What is the escape velocity of earth in mach?
78749
the escape velocity is around 11 km/sec, or 25000 mph.  this number is computed without considering the effect of air resistance.  \n\nthe mach number is defined as the ratio of an object's speed to the speed of sound.  however the speed of sound changes with altitude, and so therefore the mach number changes with altitude even if the velocity is the same!\n\nhowever if you are launching some kind of projectile from the earth's surface, the muzzle velocity will be the escape velocity and that has a well-defined mach number.  the speed of sound in the atmosphere at the surface of the earth is around 340 m/s.  11 km/s is then around mach 32.
680534
Use the equation from the link below.  You will find that \nv/c = 0.14365, or\nv = 0.14365 c\nor\nv = 43,100 km/s\n\nThe object is moving away from the observer at this rate.\n\nBy the way, you probably won't get credit unless you can work out the math steps I left out.
316157
The correct formula to use is one of the simple equations of uniform linear motion.\n\nv² = u² + 2as\n\nu = initial velocity = 0\na = acceleration = 1.7 m/s²\ns = height = 500\n\nWhen rearranged\n\nv = sqrt(2as) = 41.2 m/s on the moon\n\nThe apple on earth will approach terminal velocity asymptotically due to air resistance. To prove the apple reached a faster velocity from this height on the moon is difficult without doing calculations involving air resistance. For a height higher than 500m it is trivial. The apple on the moon will reach a higher velocity before impact, whereas the apple on earth will still be limited to falling at its terminal velocity of 42m/s.
161923
Earth and all the other planets in the solar system revolve, or orbit, around the Sun due to the force of gravitation. The Earth travels at a velocity of about 107,000 km/h (about 67,000 mph) as it orbits the Sun.
433949
The Earth is rotating at about 1040 mph (~1670 kph) at the equator. The speed decreases relative to your distance to the (axial) poles, where it is effectively 0. \n\nTo find out the rotational speed where you live, simply multiply the equatorial rotation speed above by the cosine of your latitude.
1146651
You're actually right.  Furthermore, escape velocity is a function of distance from the center of mass (of the Earth in this case)\n\nEscape velocity applies to ballistic motion.\n\nV(esc.)=(2G*M/r)^1/2\n\nWhere G is the gravitational constant\nM is the mass of Earth (in this case)\nAnd r is the initial radius from Earth's center from which you are computing escape velocity.
925006
Since the magnitude of the acceleration is v^2/r, and the radius of the Earth is approximately 4000 miles (21120000 ft), you would have to solve the following equation: v^2/21120000=32\nv^2=675840000\nv=25997feet per second  or\nv=4.92 miles per second
1004785
m1*a1+m2*a2=(m1+m2)a\n    47*9.8+6*10^24*x=(6*10^24+47)*x, because final  acceleration after landing does not change for the mass of earth is enormously big. \nThen, 47x= 47*9.8 and x=9.8. Therefore, the acceleration of earth towards the student is 9.8m/s^2
301896
the earth's orbital speed is 67 000 miles per hour.
256767
I'm not sure I really understand your question.  Are you how fast a particular point on earth is moving?  If that is the case it's not an easy answer because it depends where you are on the planet.  If you are on one of the poles then there is very little rotational speed.  If you are on the equator then your velocity is much greater.  Think of it like a CD in a CD player.  Also you would need to factor in the speed of the earth as it travels around the sun sometimes spinning in the same direction the earth is traveling resulting in a greater total velocity and sometimes spinning in the opposite direction yielding a lower total velocity.  The final analogy would be to tie a CD player to a string and swing it in circles while the CD is playing.  The speed of a single point, other than the center, of the CD is constantly changing.  However, it's still not even that simple due to the earth's eliptical orbit and varied velocity.
1174553
You misinterpreted something. By your theory, when I jump up from the ground, my final velocity is equal to Earth's escape velocity, which is around 9 meters per second, I think. Probably more. But either way, that is much faster than I can possibly go.\n\nAlso, there is NO law which says that velocities higher than c are impossible, it just says that velocities EQUAL to c are impossible.
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.
730409
The sun and earth exert equal gravitational forces on each other. If you ignore all the other interacting bodies, the sun and earth rotate about a common center of mass (which is actually located inside the sun). Contrary to what one answerer said, the earth is not in a "tenuous" or precariously balanced orbit. Our velocity in solar orbit is about 18.5 miles/sec. If we added (or subtracted) a bit of speed, say 1 mile/sec, our orbital shape would become more (or less) energetic and more elliptical. We might not like the climate change but we wouldn't fly off or crash into the sun. The velocity and the distance between the bodies determines the orbit. Change one or the other and you have a different orbit. Remove all velocity and you crash into the sun; add a huge velocity increment and you escape. (Note that if you're in a circular orbit you are traveling at escape velocity * 0.707, so that tells you how much faster you need to be going to escape.) This intrinsic stability enabled early experiments in putting objects into earth orbit before precise velocity control was available. The orbits weren't beautiful (and earth orbits will always decay due to atmospheric drag) but they were otherwise stable.
981532
If you assume it is possible and of course it isn't but.... \nIf we also assume that there is no terminal velocity and that our momentum would take us past the centre and carry us to the other side. our velocity would be zero at the start and at the end and would reach maximum at the centre it would be V=(T x G) / D where V is velocity T is time G is acceleration due to Gravity (9.8 metres per second per second) and D is distance (12,756.32 kilometres). By moving the various components in this equation you should be able to find all the needed figures. The total time would be determined by found by finding the time needed to reach the centre of the Earth and then doubling it as the time needed to decelerate is the same as acceleration
1023177
It's more or less as most of the preceding answers say. But it applies to escape from any body, not just earth and not just planets. For instance, the sun also has an associated escape velocity (617.2 km/sec at the surface). And it applies to a launch in any direction away from the body, not only vertical. As the 4th answer says, escape velocity decreases with distance from the body. Interestingly, because of conservation of energy, the velocity of an object launched at escape velocity remains equal to the escape velocity that prevails at whatever distance it reaches.\nA rocket achieves escape velocity by expelling the exhaust of its burning propellant rearwards at high velocity. It is most practically done by launching along a path that is initially vertical to minimize time spent in the dense atmosphere and consequent energy loss to air resistance, and then gradually changing to a more horizontal path to minimize the loss due to gravitational acceleration. Ignoring these loss factors, the velocity of a rocket is described by the "rocket equation", \nDeltaV = Vex * log(M0/M1), \nWhere DeltaV is velocity change, Vex is exhaust velocity, M0 is liftoff mass and M1 is burnout mass.
903626
Because you are really solving a differential equation:\n\nSee the bottom of my answer for a non-calculus answer.\n\nApologies in advance if you haven't had any calculus yet, and are just working with the formulae without explanation of where they come from.\n\na = acceleration, v = velocity, h = distance, g = gravitational acceleration of the Earth,  v0, h0 = initial velocity and height.\nI use "h" instead of "d" to avoid confusion with the differentiation operator.\n\na = g = dv/dt        =>        dv = gdt  \n\nintegrate to get\n\nv = gt +v0 = dh/dt      =>     dh = gtdt +v0dt\n\nIntegrate to get \n\nh = .5gt^2 + v0t + h0\n\nNote the coefficient ".5g" for the t^2 term. This is due to the general form for integration of a polynomial - specifically the  integral of t is 1/2t^2.\n\nNon-calculus answer: (note that this only works for this specific case of constant acceleration).\n\nImagine you start at v =0. Drop a rock. How fast is it moving after 1 second?  v1 = gt = 9.8 m/s\n\nHow far has it travelled in 1 second? \nd = vt; but what is v? v was zero when we started and 9.8 after one second but it changed the entire time.\n\nIn this case we see that the velocity started out at 0 and ended at 9.8 m/s. So the average velocity is 1/2 of (0+9.8) = 4.9 m/s.\n\nUsing the average velocity we see the rock has dropped\n\nd = vt = 4.9*1 = 4.9 meters.\n\ngeneralizing we see that d = 1/2(0+at)t = 1/2at^2
303287
At 90 mph you could escape the Earths gravitational pull, assuming that you have some sort of rocket constantly pushing you through the gravitational field.\n\nIt's not a simple thing to work out, as once you leave Earth the planets are moving, so the quickest you could do the journey in would be making sure you leave Earth at the point when Mars is coming towards you, in October 2005, Mars was approching Earth at a speed of 23,500 mph!!!\n\nBut, if you go off just the shortest distance possible, and the longest distance possible between the two planets, then you are talking a journey time of approx. 42 and 317 years respectively.