I have a question regarding the Newtonian claim that a bullet fired at the exact escape velocity of Earth would indefinately follow an orbit around it, without ever touching the ground. The solution given is that, as the bullet travels, the Earths curvature causes the ground to 'continuously bend away', leaving the bullet no other option as to continuously free-fall. Then here's the question: if gravity is considered to be an acceleration (app. 10 m/s^2), wouldn't that bullet keep accelerating up to the point where it surpasses the escape velocity? Terminal velocity notwithstanding, since most of the thought experiments involved usually consider a 'perfect' sphere with no atmosphere ('perfect' vacuum)... After all, when simply dropping a bullet towards the ground, its speed is determined by the height from which it is dropped, the higher the 'drop-point', the longer the bullet is 'influenced' by gravity (ie, the longer the gravitational force can act upon it, and thus the higher its speed): a bullet dropped from 10 metres high (in a vacuum), takes 1 second to reach the ground, and if dropped from 30 metres, it'll take 2 seconds (since its speed would be 20 m/s during the second time-interval)... Or am I missing something here?

I was trying to figure this out, and tried to calculate how long it would take a projectile to cover a distance of 22 kms, given that it cannot travel faster then 11,2 km/s - which is the escape velocity at the surface of the Earth, and if it were travelling faster then that, it would leave the Earths gravitational pull and never hit its surface... So I calculated the 'extra curvature drop' to be an additional 38 metres (on top of the initial 10 metres; that was the initial gedanken experiment: drop a bullet from 10 metres high and at the same time fire a bullet from the same height: both bullets will hit the ground at the same time, according to Newton). So the fired bullet would need to drop 48 metres, but I can't see how it can do that in the 1 second it takes the dropped bullet to reach the ground, not without surpassing the escape velocity?

In case you're wondering, I got the 38 metres by taking the 22 km it would travel, divided it by the app. 6378.136 km of the Earths radius: that gives app. 0.0034492836 (this would be the cosinus if the Earth can be seen as a circle with radius equal to 1 - so 22 km would be 1/0.0034492836th part of that radius).

Sin(a) = sqrt(1 - cos(a)^2), so the difference in height due to the curvature should be app. 99.999405120 % of the radius, or app. 38 metres.
(I think )...

if gravity is considered to be an acceleration (app. 10 m/s^2), wouldn't that bullet keep accelerating up to the point where it surpasses the escape velocity?

Acceleration is a change in velocity, and velocity is a vector quantity. In other words, a change in direction is an acceleration. An object in circular orbit, for instance, is constantly accelerating but not getting faster.

a bullet dropped from 10 metres high (in a vacuum), takes 1 second to reach the ground, and if dropped from 30 metres, it'll take 2 seconds (since its speed would be 20 m/s during the second time-interval)... Or am I missing something here?

Yes, you are. You have erroneously applied the acceleration.

Gravity on the Earth's surface applies an acceleration of 10ms^-2, as you say. That means that after one second it will be travelling at 10m/s. A bullet that falls for one second will hit the ground at 10m/s, but it will not be travelling that fast all the way. A bullet dropped from 10m high will have to accelerate from 0-10m/s/s, and will therefore have an average speed during its fall of less than 10m/s, hence will take more than 1s to hit the ground. Similarly, a bullet that falls for 2s will hit the ground at 20m/s/s, but will take longer than 2s to fall 20m because its average speed during the fall is less than its final speed. What you have done is assume that the bullet travels at 10m/s for the first second, then travels at 20m/s for the second, whereas it actually accelerates to 10m/s in the first second and accelerates to 20m/s in the next second.

I hope that made sense.

This page:

http://www.glenbrook.k12.il.us/gbssc...Kin/U1L1e.html

has a table in it that shows how far a falling object travels in a given time interval.

__________________
"The very powerful and the very stupid have one thing in common: They don't alter their views to fit the facts, they alter the facts to fit their views." The Doctor, Doctor Who: The Face of Evil.

<<a bullet fired at the exact escape velocity of Earth would indefinately follow an orbit around it>>

I think you're getting mixed up with ORBITAL velocity. A bullet fired with the ESCAPE velocity wouldn't follow a closed orbit, it would go into interplanetary space and become a satellite of the sun.

What would need to happen is the bullet would have to cover enough ground so that the curve in the earth would make up for the fall towards the earth. and you would be correct then that the more time the bullet was in the air, the faster it would fall toward the ground and therefore the faster it would have to travel to make up the difference. the bullet would have to continue to accelerate until it reached it's terminal velocity, at which point the bullet would need to keep the same speed.

of course this would also assume that the earth was smooth, else you'd hit a hill or mountain. Or, with my luck, i'd intercept the bullet with my face. Ouch.

[Edit] Note: the speed of the bullet would not be the same as the acceleration of gravity. The speed would have to be whatever speed required so that height from ground = height from ground + drop from gravity. This is dependant on the earth's curvature

__________________
I'm like one of those idiot savants...well, except for the savant part.

I think he means a bullet fired above the atmosphere. Wind resistance would spoil the whole effort. To clarify the way I see his questions, wouldn't objects in orbit accelerate due to 10m/sec/sec? And if they did accelerate, wouldn't that throw the objects out into deep space? And since this doesn't happen, why aren't objects in orbit accelerating?

To provide my own answer, acceleration is a relative term. And relative to the earth's plane, the object is accelerating, but the acceleration matches the earth's curvature. I.E. The faster the object goes, relatively speaking, the more the earth curves.

Wow, did that ever not sound right!

Well, exclude wind resistance as this is a theoretical question anyway, but i guess i should have said the bullet would have to increase velocity to match the gravitational "fall". which i guess would be the same thing as a satelite in space, only would have to travel faster due to being closer to the earth therefore being more affected by gravity.

again, to sum up, given just the first second of travel, if the object falls at 10m/s/s, it would have to travel far enough that the earth would have curved downward exactly 10 meters after one second of travel to stay at the same "height", etc. etc. as the object is subjected to gravity.

__________________
I'm like one of those idiot savants...well, except for the savant part.

why aren't objects in orbit accelerating?

They are, constantly. Acceleration is a change in a vector, not necessarily in speed. You can accelerate without going faster, and this is precisely what an object in orbit is actually doing.

__________________
"The very powerful and the very stupid have one thing in common: They don't alter their views to fit the facts, they alter the facts to fit their views." The Doctor, Doctor Who: The Face of Evil.

Simple matter of orbital velocity equation:
v = SQR(GM/r)

G = 6.673E-11

For earth M = 5.976E24kg r = 6.378E6m
:. v = 7907m/s

For the moon M = 0.073E24kg r = 1.738E6m
:. v = 1674m/s

If you stood on a mountaintop on the moon and fired your bullet horizontally at 1674m/s you would then have 108.7min to duck or get out of the way before it hit you in the back of the head.

__________________
Does earth plug a hole in Heaven or Heaven plug a hole in Earth? -Peter Gabriel

Mister Earl, LayMan,

An object in circular orbit is moving forward at exactly the
speed needed to balance the acceleration toward the center of
the planet. That acceleration is perpendicular to the direction
of motion, so there is no change in the speed, only in the
direction of motion.

If a satellite is moving slightly faster than circular orbit
speed, it rises away from the planet, so the acceleration is no
longer perpendicular to the direction of motion, but slightly
backward. This slows the satellite. When its speed away from
the planet reaches zero, it is moving at less than circular
orbit speed, so it starts falling toward the planet, and speeds
up again. It is in an elliptical orbit.

Here is my web page on orbital speeds. It may answer some of
your questions:

http://www.freemars.org/jeff/speed/

Notice the little animation of the satellite in orbit. The
orbit in this case is being circularized by friction with the
atmosphere at each perigee. When the orbit is circularized,
the speed becomes constant.

-- Jeff, in Minneapolis

__________________
http://www.FreeMars.org/jeff/

"I find astronomy very interesting, but I wouldn't if I thought we
were just going to sit here and look." -- "Van Rijn"

"The other planets? Well, they just happen to be there, but the
point of rockets is to explore them!" -- Kai Yeves

Thanks everyone, I knew I was missing out on something...

Still remains pretty confusing, though: I know there's a difference between an objects speed, and its velocity. However, I still fail to see how an object can be said to have a constant acceleration, without having its speed/velocity changed? Quote from the website given by Jason:

"The Meaning of Constant Acceleration

Sometimes an accelerating object will change its velocity by the same amount each second. As mentioned in the above paragraph, the data above show an object changing its velocity by 10 m/s in each consecutive second. This is referred to as a constant acceleration since the velocity is changing by a constant amount each second. An object with a constant acceleration should not be confused with an object with a constant velocity. Don't be fooled! If an object is changing its velocity -whether by a constant amount or a varying amount - then it is an accelerating object. And an object with a constant velocity is not accelerating."


Hello Jeff, I liked the animation. But you are talking about an elliptical orbit, the gedankenexperiment (That's right, kids, don't physically try this at home: guns and bullets ain't toys, there dangerous!!) proposed by Newton suggested that a bullet fired at a straight line (not upwards) would hit the ground at the same time as a bullet which was not fired away but simply dropped from the same height (as described here: http://www.lightandmatter.com/html_b...ch06/ch06.html)...

The key factor that puzzles me there is the fact that the bullet is said to start falling as soon as it leaves the muzzle of the gun. So in order to prevent the bulllet from hitting the ground, you'd need to fire it with enough speed so that the curvature of the Earth would continuously 'recede' away from it. But the way I see it, that would also leave the bullet continuously under the influence of gravity. What would keep the bullet at a constant velocity, needed to stay in orbit (its velocity can not increase, since then it would 'fly away' outwards, and if it decreased, it would eventually hit the ground), if the effect of gravity is an acceleration?
That actually raises another question I've been meaning to ask: are orbits truly elliptical, or more egg-round? After all, only a circle has every point on its circumpherence at exactly the same distance from its center. That doesn't go for an ellips, with the heavy mass at one of its 2 "focal points" (don't know the exact english word for it)... That way, the ellips can be divided into 4 quadrants, where an orbiting body would travel along the direction of attraction in 2 quadrants (causing acceleration), and in opposite direction towards the attracting force in the other 2 (causing deceleration).

Fazor, the 'experiment' usually considers a 'perfectly smooth' planet/sphere, but also a perfect vacuum (no atmosphere). I always thought that the effect of terminal velocity was caused by air resistance, which wouldn't be a factor here... Or would an object falling towards a heavy mass in a perfect vacuum still atteign a terminal velocity at which it can no longer accelerate (ie, increase its speed)?

I'm really new to all of this, but here is the way I'm seeing it. Gravity is pulling the bullet "down". Down changed direction from the perspective of a sphere. If we were to freeze the sphere in the example in time and space, then place a reference point, like the Sun, to use as a marker, "down" would initially be in the direction of the sun (for example) When the bullet has traveled 25% of the way around the sphere, "down" is now 90 degrees away from the sun-sphere line. This would mean that for every unit of distance the bullet covers, "down" changes slightly.

Now, the bullet is traveling at a constant speed in a direction perpendicular to "Down". When it's fired, it is traveling at 90 to the sun-sphere line. 25% of the way around, it is traveling directly toward the sun. This would mean that while the 10 m/s is being applied to the bullet, the direction of the force is rotating around it, which keeps it a constant distance from the surface.

Does any of that make sense/sound right?

__________________
I'm not evil.
An evil person would do the things I think up.

Wow are people ever good at becoming confused by trivial concepts.

Try tertiary education. If that fails, try secondary education.

But just incase i can help,
That is correct. If both bullets start with no vertical component of velocity and they are at the same height they will hit the ground at the same time. the horizontal component has nothing to do with the time it takes to hit the ground.

However, this is different when we consider the earth is actually a sphere, and as such the bullet can have a horizontal velocity sufficient that as it falls the earth curves away from it and it never gets any closer or further away to the earth - it is in orbit. If the bullet was any faster it would get higher, if it was any slower it would eventually hit the earth.

Also, although it has been mentioned before i feel i need to reiterate it:
Velocity is not the same as Speed! Velocity is a vector and has magnitude (speed) and direction.
An acceleration is a change in velocity with respect to time. Therefor an object changing direction is accelerating even if its speed remains constant!

So an object going in a circle is accelerating! In such a case the acceleration is always normal (perpindicular/at right angles) to the direction of motion (the acceleration is towards the centre of the circle)

you can calculate this acceleration using a = v^2 / r
where a is acceleration
v is velocity (speed)
r is the radius of the circle

I hope i've kept that simple and that it helps you understand what is happening.

However, I still fail to see how an object can be said to have a constant acceleration, without having its speed/velocity changed?

The key here, as I have tried to explain apparently unsuccessfully, is that speed and velocity are not the same. Velocity is a vector quatity while speed is scalar. That means in layman's terms that velocity has a directional component while speed does not.

Let's say an object moves 1km due north in one hour, then reverses direction and travels back 1km due south in another hour. It has travelled 2km in 2 hours, and had a speed of 1km/h for the whole journey. However, if you describe it in terms of velocity the first half of its journey had a velocity of 1km/h, but the second had a velocity of -1km/h because the direction of travel was reversed.

An object that is changing direction is changing its velocity but it does not have to be changing its speed. An object moving in a circle is therefore constantly changing its velocity because it is constantly changing direction in order to follow a closed loop.

Since acceleration is defined as a change in velocity an object moving in a circle is constantly accelerating, but it is only changing direction and not necessarily changing its speed.

Common usage of scientific terms does lead to this apparent contradiction when the terms are used in their proper context. We are mostly used to using the term acceleration to describe a change in speed. Worse, its most common use is an increase in speed and a whole new word, deceleration, has been invented to describe slowing down. Commonly, an object travelling at a constant speed, whether it follows a straight or curved path, is not considered to be accelerating. However, in its proper sense acceleration describes any change in velocity.

An object speeding up is accelerating.

An object slowing down is accelerating, negatively.

An object changing direction, regardless of its speed, is accelerating.

Therefore an orbiting satelite or bullet is constantly accelerating because its direction and hence its velocity is changing all the time as its path is pulled from a straight to a circular one.

I hope that makes it clear.

__________________
"The very powerful and the very stupid have one thing in common: They don't alter their views to fit the facts, they alter the facts to fit their views." The Doctor, Doctor Who: The Face of Evil.

I'm getting there, I think... So: any change in direction also implies a change in 'travelled distance'. The shortest path between 2 points is a straight line, any change in direction would deviate from that straight line and would effectively increase the distance the object would have travelled if it hadn't changed direction. So even though the speed doesn't change, the acceleration would lie in the fact that the velocity would be different. Is that correct?

I have absolutely no idea what you mean by all that but it may be time to introduce you to the concept of displacement vs distance.

I think you are struggling to understand the concept of vectors and scalars. Try a quick google search, you should be able to find a lot of helpful information.

To say the least... But I'm googling away here, maybe I'll get it through my head eventually... Thanx!

So: any change in direction also implies a change in 'travelled distance'.

Distance is irrelevant.

The shortest path between 2 points is a straight line, any change in direction would deviate from that straight line and would effectively increase the distance the object would have travelled if it hadn't changed direction.

No. An object with a speed of 10km/h will travel 10km in one hour whether that 10km is a straight line, a line with a kink in it, a curve or a closed circle. If an object has travelled a greater distance in a certain time on a curve than it would have done in a straight line then it has changed its speed.

So even though the speed doesn't change, the acceleration would lie in the fact that the velocity would be different. Is that correct?

That is correct, but you've got hung up on distance for some reason. It is purely direction.

__________________
"The very powerful and the very stupid have one thing in common: They don't alter their views to fit the facts, they alter the facts to fit their views." The Doctor, Doctor Who: The Face of Evil.

Well, it is really hard not to think about a flying bullet in terms of the distance it covers (too me, anyway...).

Perhaps another way to think about this is that if you fire a bullet parallel to the ground at the escape speed, then the direction it is accelerating is perpendicular to its current velocity. Accelerations don't increase speed unless they have some component along the direction of the motion, so a perpendicular acceleration will only rotate the direction it is moving a little, without increasing speed. Once it moves a little bit, you'll see that the acceleration is now pointed slightly backward relative to the motion, so from that point on, the "falling" that is going on will actually reduce the speed, not increase it. So when you put it all together, you find not only that the speed does not increase past the escape speed, but actually it is only slowing down. However, the bullet is also getting farther from the surface all the time, so the escape speed is also getting smaller. The two remain exactly equal the entire motion-- the bullet is always "barely escaping".