity is the velocity which is needed if there is no more outward thrust. If you can maintain a net zero or net outward force (gravity versus thrust), the speed value doesn't matter. Gravity is always present until infinity so you always need some thrust if you're below escape velocity. $\endgroup$ – Bill N Jan 1 at 16:06

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