What are the three speeds to get to the orbit (1 and 2) and then to the space (3)?
539111
There is a problem with the previous answer... the earth's escape velocity is around 25,000MPH... that number is based on an object without any propulsion, mankind has yet to develop a rocket or other propulsion system that can reach that speed... just wanted to clarify that ;)  we can't hit mach 30 (mach I is 720mph, 22,500mph is about mach 30) yet, but we can leave the earth's gravity field because of continuous propulsion (lots of rocket fuel)
145934
You need to go from 0 to 11.2 km/s (earths escape velocity) and you need to overcome earths gravitational pull, which is 9.82 N/KG, in order to do so. You are constantly being accelerated downward with one G (9.92 m/s^2). If you would create a force upwards of one G you would hover (0 G). Let´s say you had a rocket engine that could thrust your spaceship upwards with 2 Gs. How long would it take to reach escape velocity? You use the equation t=v/a where v is 11200m/s and a is 9.81m/s^2 and get 1141.7s. You need to accelerate for 19 minutes to reach escape velocity. Would be sweet if you could do that but it is a long time to keep an engine running. Let´s accelerate twice as fast. Put a as 20m/s^2 and you´ll get 560 seconds or 3.3 minutes. \nOK. So now we´re cooking. How much thrust do you need to get this kind of acceleration? It´s another simple equation: F=ma. We have a (20m/s^2). But you also need m that is the mass of the spacecraft. This we can only guess but lets say a tonne (1000 kg). Then it is pushing downwards onto the launch pad with the force of 1 G wich is 9810 N (9.81x1000). We need to overcome this by adding another G to our thrust so a is 3Gs or 29.81m/s^2 (20 + 9.81). Thus we get 29810 N of thrust for 560 seconds to kick that 1 tonne payload into outer space. That´s 40.5 horsepowers to you non-metric folks (go metric already. A pound isn´t alwasy a pound is it? So measuring thrust in pounds isn´t valid). "But wait. Doesn´t that sound a little too sweet? 40.5 hps? My old Dodge can do better than that". That is correct. You also need fuel to make thrust. The fuel adds weight to the spacecraft which means we need more thrust. Whch means the need for more fuel. Also the craft is exposed to alot of drag from the air. More fuel. More weight. The farther you get from earth the less the value of a G gets but the spacecraft uses up fuel which makes up for the amount of force needed for acceleration. And so on. So this little thought experiment can only take us so far but I´ll stop here anyway. Now you got an idea atleast.
620442
90ft from home to first, 90ft from first to second, 90ft from second to third, and 90ft from third to home again...\n90+90+90+90=360
626127
V = (v + u) / (1 + uv/c²)\n\nTry that. The velocities are quite high so relativistic. Haven't tried it yet though.\n\nNevermind, the velocites aren't nearly high enough to be relativistic. If it was km/s it would work.
864791
A.) 3.02891 seconds\nB.) 290.1 m/s\n\nTrust me... =)
1102478
It varies between 30.287 km/s and 29.291 km/s with 29.783 km/s being the average\n\nThe reason for the varying speeds is that Earth's orbit is not perfectly circular.  At some times, it closer to the sun, causing the orbital velocity to become faster.  As it moves further away from the sun, the velocity gradually decreases to the minimum before beginning acceleration again.
664808
The shuttle usually operates at an orbital altitude of between 200 and 350 miles (careful! NASA often uses nautical miles for shuttle statistics), depending on what the mission and payload requirements for the flight are. A circular orbit at these heights implies an orbital speed of about 17,000 miles per hour. If the orbit is elliptical, the speed will be slightly higher, but not by much.
752738
The 120N force on the astronaut accelerates him/her at a rate of 120/79 m/s^2. In 0.42 s, this would accelerate the astronaut to a velocity of 0.42 (120/79) m/s.\n\nSimilarly, the satellite accelerates in the opposite direction to a velocity of 0.42 (120/660) m/s\n\nThe relative velocity of the astronaut and the satellite is \n0.42 (120) (1/79 + 1/660) m/s\n\nMultiply this by 84 seconds (which is equivalent to 1.4 min), and you'll have the requested distance. \nNote: This is the distance the astronaut will be from the satellite IN ADDITION TO the length of the astronaut's arms. We started counting the time when the astronaut had straightened his/her arms.
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!
256497
around 1/6th that of earth so around 1.64m/s
933881
t1=s1/v is time before yodeling, s1=20 miles, v=168mph; s2=t2*v2 is path of yodeling, t2=3/60 hours, v2=98mph; s3=t3*v is path of the bus after yodeling, t=t1+t2+t3=1 hour or t3=t-t1-t2, s=s1+s2+s3 is total path of silent+yodeling+shut_up phenomena; thus u=s/t is average speed of the bus;\nu=(20 + (3/60)*98 + (1-20/168-3/60)*168) / 1=\n=164.5mph
471913
The daily path of an object on the celestial sphere, including the possible part below the horizon, has a length proportional to the cosine of the declination. Thus the speed of the diurnal motion of a celestial object is this cosine times 15 °/hr = 15'/min = 15"/s, i.e. (compare angular diameter):\n\nup to a Sun or Moon diameter every two minutes \nca. four seconds for the largest planet \n2000 diameters of the largest stars per second
984027
An ion engine will get you there the fastest. You get the biggest delta-vee per kg of exhaust. \n\nProblem is, in general you only get one encounter with a specific planet once each solar orbit, and you can only get a kick so large each time. I suppose you can hypothesize a super convenient planetary alignment that could shave off some time, but you might spend a long time waiting for that particular configuration.  A lot of really smart people spent a lot of time developing the computational tools to come up with the VEEGA and VVEJGA options.  I doubt the population here is going to come up with anything a lot better.\n\nYou can read more about the DS1 ion propulsion experimental system here:\nhttp://nmp.nasa.gov/ds1/tech/ionpropfaq.html\n\nThe first science mission using ion propulsion is the Pluto bound New Horizons.  Thanks to this tech, it will reach pluto in 10 short years and reaches Jupiter for a gravity assist in only 1 year.\n\nhttp://pluto.jhuapl.edu/mission/mission_timeline.html
449850
3,You can walk in X axis direction(horizontal),the earth can move you also in Y direction(by turning or wind),and the solar system can make up for the Z direction(up or down).But you would not FEEL the Y and Z even if present they are small so you really would feel only one but three would be there.three=X,Y,Z. \n\nOR If we'd blow you up with a bomb you'd FEEL all three at the same time!!! probably momentarily too!!!\n\nthe result would be you'd travel a distance(AND a direction represented by XYZ) in a strait line to somewhere.\n\nPearlsawme:how would you explain to somebody where to go if you had to give directions. Force=f(x)+f(y)+f(z] if i give values to XYZ you'll know where to go.The relationship by which the alignment or orientation of any position with respect to any other position is established.\nApril 16:GOOD POINT YOU WIN Pearlsawme but representing the direction as to take XYZ in consideration.Also just to be clear to others,lets not forget that a force implies an acceleration, f=ma,after the initial impulse you would still move in that direction without any additional force.Example tennis ball in space.\n\nChinu:your explanation sounds logical(?) enough but i have to wonder if parallel dimensions really exist.Personally being a realist i think they are seeing some sort of image or a fast jump of the electron(or both)...but I'm not qualified to say.
696298
1st way :-\ndistance = speed x time = 55 mph x 3 h = 165 m\n\n2nd way :-\ngiven that d = 55t , so on putting t = 3hrs , we get d = 165 m
24866
The Earths orbit is 93 million miles from the sun, so the diameter of the earths orbit is 186 million miles.\n3.14 x 186 million = 584 million miles for 1 orbit. This is an approximate value.
819169
impulse = change of momentum ( delta P) = Fx( change in time)\n\nAstronaunt's initial momentum (P) relative to the satellite is zero because they are in contact.\n\nSo his new momntum\n\nPf-Pi=Fx(change in time)=90x0.54=48.6 kg*m/s\n\nhis velocity = 48.6/76.0= 0.639 m/s\n\nthe satellite also has P= 48.6 kg*m/s (conservation of momenum)\n\nso the velocity of satellite = 48.6/635 = 0.0765 m/s\n\nas a result they are drifting apart at \n\nv=0.0765 + 0.639=0.7155354331 m/s\n\nso at t=1.1x60=66 sec they are \n\nd=66x0.7155354331=47.2 m  apart