When would a rocket return that was at escape velocity

[zanick]

we all know escape velocity is a velocity in which a body can escape orbit around the Earth and fly off into space. this needs to happen in space our just out of our atmosphere, due to drag that could bring an object back into orbit and cause it's velocity to degrade to a point where it would fall back to earth. So, what if a rocket is shot straight up, instead of horizontally at escape velocity (near 22,000mph) and as it enters space, the rockets are cut off and it flies off into space.. … will the pull of the Earth's gravity which would go down with distance, be enough to slow it down and pull it back to the earth? Or would it get far enough away to where gravity of other large bodies would have more influence . I just read that the pull of gravity to the sun, acting on the moon, is GREATER than what the Earth acts on the moon... so maybe if you sent off this rocket , you would want it to be done when the sun is behind the earth... This also means, the LaGrange point, would change with the positions in relation to the sun, so I guess the forces would change too. in a thought experiment of just the Earth and no external forces... a rocket coasting at 22,000miles/hour pointed away from the earth, would eventually slow and return back wouldn't it??

 

[DaveE]

Escape velocity is the speed required so that the rocket will not orbit or return to the earth. So, a rocket beyond escape velocity will travel away from the Earth forever. Of course it is more complicated if you add other planets, etc. I think people can be kind of sloppy with this in the media. For example, theoretically, you can get to the moon (or even Mars) without reaching escape velocity. You just have to get far enough from the Earth that the other objects gravity is stronger than the earth's.

 

[phinds]

I think you've lost track of the DEFINITION of "escape velocity"

EDIT: I see Dave beat me to it

 

[zanick]

Escape velocity is the speed required so that the rocket will not orbit or return to the earth. So, a rocket beyond escape velocity will travel away from the Earth forever. Of course it is more complicated if you add other planets, etc. I think people can be kind of sloppy with this in the media. For example, theoretically, you can get to the moon (or even Mars) without reaching escape velocity. You just have to get far enough from the Earth that the other objects gravity is stronger than the earth's.

I understand, i think... but isn't escape velocity different for satellites at different distances from earth? or does it just put the object in a higher and higher orbit, until you reach the magic "escape velocity " ? Also related to escape velocity, but can't you just take off from the earth, straight up, with enough thrust to keep moving and you will get to the moon . It might take a lot of fuel, but you will get there, right? on that same thread, if you keep on traveling away from the Earth , won't you get to some point where you won't return, based on gravitational pull from another planet or star?
Again, to my original point, isn't "escape velocity" tied to breaking the centripetal force for circular .elliptical orbits, and if you shoot a rocket straight up vs going into orbit, wouldn't that change the escape velocity? I mean, if you had enough fuel , couldn't you inch your way out into space and beyond?

 

[Vanadium 50]

but isn't escape velocity different for satellites at different distances from earth?

Those satellites are in orbit, so they haven't escaped.

 

[zanick]

Those satellites are in orbit, so they haven't escaped.

I get that... but if they were to turn on the thrusters and try and "escape" would the escape velocity be different?

 

[phinds]

I get that... but if they were to turn on the thrusters and try and "escape" would the escape velocity be different?

Yes. The escape velocity is a function of distance from the center of the Earth.

 

[DaveE]

isnt escape velocity different for satellites at different distances from earth?

Yes. Because the force of gravity (or better yet, the potential energy from gravity) is less, the velocity (i.e. kinetic energy) required to escape will be less.

does it just put the object in a higher and higher orbit, until you reach the magic "escape velocity " ?

You only "reach" escape velocity by putting kinetic energy into the rocket (like burning fuel to increase velocity). At any point in space there is an escape velocity from a gravitational source. If the rocket's velocity is below the escape velocity, the rocket will end up in a higher orbit.

cant you just take off from the earth, straight up, with enough thrust to keep moving and you will get to the moon . It might take a lot of fuel, but you will get there, right? on that same thread, if you keep on traveling away from the Earth , won't you get to some point where you won't return, based on gravitational pull from another planet or star?

Yes. The concept of escape velocity really only makes sense for simple gravitational systems. You can have enough velocity to travel to a place where the moon's gravity is stronger than the Earth's gravity. Then your rocket may hit the moon or orbit it However, you will not have escaped the Earth's gravity (after all the moon is orbiting the Earth too).

Again, to my original point, isn't "escape velocity" tied to breaking the centripetal force for circular .elliptical orbits, and if you shoot a rocket straight up vs going into orbit, wouldn't that change the escape velocity?

Yes, velocity is a vector that has direction. Direction matters. For example, velocity towards the Earth isn't as useful for escape as velocity away from the earth.

I mean, if you had enough fuel , couldn't you inch your way out into space and beyond?

Yes

 

[zanick]

Thanks... that makes it a little more clear!

 

[Nugatory]

Yes, velocity is a vector that has direction. Direction matters. For example, velocity towards the Earth isn't as useful for escape as velocity away from the earth.

Clearly there's a problem if your trajectory intersects the surface of the Earth (you crash into the ground instead of escaping) or passes through too much atmosphere (you lose kinetic energy to friction with the air). But in the somewhat idealized case of a projectile being launched from a given height on an airless planet, the escape velocity is the same in all directions - aim straight up or aim to just graze the horizon and you'll escape in some direction.

 

[jbriggs444]

Clearly there's a problem if your trajectory intersects the surface of the Earth (you crash into the ground instead of escaping) or passes through too much atmosphere (you lose kinetic energy to friction with the air). But in the somewhat idealized case of a projectile being launched from a given height on an airless planet, the escape velocity is the same in all directions - aim straight up or aim to just graze the horizon and you'll escape in some direction.

Also works for downward angles if you idealize the planet as a point mass above which you are launching or as a spherically symmetric fluid ball which is magically free from viscosity.

 

[Janus]

I get that... but if they were to turn on the thrusters and try and "escape" would the escape velocity be different?

The escape velocity at a given distance would be equal to the circular orbital velocity at that distance time the square root of 2. So a satellite in a circular orbit would need to increase its velocity by ~41.4% to escape.

To get back to your original question of launching at escape velocity. Assuming an ideal case, as the rocket climbs, the Earth's gravity robs it of speed, but at the same time, the increasing distance it is putting between it and the center of the Earth reduces that pull of gravity. It works out that the speed lost in attaining a given distance is exactly equal to the difference in escape velocities between launch point and that distance. In other words, when the rocket reaches any given distance, it will be traveling at escape velocity for that distance from the Earth.

I just read that the pull of gravity to the sun, acting on the moon, is GREATER than what the Earth acts on the moon... so maybe if you sent off this rocket , you would want it to be done when the sun is behind the earth... This also means, the LaGrange point, would change with the positions in relation to the sun, so I guess the forces would change too.

That bit about the Gravity of the Sun pulling on the Moon more than the Earth's gravity does, can be a bit misleading. The Earth and Moon share almost the exact same trajectory around the Sun, so the Sun accelerates both pretty much equally. There is a small difference as the Moon orbits the Earth, since their relative distances to the Sun change. It is the difference in the acceleration the Sun exerts on the Moon and Earth and how this difference relates to the gravitational pull the Earth exerts on the Moon that allows the Earth to keep the Moon in orbit, and not the direct comparison between the Sun's and Earth's gravitational pull on the Moon.

The same thing applies to a rocket leaving the Earth. Two things are happening: One is that the grip of the Earth is weakening. The other is that as the distance between rocket and Earth grows, The difference between the acceleration due to the Sun's gravity on the two increases. When this difference exceeds the Earth's gravitational pull, the rocket will transition into an independent orbit around the Sun. Thus with the help of the Sun, the rocket doesn't have to attain quite the full escape velocity in order to not return to the Earth.

 

[rcgldr]

The difference between the acceleration due to the Sun's gravity on the two increases. When this difference exceeds the Earth's gravitational pull, the rocket will transition into an independent orbit around the Sun. Thus with the help of the Sun, the rocket doesn't have to attain quite the full escape velocity in order to not return to the Earth.

Depending on what orbit around the Sun that the rocket transitions into, couldn't the rocket's orbit around the Sun eventually put it on a collision course with the Earth (despite being at or above escape velocity at the time of collision)?

 

[Nugatory]

Depending on what orbit around the Sun that the rocket transitions into, couldn't the rocket's orbit around the Sun eventually put it on a collision course with the Earth (despite being at or above escape velocity at the time of collision)?

Yes, but we don't consider that case a failure to escape the Earth's gravity. It's not the Earth's gravity that brings the rocket and the Earth back together, it's just bad luck that puts them both at the same place at the same time some orbits later.