A question on escape velocity

My understanding of escape velocity is that it applies to a projectile i.e. an object not under it's own power such as a bullet. This would not apply to an object under its own power such as a rocket. To me, this is totally logical.

Confusingly, NASA states that the shuttle must reach escape velocity to leave the Earths atmosphere. www.nasa.gov...
What have I missed? Are they launching the shuttle from a giant rubber band? I've heard "everything's bigger in Texas" but come on..............

This leads me to another question. The escape velocity of "black holes". Since photons are "projectiles" per se, if the escape velocity of a black hole exceeds c then clearly photons would not be able to leave once past the event horizon. Could a rocket just cruise out? For a moment, lets put tidal forces etc aside.

reply to post by OZtracized


well there is the gravitational pull of 9.8.....so here on earth it would be terminal from a given distance..I'll think about it

Escape velocity (per my understanding) is purely the amount of energy (newtons) needed to break free from the opposing force of earths gravity. Whether or not that requires a rocket attachment or not is irrelevant.

edit:
Think of the bullet having a mass and a vector (velocity with a trajectory), F(newtons)=Mass x Acceleration, so long as that is around 11 km/s (from memory) you can break free.

I don't see why the term escape velocity should be limited to objects that have no internal propulsion mechanism. If it troubles you, then regard the rocket stages of the space shuttle as a slingshot, so that it is an external power source again.

As for escaping a black hole, if something can go faster than c I guess that should be possible. But we don't know of any rocket capable of doing that.

reply to post by OZtracized


Rockets undergo continuous acceleration bullets undergo instant acceleration and then slow down.

Photons are not really particles but waves, but this really depends on the experiment you are conducting. They state light cannot escape a black hole because the velocity required to leave is faster than C.

However, black hole physics is all hypothetical, no ome really knows what they are or what they do. We can observe events around a black hole but we cannot observe a black hole itself unless they manage to observe hawking radiation.

Once light passes the event horizon then there is no way back apparently. If our sun was to collapse it would only be couple of kilometers wide and we would not be able to see it. We wpuld only be able to see the effects of it.

Originally posted by TheMindWar
reply to post by OZtracized

 

Rockets undergo continuous acceleration bullets undergo instant acceleration and then slow down.

This is true, but if a bullet could be strong enough so as not to disintegrate but be shot with enough force to go faster than 11 km/s to get past 2000 km above the earth surface, in theory isn't that then capable of an escape velocity. (obviously not capable yet with current tech, but as a thought experiment here)

From the wiki page. Which actually appears quite good in this case.

"Escape velocity is sometimes misunderstood to be the speed a powered vehicle, such as a rocket, must reach to leave orbit and travel through outer space. The quoted escape velocity is commonly the escape velocity at a planet's surface, but it actually decreases with altitude. It is the speed above which an object will depart on a ballistic trajectory, i.e. in free-fall, and never fall back to the surface nor assume a closed orbit. Such an object is said to "escape" the gravity of the planet.

A vehicle with a propulsion system can continue to gain energy and travel away from the planet, in any direction, at a speed lower than escape velocity so long as it is under power. If the vehicle's speed is below its current escape velocity and the propulsion is removed, the vehicle will assume a closed orbit or fall back to the surface. If its speed is at or above the escape velocity and the propulsion is removed, it has enough kinetic energy to "escape" and will neither orbit nor fall back to the surface."

To use your example, your magic bullet could be fired into space. As its not supplying more energy after launch it needs to achieve escape velocity at launch. The escape velocity it needs to acquire if fired from the ground is greater than if fired from a platform at 100,000ft.

reply to post by Qumulys


Well, first of all, rockets don't require to continuously accelerate. Once they reach a certain speed, they can just maintain that speed and fly all the way to space, then slow down and go into orbit. They do need continuous thrust though to maintain speed. And there is this:

en.wikipedia.org...

reply to post by OZtracized


TheMindWar and justwokeup already answered the technical part of the question. I just want to add that the NASA page you linked, which is aimed at young kids, is wrong. It would only be true if the shuttle's engines shut off immediately after launch. The shuttle can go at a proverbial snail's pace and, as long as it has continuous thrust (providing continuous acceleration), it will leave Earth's orbit.

reply to post by -PLB-


I was just replying to the above poster who mentioned the rockets. I agree with you, Rockets actually start reducing thrust as the escape velocity need decreases so as to conserve energy. Mind you, its been 20 years since I was doing this stuff in uni, and I've forgotten most of it!

To clarify in my above thought experiment bullet, the bullet would need to sustain a speed that does not go below 11km/s. (btw, I'm happy to be corrected if I'm wrong)

reply to post by Qumulys


Actually, all the bullet needs is an initial, just-fired speed of 11 km/s (straight up). After that, no matter how much it decelerates (since it will be "decelerating" at -9.8 m/s^2), it will still keep going fast enough to escape Earth's gravity.

The shuttle is under main engine power until it reaches its orbital insertion point. The shuttle never has to reach escape velocity, since it experiences continuous thrust. It just happens that, as it climbs in altitude, the escape velocity decreases until the shuttle's speed reaches the Earth's decreasing escape velocity. This equilibrium is what we call an escape (or parabolic) orbit.

reply to post by CLPrime


Ahhh, I see! Thank you for clarifying things for me!

Can I ask, my understanding is that wouldn't the initial force applied to the bullet actually have to be slightly greater than 11km/s due to atmospheric friction resistance though?

reply to post by Qumulys


Yep...in real life, we have to take drag into account, which causes both resistance to acceleration and extreme heating. That's why the shuttle is as streamlined as possible.
A bullet, though, is small and streamlined enough that it probably wouldn't make much difference. As you say, it would only have to be slightly more than the escape velocity. But, in practical situations, it certainly matters.

reply to post by CLPrime


Cheers, thanks again, stars for your helping me! I love seeing this back-n-forth discussion on ATS that helps to further education

Now, I'm thinking more on this though, once the bullet has left earths gravity, if it was given (lets discount friction) the required 11 km/s, at what speed would it be going at once "free" (as in not in orbit, but breaking out)? (lets say the bullet is 1 kg if that matters and to keep it simple)


Am I correct along this line of thinking? Lets say it breaks orbit at 2000km, therefore it would have taken 182 seconds. So is that 11,000 m/s - (182 x 9.8^2) ?
(nope, thats giving me approx -4000m/s, what am I doing wrong?)

reply to post by Qumulys


If the bullet is given an initial velocity, straight up, of 11 km/s, its velocity will tend to 0 at infinity - meaning, the further it gets from Earth, the slower it will get until, at an infinite distance away, it will come to a stop. The bullet would be moving veeeerrryyyy slow long before that, though. And, of course, infinity is impossible to reach, so the bullet's velocity will never truly reach zero...but it will approach it (the magic of asymptotes).

ETA: that calculation won't work since the acceleration of -9.8 is for surface gravity. It decreases with distance. If the bullet is given an initial velocity equal to Earth's escape velocity, then it should (I can't say it will, 'cause I haven't actually done the calculations) stay equal to the escape velocity at altitude - meaning: the escape velocity decreases with altitude/distance, and the bullet's velocity will follow that out to infinity (where the escape velocity is 0).

reply to post by CLPrime


Hmmm, so technically I think your saying the bullet is racing up and slows down and slows and slows and is either going to stop and stay in orbit if exactly right, accelerate back to earth, or if going just say 0.0000001 m/s faster is all it takes to break free, although I'd say it would never stop at infinity, given an infinite time frame, its bound to bump into something eventually! Ok, its late and I'm getting silly! Cheers

(edit, and of course, forgot about surface gravity reducing. duh! I wonder how accurate then it could be calculated taking into account multiple gravitation points such as say the moon's increasing gravity upon it... The maths has gone beyond me now and I pass the baton to the pro's! But a very education night for me

Originally posted by Qumulys
reply to post by CLPrime

 

Hmmm, so technically I think your saying the bullet is racing up and slows down and slows and slows and is either going to stop and stay in orbit if exactly right, accelerate back to earth, or if going just say 0.0000001 m/s faster is all it takes to break free,

To enter orbit requires tangential velocity (velocity perpendicular to "straight up"). An object moving straight up will never enter orbit. If it's moving exactly right, it will keep going forever, slowing down to 0 at infinity. If it's moving faster, it will do the same thing - it will just be moving faster along the way and come to a stop "further out" into infinity. If it's moving slower, if will stop at some altitude and fall back to Earth. It will never enter orbit.

although I'd say it would never stop at infinity, given an infinite time frame, its bound to bump into something eventually!

Ok, its late and I'm getting silly! Cheers

That's pretty much a guarantee.

The bullet will be pulled into solar orbit once the sun's gravitational pull is greater than the earth's when the bullet reaches that particular distance from earth. One then would have to calculate the speed necessary for that bullet to be going at around 1AU from the sun. I know one thing (about), that if that bullet was shot from the surface of the sun it would have to be shot at around 1.3 million mph to defeat the solar pull.

Now if that bullet was shot from earth towards the moon, it should have momentum left to defeat lunar orbit, unless it was a direct collision trajectory, so hypothetical in a way, so if it missed the moon I believe the sun would capture it into solar orbit before it got to Jupiter. But at the right speed, in the right kind of solar orbital trajectory, and if Jupiter was going to cross it's orbital trajectory around the sun, Jupiter would capture the bullet into its orbit or collision. Thats sort of how they are going about getting JUNO to Jupiter, but they are first going to get an accelerated boost from an earth slingshot to gather the proper momentum to swing out to Jupiter in its solar orbit.



That flyby of earth is going to be controlled to miss earth by a mere 300 miles, after a thrust maneuver out at the distance of Mars's orbit.!!!
Not only that but they are also going to enter a polar orbit around Jupiter after an extreme braking thrust maneuver, as Jupiter's gravity will speed the craft to near 160,000 mph, and Jupiter is only going about 29,000 mph around the sun.
The further from the sun you are the slower your orbital speed is, as for the planet Mercury in its rather extreme elliptical orbit reaches near 110,000 mph at its perigee.

Originally posted by Illustronic
The bullet will be pulled into solar orbit once the sun's gravitational pull is greater than the earth's when the bullet reaches that particular distance from earth.

How so? Isn't the bullet already in the suns orbit when it is shot from earth? So any movement away from the sun will already be at escape velocity from the sun.

reply to post by -PLB-


Earth's gravity defeats the suns at a particular proximity, quite far away since the huge moon is captured by earth's gravitational pull over the suns, but actually its a balance of both the sun and earth's gravity, and the sun is pulling the moon further away from earth. But the earth will ultimately keep the moon from a free orbit around the sun sans earth, one day billions of years from now the moon and earth will reach an equilibrium distance from each other, being the sun stays put. At that time a day will equal a month, and it will be about 55 days for the moon to orbit the earth and a day will last the same amount of time, 55 days long. so an earth year will only be about 6.6 days long, life would have been extinguished a few billion years before that, in fact the sun wont last that long.

Gravity emanates from the center of a mass, and never becomes totally absent, yes gravity travels forever. If there were only 2 objects in the Universe, they would eventually collide, due to gravity, no matter how far away they might be.

In the absence of a greater gravity one never truly escapes it unpowered. Its just that there are so many bodies in the Universe that a greater gravity will eventually overrule.