Ever try walking at 25,000mph?
No, the weight of an object does not affect the acceleration of gravity. All things fall to earth at 9.8m/s[sup]2[/sup]. Things like wind resistance do affect the speed, but there isn’t much difference in wind resistence for a marble and a cannonball, both being pretty solid spheres.
In order to throw a cannonball up at 30mph, you will need to accelerate it to that speed. Acceleration always requires energy, and the more stuff you’re accelerating, the more energy you need. Thus, you need to be a lot stronger to throw a cannonball at 30mph than a marble.
Well, no, but I’m not streamlined, nor do I have a solid fuel rocket booster up my ass.
Perhaps I should have said “a minor factor, compared to overcoming Earth’s gravity”…
Why are you so sure that the be cannonball will come down first? Certainly they would act the same with no air resistance, and I imagine with air resistance, the marble would be slowed first. No, the weight of the object makes no difference. Escape speed is give by:
v[sub]ESC[/sub] = sqrt(2 G M[sub]Earth[/sub] / R)
You can never be free of the Earth’s gravitational pull - it attracts every object in the universe.
However, as scr4 and Master of Ceremonies said, escape velocity means that the Earth’s gravity will never be enough to stop you and pull you back. The attraction will continuously slow you down, but it will never slow you to zero.
Orbital velocity and escape velocity are theoretical and discount the atmosphere. For example, orbital velocity at sea level is approx 17,680 mph. If you launch tangentially, the gravity will pull you down in an arc that is the same radius as the Earth itself, in other words you will fall forever, but never hit anything.
Orbital velocity = Escape velocity/root of 2
Golly. Lots of things said here. 
The equation for escape velocity is: ev = square root(G *M / R)
where G is a constant, m is the mass of the object pulling you down, and r is your distance from it.
The higher you go, the bigger r is, and the lower the escape velocity. Plugging and chugging for the Earth, you get about 11 km/sec, or 7 miles/s. That is the initial velocity you need to go up and never come back. Don’t confuse this with “escaping Earth’s grvaity”; since gravity never completely goes away. All it means is that gravity will slow you, but it cannot slow you fast enough to stop you. Once you reach escape velocity, you will always have some velocity away from the Earth.
Now, imagine being in a balloon. You can rise at, say 5 mph until you are quite high. The record for a balloon is well over 100,000 feet. In principle, you could build a ballon with a vacuum inside and rise higher, never faster than 5 mph. Then you launch a rocket. That rocket will have to go slightly slower to escape than one launched from the ground.
Or, you could simply climb an infinitely long ladder at whatever speed you chose until you reached a distance where the escape velocity from the Earth is slower than you can jump. Jump, and you’re gone. Bye bye! And you never had to go 7 mps.
And this is all different than escaping the atmosphere. At 100,000 feet, the atmosphere is a pretty good vacuum!
I think that an evil force deleted a character from your post. You actually typed the following, right?
ev = square root(2 G *M / R)
Or else you meant G as twice Newton’s Constant. 
The atmosphere is not insignificant. Because anything going 25,000 mph in the atmosphere would burn up. Think about what happens when the shuttle comes back down again. Just braking the speed off requires exotic ceramics to keep the thing from burning up. And think about how much energy is being shed. That’s how much you’d lose if you tried to go the other way and shoot a cannonball at 25,000 mph up.
Rockets burn fuel more slowly and in stages so that they can get past the major parts of the atmosphere and still be accelerating so that that the energy lost to friction in the atmosphere is minimized.
It seems that some basic physics is needed. The 25000 mph is a separate issue from whether or not the cannon ball will come down before the marble if both are launched at the same velocity. You have to talk about one subject at a time in order to figure things out.
One of the principles develped by Isaac Newton is that if you push on something, that is exert a force (we’ll call forces F) on it, it will accelerate. That is it’s velocity will change by either increasing or decreasing in magnitude or changing direction, or both.
For now we will stick to motion in a straight line so the only thing we pay attention to is the magnitude of the velocity. Newton devoped an equation that relates force, F, to acceleration, a, using the mass, m, which is a measure of an object’s inertia, as the proportionality constant. That formula is
-
F = m*a, the symbol * signifies multiplication.
Gravity pulls objects toward the center of the earth with a force which equals their weight. That is, the weight of an object is the force with which the earth and an object on the earth pull toward each other. Letting w[sub]o[/sub] stand for the weight of the object and m[sub]o[/sub] for its mass.
- w[sub]o[/sub] = m[sub]o[/sub]*a
Newton also provided us with a formula for the force of gravity or the force with which two objects attract each other. That formula for the earth and any object whatever is:
- F = G*m[sub]o[/sub]*m[sub]e[/sub]/r[sup]2[/sup]
G is the gravitation constant which has been measured and is known.
m[sub]o[/sub] is the mass of the object
m[sub]e[/sub] is the mass of the earth
r distance between the center of the earth and the center of the object.
If we want to compute the acceleration of an object caused by the gravitational attraction of the object and the earth we can substitute the force from the gravity equation for w[sub]o[/sub] in equation 2):
G*m[sub]o[/sub]*m[sub]e[/sub]/r[sup]2[/sup] = m[sub]o[/sub]*a
Since the quantity m[sub]o[/sub] appears on both sides of the equation we can cancel it on both sides leaving:
G*m[sub]e[/sub]/r[sup]2[/sup] = a
Everything on the left side is a constant, that is G, the mass of the earth and the distance between centers, r. So the acceleration of an object resulting from gravitational attraction is a constant irrespective of the mass of the object.
OK so far?
Before Newton came along, Gallileo had provided the world with some equation for the motion of objects. One of them is that the velocity is equal to acceleration time time. Another is that the distance traveled equals velocity time time. That is:
-
v = a*t.
-
d = v*t
Now it should be obvious that if two objects start at the same velocity, like the cannon ball and the marble, and have the same acceleration the time it takes them to stop will be the same and the distance the travel will be the same.
Just in case it isn’t obvious here is how to figure it out. We proved above that the acceleration is the same for all objects under the conditions of our problem. So:
v[sub]1[/sub] = a*t[sub]1[/sub], where v and t are the velocity and time of the marble. Also
v[sub]2[/sub] = a*t[sub]2[/sub], where v and t are the velocity and time of the cannon ball.
Now we start them with the same velocity so
v[sub]1[/sub] = v[sub]2[/sub]
and we can now equathe the right hand sides of the two equations"
at[sub]1[/sub] = at[sub]2[/sub]
The a’s cancel and t[sub]1[/sub] = t[sub]2[/sub]
and by equation 5) if the v’s are equal as assumed and the times are equal by computation then the distances must also be equal.
So both will rise to the same height if they start at the same velocity.
Once you understand this thoroughly you are prepared for the next step.
Sure? GR states that massive objects bend spacetime near them.
This has already been addressed in other ways but you can disprove this with a nice thought experiment (I think Galileo did this one before his real experiment):
Tie a light object and a heavy object together with a string of negligible mass that we will also assume is unbreakable and won’t stretch.
Drop the 2 objects simultaneously. If the heavier object falls faster, then it will be slowed down because of the drag from the lighter object (which wants to fall slower) and the combined object; light, heavy and string, will actually fall slower than the heavy object alone: a contradiction since the combined mass is heavier than the heavy object alone.
They proved it conclusively in the Apollo missions to the Moon when one of the astronauts dropped a hammer and a feather and they landed together.
Could the degree of bending decrease asymptotically? i.e: get ever nearer to but never theoretically reach zero?
Disclaimer: I am not a physicist and I even have doubts about my spelling of the word “asymptotically”.
There is some kind of mistake here. I am not certain where it lies however. From what I remember of physics, if a force is applied in one vector, say 9.8m/s^2 down (toward the center of the Earth), then the only way to make it go up, is to supply a greater force say 13m/s^2 in the upward direction. If this force is supplied initially, then barring anything else from happening (like a sudden downdraft, windsheer, etc) then all that has to happen is that the force maintain a 9.8m/s^2 acceleration upward to compensate the 9.8m/s^2 acceleration due to gravity. This in turn means that it is possible to go up at say 5 mph. What will actually happen though is that as the object gets farther away, the force of gravity weakens, so that the object requires less and less energy to maintain its 5 mph speed. So I think it would be possible to get away from the planet at a very low speed.
Oops, yes, I forgot the 2 in the equation. What I wrote is the velocity for a circular orbit. The factor of root(2) gives you escape velocity.
Also, gravity goes as (1/r)^2, so classically gravity goes on forever. There are space-time issues in relativity about this, dealing with the finite age of the Earth, so that its gravity only extends for 4.5 billion light years, but that involves simultaneity and other headaches I won’t pursue.
Gee that is all well and good (and great fodder for my headache) but suppose we had a Dyson’s ring orbiting asynchronously around the equator of the earth and we were to lower an rope of infinite tensile strength from which a vehicle can raise itself from the surface to the ring which is just outside the gravitational pull of the earth. does the vehicle still need to travel at escape velocity to get to the ring or can it just plod along leisurely?
Here is a simple explanation :
There are two scenarios for a body “escaping” earrth’s gravity:
Scenario 1 : The only energy available to the body escaping earth is its initial kinetic energy. So as the body keeps going it gets slower and slower but the de-acceleration is such that it never slows down the body sufficiently to stop and then turn back. This initial velocity corresponding to this kinetic energy is called the Escape Velocity
Scenario 2 : The body has its own kinetic energy source. In this scenario the body can move at what ever velocity it chooses as long as it can provide the force (and hence the energy) to overcome the ever changing gravitational force on it.
Hope that simplifies.
<pedantic irrelevance> Of course, you’d have to worry about avoiding the sun, given that the radius of the Earth’s orbit is 192 million miles. </pi>
Actually, that suggests an interesting question, though not related directly to the OP. If I’m 5 billion miles “above” the Earth — that is, perpendicular to its orbital path, outside the plane of the solar system — won’t I basically get pulled straight toward the Sun, which has a much bigger gravity well?
You know, I think a much better question would be along the lines of, 'Why can’t we go back into the earth’s atmosphere slowly?" Knowing you won’t burn to a crisp would be much more useful knowledge, IMHO.
I’ll take a shot at that too :
Most spaceships are on a orbit around the sun. We on earth right now are moving at the speed of approx 2681mph (or 1676kmph) if you consider only the rotation around the axis. So a Geostationary station would be moving much faster, even if its not geostationnary it will still be moving at a considerable speed.
The atmosphere of the earth rotates with it, but the top layers have very strong winds.
So when the spaceship enters back, it has a high velocity compared to the earth’s atmosphere. To match it to the speed of atmosphere, you’ll need reverse boosters or things like that (which will act as load during take off ). Much Cheaper to design one with a sturdier nose cone.
Again, it’s more efficient to come down fast. A spacecraft in orbit is travelling at 1600mph, which is a lot of kinetic energy. To get down to earth you need to slow down to zero somehow. You could do it with a rocket engine and then descend slowly, but that takes a lot of fuel. It’s easier to just crash into the atmosphere and let the friction slow you down.
Re-entry isn’t that dangerous anyway. Nobody has burnt to death during re-entry. The only fatal accident during landing was the Soyuz 1, and that was a parachute failure.