I know there’s a formula for this, but anyway. . .
Let’s say we launch a rocket at only 90-95% of escape velocity. How far would it travel before it fell back to Earth?
In a similar vein:
I have the notion that escape velocity is generally a concept applied to a single astronomical body as if it was the only significant mass in the universe, not accounting for other nearby bodies. What’s the difference between the earth’s ‘theoretical’ escape velocity, and the velocity required to get to a point where the sun can ‘capture’ a rocket away from the earth’s gravitational influence??
The escape velocity is the speed which an object needs to never fall back. That is, it goes “infinitely” far away. At a speed just short of the escape velocity, the object would reach just short of “infinity”.
To get an actual number, compare the object’s kinetic energy to its gravitational potential energy. Convert all of its kinetic energy into potential energy and see what distance you end up with. Simple first-year university physics.
chrisk, there’s no simple solution to the three-body problem. In general, if the sun’s gravitational force is greater than the earth’s, then the object will be orbiting the sun.
Depends on where you aim it, but chances are that it will enter some sort of orbit, so… really far.
If you work through the math that Pleonast alluded to, you find that a rocket launched at a velocity of f v[sub]esc[/sub], where v[sub]esc[/sub] is escape velocity, will attain a height of
h = R[sub]e[/sub] f[sup]2[/sup] / (1 - f[sup]2[/sup])
above the surface of the Earth, where R[sub]e[/sub] is the radius of the Earth. As you can see, this increases without bound as f approaches 1.
Just to clarify an issue that has come up on this board in similar discussions (but does not appear to be a point of confusion in the above posts), here “launch a rocket” means to apply thrust until it reaches that speed and then stop all thrust. If thrust is continually applied, then the rocket can go an arbitrary distance at speeds lower than escape velocity as long as it still has fuel.
A matter of some debate, since by that standard, the Moon is orbiting the Sun. Which I don’t have a problem with, but some folks do.
Earth’s escape velocity is 11.2 km/s; launching at escape velocity (v[sub]e[/sub]=√(2GM/r)) in any direction is gives you an infinite height, i.e. a parabolic orbit. This can be easily derived by setting the gravitational potential energy (GMm/r) at the initial state to be equal to the kinetic energy (1/2mv[sup]2[/sup]) at the “final” state and assuming potential energy at the "final state to be zero.
To find terminal altitude at sub-escape velocities, you do hte same thing, except that instead of setting potential energy at the final state to zero, you set it to the terminal altitude, h. (h is actually the distance from the Earth’s center of mass, not the AGL.) So you get:
**U[sub]i[/sub] - V[sub]i[/sub] = U[sub]f[/sub] - V[sub]f[/sub]
½mv[sup]2[/sup] - GMm/r = -GMm/h**
Dividing out m and solving for h gives:
h = GM/(GM/r - v[sup]2[/sup]/2) = 2GMr/(2GM - v[sup]2[/sup]r)
Substiting in escape velcoity, v[sub]e[/sub] multiplied by a fractional constant, c for v:
h = 2GMr / (2GM -c[sup]2[/sup]2GM) = r/(1-c[sup]2[/sup]
So given a starting distance, r = 6,400km,
For c=0.90, h=33.7km
For c=0.95, h=65.6km
You can plug and chug other numbers yourself.
[On preview: MikeS, I think your solution is wrong; you’ll find that if you have f<=√2 your final altitude is underground.)
In the general solution to the N-body formulation there’s really no problem with having one body in orbit of two or more other bodies, or indeed, having the bodies “in orbit” of each other. Strictly speaking, they’re all in a very complex and highly perturbative orbit of their common center of mass (barycenter), but modeling this solution in closed form is impossible (although it’s pretty easy to simulate it iteratively using any commonly available math tool like Matlab or Mathematica, or even just a few lines of code). For satellites and the like the fact that the satellite weighs so much less than the Earth (ditto for the Earth with respect to the Sun) that we can assume that the influence of the smaller mass is minimal and that it orbits the larger one with only minor corrections. The Moon- Earth ratio is just about big enough to challenge that assumption.
In the case of an Earth-orbiting satellite, the Earth’s local gradient is so dramatic compared to the change in the Sun’s attraction that you can ignore the effects of the Sun for a simple, back-of-envelope calculation. For a much longer orbit or one that passes within the influence another body (like the Moon) then you’ll have to account for other effects; a satellite that passes through the L4 or L5 points, for instance, can easily achieve escape even if it otherwise lacks escape velocity. In any case, the Moon is indeed in orbit of the Sun, as well as the Earth, and has an essentially indepdent escape velocity relative to each.
Stranger
Thanks. What originally prompted this was I was wondering about the moon shots, and if with a launch of something less than (but close to) escape velocity just how far would it go. Apparently not far at all. I know that if it were possible, then those rockets would never have needed to achieve escape velocity to make it, and that that would have been determined a long time ago.
Lunar injection velocity was around 10.8km/s, so close to (~0.98) Earth escape velocity at that orbital altitude (~100km). (I’m using numbers from Apollo 14 but other missions should be very similar.) One thing to bear in mind, though, is that in the case of a Lunar mission, the craft is being directed toward the Moon, or rather on a sort of stretched ellipse that intersects where the Moon will be, so that the further along it gets the more the Moon’s influence pulls on it. It’s not quite the same as just launching a capsule in any arbitrary direction; the idea is to make it go as slow as reasonably feasable (given the mission parameters, i.e. how long you can safely keep the crew alive) so that the Moon pulls it into an intercept with as little additional velocity as possible.
Think of driving up a hill when you know that there’ll be a stop sign at the bottom on the other side, only you can only use the accelerator or brakes for a few seconds at a time. You accelerate hard at the start of the hill sufficient to have minimum velocity as you crest the hill, and then let gravity pull you down coasting, then hit the brakes at the bottom. Actually, on a lunar injection, you use the “brakes” twice; once to slow down into a capture orbit, and a second time (about halfway around) to circularize the orbit so that you don’t have this extended ellipse.
Stranger
As typically used, “escape velocity” means earth escape velocity with respect to some other body. That number varies. For example earth escape velocity for a lunar destination is less than earth escape velocity for Mars or the outer solar system. Very roughly earth escape velocity for a lunar destination is about 24,250 mph (10.84 km/sec). A Mars destination requires about a little more, I think about 11.3 km/sec for a nine-month transit time. More details:
Less frequently the term means escape velocity from other bodies. E.g, each Apollo Command/Service module had to achieve lunar escape velocity to return to earth, roughly 3131 mph (1.4 km/sec). More details: Escape velocity - Wikipedia
Re what happens if a vehicle fails to achieve earth escape velocity, this has happened several times. The early unmanned probes Pioneer 1 and 3 slightly failed to achieve this speed, and fell back to earth within several hours. Pioneer 3 reached a maximum altitude of about 107,000 km.
However you can fail to reach earth escape velocity and not fall back to earth. On Apollo 12 the Saturn S-IVB third stage was supposed to escape the earth-moon system, but an engine failure left it in a high earth orbit, which eventually transitioned to a solar orbit. More details: Apollo 12 - Wikipedia
Contrary to popular belief, no vehicle is ever outside the earth’s gravitational influence, or that of any other celestial body. Gravity has infinite range and even Voyager 1 (currently about 9 billion miles from earth) is still attracted by the earth’s gravity. However its kinetic energy ensures it will never fall back to earth. Voyager 1 - Wikipedia
In a simple two body example, which of course doesn’t exist, if you have less than escape velocity at any point with respect to the other body, then you simply don’t escape. You either orbit, or fall back to crash. The size of the orbit is indeterminate with the given data, since “escape velocity” is a variable dependant on distance.
Tris
>A matter of some debate, since by that standard, the Moon is orbiting the Sun. Which I don’t have a problem with, but some folks do.
Yes, I like this, the Moon is orbiting the Sun. It’s 400 times closer to us than the Sun is, and we see it progress around the sky 12 times in a year, so from the point of view of someone way north of the solar system, its path is a circle whose radius changes by 1/4% in about 11 cycles. That would look like a cyrcle.l
Don’t think so. My formula was for the height above the Earth’s surface, not the distance from the centre of the Earth.
I stand corrected; your equation is correct for altitude from the Earth’s surface. This doesn’t seem to be my day for getting anything right.
Stranger