what is the speed of a rocket? HELP ME!!!?
957293
Rockets go at many different speeds depending on the application. A good example is the speed a rocket needs to go when leaving the earth and going into earth orbit. This is called "escape velocity". For rockets leaving the north pole where the spin of the earth is not a factor, the escape velocity is 11.2 km/sec (approx. 25,000 miles per hour).\n\nOnce out of earth orbit, rockets attached to spacecraft may increase speed to reach distant planets. Speed will stay the same until the rocket is ignited. Once the rocket is burning, the speed will continue to increase. Small rocket engines, like "ion engines", with very low but continuous thrust will keep the rocket accelerating for a very long time.
218504
Minimum speed that a spacecraft has is always gerater than the escape velocity i.e. over 11 Km/Sec. This means you can reach to USA under 20 mins or so!
368139
If the object has onboard power, then the speed is not really relevant to whether it will leave the atmosphere, just when. If you just fire an object upwards and it has no self-propulsion, however, then you need to calculate the change in gravitational potential energy (find out how far the atmosphere extends) and add the energy lost to air friction (this depends on velocity). Equate this to the object's initial kinetic energy and solve for initial velocity v. Your initial velocity will have to be larger than this.
767660
A rocket is subjected to four forces in flight; weight, thrust, and the aerodynamic forces, lift and drag.\n\nat rest I will denote the launch pressure by the symbol ps. The pressure exerts an equal force on all sides of the pressure vessel, the cone of the engine. Because the rocket is free to move along the launch mechisims, the pressure exerts a vertical force given by: \n\nFp = (ps - po) * A \n\nwhere Fp is the pressure force, po is the atmospheric pressure, and A is the cross-sectional area of the rocket tube. The net force acting on the rocket is the difference of the vertical pressure force and the weight W of the rocket: \n\nFnet = (ps - po) * A - W \n\nTo simplify the analysis, we are going to assume that the length of the launch tube is small relative to the length of the rocket. As the rocket moves, the pressure remains fairly constant inside the tube. In reality, there is a slight decrease in pressure as the volume increase, but we are going to neglect that effect to simplify this analysis. With a constant pressure, and constant pressure force, we can use a simple algebraic form of Newton's second law to determine the acceleration a, velocity v and the distance y which the rocket moves along the launch tube: \n\nFnet = m * a \n\nm * a = (ps - po) * A - W \n\nwhere m is the mass of the rocket This equation can also be written as: \n\nW * a = g * ((ps - po) * A - W) \n\nwhere g is the gravitational acceleration, which is equal to 32.2 ft/sec^2 or 9.8 m/sec^2 on the surface of the Earth. The value of g is different on the Moon, and on Mars. The weight W is equal to the mass times the gravitational acceleration. Now divide both sides by the weight to determine the rocket acceleration along the launch tube: \n\na = g * ((ps - po) * A / W - 1) \n\nFor a constant acceleration, the velocity v and distance traveled y are given by: \n\nv = a * t \n\ny = .5 * a * t^2 \n\nwhere t is the time. Using the length of the launch tube L, we can now solve for the lift off time - TLO by using the second equation: \n\nTLO = sqrt (2 * L / a) \n\nand then using this time, we can use the first equation to solve for the velocity at the end of the tube: \n\nv = TLO * a \n\nv = sqrt ( 2 * L * a) \n\nAs the rocket clears the end of the launch tube (t = TLO+), the air inside the rocket, at pressure ps, will rush out the exit and the pressure will equalize with free stream inside the tube. In reality, this will provide an additional amount of thrust T given by: \n\nT = m dot * Ve \n\nwhere m dot is the air mass flow rate out of the tube, Ve is the exit velocity of the air. For an air rocket, with no nozzle and small volume, this additional thrust is small relative to the weight of the rocket, and we will ignore this effect. \n\nThe velocity v at the end of the launch tube is used as the initial velocity in the flight equations to determine the flight trajectory of a rocket.
652370
Okay, I've thought about this and it's not possible to determine the rate of acceleration from the given criteria.  It is an "average" of 60 miles per second, BUT how many miles in the first second compared to the number of miles in the final, fifth second?  If that is known, you can project MPH from seconds 6 to 10 or the additional 1/4 mile.  How do you calculate the rate of acceleration in your theory?  How fast is the vehicle traveling at each point of the 1/4 mile distance?
573026
All 3 of Newtons laws of motion apply to a rocket, no matter the environment. Actually, All the laws of physics apply to a rcoket no matter the environment...it is not as if just on law controls how something opperates.\n\nNewton's 3 Law as well as the law of conservation of momentum work well to describe how a rocket engine accelerates a rocket.  These two laws apply both in an atmosphere and in space.
942655
Assuming that I understand the geometry correctly, the combine package-rocket will move off the cliff with only a horizontal velocity component. The package will then fall under gravity traversing the height of the cliff (h) in a time T given by\n\nh = 0.5*g*T^2\n\nHowever, the speed of the package-rocket system must be sufficient to cross the river in that time\n\nv2 = L/T\n\nConservation of momentum says that\n\nm1*v1 = (m1 + m2)*v2\n\nwhere m1 is the mass of the rocket, v1 is the speed of the rocket, m2 is the mass of the package, and v2 is the speed of the package-rocket system.\n\nExpressing v2 in terms of v1\n\nv2 = m1*v1/(m1 + m2)\n\nand then expressing the time in terms of v1\n\nT = (m1 + m2)*L/(m1*v1)\n\nsubstituting T in the first expression\n\nh = 0.5*g*(m1 + m2)^2*L^2/(m1*v1)^2\n\nsolving for v1, the speed before impact is given by\n\nv1 = sqrt(0.5*g/h)*(m1 + m2)*L/m1
117189
Depends.  (How many times does that answer show up?)\nThe rocket equation is\nV = Vex * log(m0/m1)\nwhere \nV is final velocity\nVex is the rocket exhaust velocity\nm0 is liftoff mass\nm1 is burnout mass\nThis equation says (e.g.) that you need a ratio of liftoff to burnout weights = e (2.718, base of natural logarithms) to achieve a V = Vex.\nThe equation is used as-is for a single-stage rocket.  For multistage (serial not parallel burns) rockets you sum the V values for each successive stage.\nVex of the Shuttle main engines, which are pretty state of the art for efficiency, is about 15000 ft/sec.  (You could do better but you'd be working with much more exotic, toxic and pricey chemistry than Shuttle's good old LH/LOX system.)\nSo the answer to the first question is: Every time you increase the m0/m1 ratio by a factor of 2.718 the burnout velocity becomes 15000 ft/sec larger, so there is no theoretical maximum speed.  However, you say "using current designs".  If in addition to using the Shuttle main engines this means using existing vehicles, propellant tanks, etc. then a lot of work would be needed to get the answer.  The Shuttle itself cannot go beyond low earth orbit, so you'd have to piece parts together to make the fastest practically achievable interplanetary (IP) vehicle.  You need ~35000 ft/sec to escape earth's gravitational field.  This means you need an m0/m1 of about e^2.3 or 10 just to get into IP orbit.  Now you get into tradeoffs.  How fast do you want to get there?  Slower means less rocket fuel needed, but more mass is needed to support a longer trip which means more fuel.  How comfortable must it be?  There are too many questions, too many trades, too much engineering to cover it all here.  I hope this gives you a start.
531134
Depending on the resources you have available you could do an experiement with rockets (small ones foudn at the store). With this you could measure the time in the air, distance traveled, speed would be difficult but you could get an average if you had the distance and time. And by changing the wieght/structure of the rockets would give you measurements on mass. You would have to keep the angle at which you launched them at a constant. (keep in mind the wind factor if out doors)\n\nOther wise if you are looking to only hit one of those topics I suggest testing the mass of objects using water displacement.
298421
The origin comes from classical equations of physics where the position, velocity, and acceleration of an object are given by equations in terms of time, specifically 't'. So when the rocket is airborne it's data points are tracked against 't' or mission time. So, for example, a rocket may pass through the upper atmosphere at t = 1,256 sec. This value is exactly zero upon liftoff though, meaning it has a negative value before liftoff. Hence you hear "t minus 10 seconds" or "t = -10 sec."
512134
A. First, do a quick magnitude check to see if you can use Newtonian physics or need to use special relativity. Your Newtonian velocity would be \nv = at \n  = 1g x seconds in a year  \n  = 10m/s x 3.15x10^7 seconds \n  = 3.15x10^8 m/s.\nwhich is greater than the speed of light (= 2.99x10^8 m/s), so you do need to use relativity.\n\nThe relativistic equation for velocity after constant acceleration is (source 1):\n v = c tanh(aT/c)\nusing aT from above, \n v = c tanh(3.15x10^8 / 2.99x10^8)\n   = 0.78c\n   = 2.34 x 10^8 m/s\n\nB. Now that you have v, use the special relativity energy equation to calculate your energy (see source 3)\nE = (gamma) x mc^2\n(gamma) = 1/sqrt(1- v^2/c^2)\n(gamma) = 1.61\nE = 1.61mc^2\n\nThis energy includes the rest mass, mc^2, which we don't need to add (it was there before we started), so the kinetic energy is\nKe = 0.61mc^2\n   = 5.48x10^19 Joules\nOf course, no propulsion systems are 100% efficient, so it would take more energy than this.  Rockets, in particular, are very inefficient, and would take many orders of magnitude more energy.\n\nC.  Nothing we have even comes close. For comparison, this is about 100 times the entire world's energy production in 2002.
577594
12.66 mph\n\nor 12.65558195 to be (more) exact!
594787
The speed depends on what you want to accomplish.  If a rocket pushes a craft slower than 5 miles per second (mps), the craft will fall back to Earth.  If the rocket gets the craft to 5 mps, the craft can orbit Earth in more or less a circle.  If a rocket gets a craft to 7 mps or more, the craft will leave Earth all together.
1135961
in order to solve this problem, you need to work part (b) first before (a)\n\nb): you know gravity (g=32.2 ft/sec) acts opposite the rocket, so calculate, using:\n\nv_final = v_initial - 2*g*y_1\n\nyou know that the final velocity is zero (max height) and you are given the inital velocity. the negative sign is to account for the opposite direction of gravity. move the equation around to solve for y_1 (height from tower to max. height). then add 100 feet (to account for the tower) and you have your answer.\n\na): since you know your maximum height (from part b), substitute that answer for h. [*NOTE: remember to subtract 100 feet from your answer in part b, because t=0 at the top of the tower] You now have a quadratic equation and solve for t using the quadratic formula (you should know what that is). your answer will be the answer!\n\nThis prob is easy only because it worries upto the max height. if we had to worry about the falling down part, well thats another story.\n\n** i already worked out this prob, so if you want to ask more questions/verify your answer, just post an additional response and ill modify my question to answer it. GL!
895548
Minimum speed for what?\nIf you want to leave earth orbit then you need to be over the earth escape velocity.
943563
I have no clue of any, but I know that the escape velocity of Earth (the speed of which it takes to escape Earths gravity) is 25,000 mph. Try launching six and a half tons 25,000mph!=)
750537
acceleration = v^2/r or v*v/r (this is m/s^2 so divide by 9.81 to get g's)\n\nwhere v is the velocity\n\nr is the radius\n\nBe careful, think about the one g the pilot feels from the pull of the earth. ;)