anonymite

Hmm...

As far as my understanding of it goes, I think he's asking would a ship be able to leave the earth's gravity by maintaining a constant speed of 200 mph, regardless of force of gravity? I would think the answer to that question would be yes.

As far as I know, the top speed achieved in space shuttles is only about 17800 mph, and this is obviously not the initial speed when it takes off. I would guess then, if you could put a constant acceleration on an object greater than but in the opposite direction of gravity, the net force acting on it would be the differential in the direction leaving earth. I don't know if you could have enough fuel in the tank for a 200 mph trek, but I think you could leave earth at less than escape velocity as long as there's a continuous force acting on you that counteracts gravity.

Shadowy Man

Yes, the Space Shuttle goes up relatively slowly, at least compared to pretty much any other rocket. Remember, there is a human crew that needs to survive the launch!

From an energy conservation standpoint though, it seems like you'd need to be going faster than 200mph at some point or you won't leave the Earth.

capnkirk

Actually, for the case cited, there is an easier way to calculate it.

Think of it this way: How far away from Earth would you have to be for Escape Velocity (as defined above, the initial velocity required to escape Earth's gravity w/o application of additional energy) to equal 200?

If you start with the baseline example (EV starting at Earth MSL), as distance traveled increases, instantaneous velocity decreases. Further, for this particular curve, EVERY POINT on that curve represents the MINIMUM EV required for that particular distance from Earth.

So for any "sustained" velocity relative to Earth, there is a point on the baseline curve corresponding to the required distance one would have to have separated from Earth such that no additional energy would have to be added in order to escape.

A MORE THOROUGH EXPLANATION:

Newtonian physics defines the relationships between mass, energy, velocity, and inertia. Velocity and inertia are vector quantities (they have a direction component as well as a size or magnitude component). the Kinetic Energy (E) of a body is equal to its Mass (M) times its Velocity (V). Kinetic Energy (of an object) is also referred to as its Inertial Mass. So for an object of given mass, its Velocity and Energy are directly proportional. In other words, whether we talk in terms of velocity or inertial mass or kinetic energy, we can convert from one term to another simply by applying a multiplier (This is one of the things that confuses laymen when reading about this subject in that the need for the multiplier is implicit, and remains undeclared. Do NOT however presume from this that the EV vs. distance relationship is linear. It's not. What I am saying is that the "shape" of the curve does not change whether you express EV in units of Velocity, kinetic energy, or inertial mass.).

A projectile launched straight up from Earth begins with a finite amount of kinetic energy (Inertial Mass, Velocity). For the entire time of its flight, Earth's gravity drains energy away from it. The drain is worst at earth's surface; as the projectile rises and its distance from earth's surface increases, the drain decreases. It is in effect a race with gravity roughly equivalent to coasting up a hill. If the projectile starts with enough energy, it will never lose all its energy and will continue to get farther from earth forever. EV is the minimum value (expressed as velocity) to accomplish this, relative to the default starting point.

As one moves farther from the starting point, the EV from that point decreases. So, irrespective of the means used to get to that point, a body moving straight up will require exactly the same EV from that point as a body that started with the default EV and coasted to that point.

I hope that this little tutorial helps you to visualize the relationships between EV, V, M, E, and distance as they apply here, and from that can understand how to relate your various questions to the curve that Shadowy Man posted.

For Example: Easy Be's posting concerning the eventuality of applying a sustained acceleration of 1.00001G. That acceleration would have to be sustained long enough for distance and velocity to accrue to a point that it exceeds that which the baseline "coasting" object would have at that same distance.

True,but... The pertinent questions here become, how long would this acceleration need to be sustained to guarantee escape, and how much energy would be consumed in the process? The efficiency of this system would be extremely low (.01%) because the energy required to produce 1G (per time unit) serves only to preserve the velocity already accrued, leaving only .0001G for increasing that velocity. The presumption is that the 1.0001G figure is gross acceleration, rather than net. In this case, the net acceleration would be only .0001G and the time required to reach a velocity of 1ft/sec is greater than 5 minutes! To reach 200mph would take over 2 months!

Shadowy Man

Actually, I believe this is called momentum.

Demosthenes

yep, the kinetic energy is actually defined to be equal to one half of mass times velocity squared.

capnkirk

touche :notworthy ...as noted correctly by shadowy man's equations. Never argue with an astrophysicist.

Kintaro

I heard somewhere that the escape velocity was 11200 m/s.

If we count that g = 10 m/s (I know, 9.8065), it would go 11195m the first second, 11185 the next, it could go 600 km off the ground easily. Which i think clears it.

11200 m/s = 11.2 km/s = 7 miles/s = 25200 miles/hour.


It did look ridiculous in imperial measures, but converting it from metric gives it credibility.

Remember .... aww, screw it, here http://www.nasaexplores.com/lessons/01-049/


Google works wonders (speed escape velocity)

Answerer

Well, actual case is much more difficult. We have to find the escape velocity of the solar system, itself. Otherwise, rocket will fall back to Earth or the sun just after it just reach Mars or Jupiters. Come to think of it, we will need the Hohmann orbits to ensure our spacecraft stay in space without wasting too much fuels on our part.

Stephen T-B

I used to wonder why tv shots of a rising rocket show it taking off vertically and then beginning to cross the sky at an increasingly shallow angle.
This is my take: the rocket rises from a launch pad which, because of the Earth’s rotation, moves in relation to the rocket’s position in the sky. You thus have two moving points - not just one as the tv viewer is inclined to think. So, if you were looking up the rocket’s back end as it was taking off, you’d very soon find that it was no longer directly overhead. It is, in fact, travelling at a tangent to the Earth’s circumference, the angle of that tangent being greater or lesser, I suppose, according to the relative speed of the rocket as compared to the speed of the Earth’s rotation.
So a vehicle rising at a steadily and maintained 200 mph would be moving off at a very shallow tangent and would soon be overwhelmed by gravitational forces, and instead of going at 200 mph, it would be accelerating straight into Mr Harvey’s back yard, several miles away from the launch site.
Right?
(Of very slight relevance: in a snow storm I was travelling in a bus which frequently stopped. When we were stationary, the snow flakes fell vertically, but as soon as we began to move, their motion was slanted. The faster the bus went, the greater the angle of slant. I only worked out what was going on when I realised I was observing moving objects from a moving object. So if a snow flake appeared in the top left-hand corner of the window, by the time it had fallen a couple of centimetres it was in the middle of the window - and within another couple of moments it was out of sight, having passed behind the right-hand edge of the window frame. To me sitting in the bus, therefore, it appeared to be falling at a slant.)

Shadowy Man

Actually, it is because most of the rockets you've seen are probably going into orbit, so they need to turn over in order to do so.

I've seen several sounding rockets launch and they go straight up. In fact, even after the second stage burned out I could see the glow from the tail end and I had to bend my neck back so I was looking directly up; it was actually kind of unnerving. The sounding rockets will achieve an apogee of about 200 miles.