how much thrust does it take to escape earth's gravity?
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.
927827
K so...I haven't done any Grav froces recently...but I'll try\n\nIm guessing use  Net force = ma\nThe net force is the Gravitational force, Gm1m2 / r^2 is = m2 4 (pie)^2 r/ T^2\n\nfor the 4 (pie)^2 r/ T^2    r is the distance form the planet's center, to the moon, for T, find how many second 40 days is.\n\nand the other r is the planet's radius\n\nm2 cancels\n\nsolve for m1, and yo uhave the mass of the planet.\n\nI guess plug in numbers after that, into the equaltion Fg = Gm1 m2  /  r^2\n\nm2 would be the 70 kg\n\nhope thats right
139268
One interpretation of the question is, why not just travel slower, continually applying thrust, e.g., a booster that goes at jetliner speed?  The reasons why that is impractical are:\n(1) you have to use a rocket since it must work beyond the atmosphere\n(2) with a rocket you carry all your propellant with you, unlike a jetliner which obtains reaction force from pushing back combustion products plus a large proportion of air\n(3) this means you don't have energy to spare \n(4) you have to put as much total (kinetic + potential) energy into the spacecraft whether you escape quickly or slowly\n(5) you want as much rocket impulse as possible to go into accelerating the spacecraft, and as little as possible spent just fighting gravity.  (Think what a waste it would be to hover on rocket power!)  This means high thrust for a short time to maximize kinetic rather than potential energy. \n\nSo you end up taking a modified "ballistic" approach, similar to firing the spacecraft from a cannon (which is all kinetic, no potential), but not so stressful and a little smarter.  You do all the accelerating during the first few minutes, then coast free of gravity.  You shape your ascent so your path is mostly vertical until you leave the atmosphere, to minimize energy wasted on air resistance.  In this phase you are providing some potential energy, but this can't be avoided.  Then you "level off" somewhat so your thrust isn't fighting gravity so much as you accelerate to escape velocity.
256497
around 1/6th that of earth so around 1.64m/s
1068202
Zero gravity is impossible on earth.  Actually, zero gravity is impossible anywhere.  Gravity acts through millions of light-years, holding galaxy groups together, so there's no way that you can eliminate gravity this close to a major mass like the earth.\n\nHowever - you CAN have a situation that is mathematically and physically equivalent to zero gravity, and that is free fall.  Jump on a trampoline, you'll spend a few fractions of a second in free fall, which is just like zero-G.  NASA does it with the Vomit Comet, which is a modified KC-135 aircraft that flies in parabolic paths to give astronauts-in-training a freefall experience.   NASA also has a (mis-named) test site called the "Zero Gravity Research Facility."   See the last link, below, for more info about this 500-foot tall drop test site.
745944
Loosely speaking, a black hole is a region of space that has so much mass concentrated in it that there is no way for a nearby object to escape its gravitational pull. Since our best theory of gravity at the moment is Einstein's general theory of relativity, we have to delve into some results of this theory to understand black holes in detail, but let's start of slow, by thinking about gravity under fairly simple circumstances. \nSuppose that you are standing on the surface of a planet. You throw a rock straight up into the air. Assuming you don't throw it too hard, it will rise for a while, but eventually the acceleration due to the planet's gravity will make it start to fall down again. If you threw the rock hard enough, though, you could make it escape the planet's gravity entirely. It would keep on rising forever. The speed with which you need to throw the rock in order that it just barely escapes the planet's gravity is called the "escape velocity." As you would expect, the escape velocity depends on the mass of the planet: if the planet is extremely massive, then its gravity is very strong, and the escape velocity is high. A lighter planet would have a smaller escape velocity. The escape velocity also depends on how far you are from the planet's center: the closer you are, the higher the escape velocity. The Earth's escape velocity is 11.2 kilometers per second (about 25,000 m.p.h.), while the Moon's is only 2.4 kilometers per second (about 5300 m.p.h.). \n\nNow imagine an object with such an enormous concentration of mass in such a small radius that its escape velocity was greater than the velocity of light. Then, since nothing can go faster than light, nothing can escape the object's gravitational field. Even a beam of light would be pulled back by gravity and would be unable to escape. \n\nThe idea of a mass concentration so dense that even light would be trapped goes all the way back to Laplace in the 18th century. Almost immediately after Einstein developed general relativity, Karl Schwarzschild discovered a mathematical solution to the equations of the theory that described such an object. It was only much later, with the work of such people as Oppenheimer, Volkoff, and Snyder in the 1930's, that people thought seriously about the possibility that such objects might actually exist in the Universe. (Yes, this is the same Oppenheimer who ran the Manhattan Project.) These researchers showed that when a sufficiently massive star runs out of fuel, it is unable to support itself against its own gravitational pull, and it should collapse into a black hole.
110744
Short answer: usam da inverse square law Jar Jar!\n\nLong answer:\n\nNewton simplified things by picturing all of a planets mass gathered in a single point in the center of the planet and the surface as a point on a certain distance from that center of mass. So the gravitationall pull (F) on a body (m) at a distance (r) from the mass (M) can be calculated using this equation:\n\nF=G*((M*m)/r^2)\n\nwhere G is the gravitational constant (6.67 * 10^-11 Nm^2kg^-2)\nEarths (equatorial) radius is 6.378 * 10^6 meters and its mass is\n5.974 * 10^24 kilogrammes\nSo lets test this on a dude weighing 100 kg:\n\n6.668 * 10^-11*((5.974*10^24 * 100)/4.068*10^13)\n==>\n6.668 * 10^-11*(5.974*10^26 /4.068*10^13)\n==>\n6.67 * 10^-11* 1.424*10^13 \n=\n979 N (what I got in my calculator)\n\nSo earth pulls on a guy, with the mass 100 kg, with the force of 979 Newtons. That means 9.79 N/kg, earths gravitational force (g). This should also mean that a mass dropped from a small height should accelerate at the rate of 9.79 m/s^2. It´s a simple experiment to do but where I am the value of g is actually 9.82. This is because I am closer to the earths center. Earth bulges out at the equator because of its rotation. I am farther north. But still a simple experiment can be done to validate this calculation. \nSo if the earths mass was 3 times greater than g shoud be 3 times greater. Right? You betcha. And I have already shown that moving closer to earths center increases g. Moving twice as far from the center should decrease g by half. Right? Wrong! Now what? We where so close. Oh man...\ng is reduced to a quarter if we increase the radius by 2. This is because of the inverse square law, a very important law in astronomy. It means that the apparent brightness of a light source is inversely proportional to the square of its distance. This also aplies to gravitational fields. Double the distance, divide by 4.\nSo lets put this all together. You should be able to do it yourself by now. I got 734 N or 75 % of the dudes weight on earth. And the reason he doesn´t weigh more than on earth is because of the inverse square law. Eventhough the new world is three times as massive we still moved twice as far away from the center. So take the weight of your astronaut, multiply by 3 and divide by 4.
146292
11.2 km/s with the following caveats:\n\n"Due to the atmosphere it is not useful and hardly possible to give an object near the surface of the Earth a speed of 11.2 km/s, as these speeds are too far in the hypersonic regime for most practical propulsion systems and would cause most objects to burn up due to atmospheric friction. For an actual escape orbit a spacecraft is first placed in low Earth orbit and then accelerated to the escape velocity at that altitude, which is a little less, ca. 10.9 km/s. The required extra velocity, however, is less because the spacecraft has already been accelerated to about 8 km/s."
1042618
weight = mass x gravity\n\nso...\n\nweight / gravity = mass\n\nthe gravitational constant on earth is approx. 9 m/s^2
78972
Very large rockets produce enough thrust to overcome the Earth's gravity. Eventually, they go so high up, that even with the rockets turned off, the Earth's gravity no longer pulls it back down.  Everytime you jump up, you're essentially 'leaving' the Earth's gravity, so think of it as 'jumping' with so much force that you keep going right into space.  Sorry for the watered-down explanation. You could probably get a more scientific responce by typing 'rocket propulsion' into a search engine.
365720
Because there are two basic ways to leave the planet.  One is to achieve escape velocity.  Throw a ball up into the air, it goes up and then back down.  Throw is hadrer (faster), it goes higher and then comes back down.  If you could throw it hard (fast) enough it would not come back, it will have left the earth.\n\nNow this speed is really fast, and very few things could withstand suddenly being hurled skyward at the speed, not to mention the force it would take to do so.  That is why we launch things by applying continuous force to them.\n\nBy constantly pushing against the earth we can keep accelerating upward against the force of gravity, and manage to leave the planet at a more gradual rate of increasing speed.
862938
Simply? Gravity is the force responsible for accelerating you towards the earth's center. Thus, when considering the forces exerted during launch, there is a constant m(9.8m/s^2) force directed straight down to the center of the earth. Dividing all forces out by mass, then no matter what acceleration into the heavens the rocket is given, it will face a near-constant 9.8m/s^2 (~ closer to 8.6 at altitude (not that much different from on earth)). What this means pertains to energy.\n\nThe rocket must get form one level to another, so the energy well decreases, in the sense of its kinetic energy must at least be (GMm/r1 - GMm/r2), where r1 is on the ground and r2 is at altitude. If you didn't factor in gravity, no force would be acting downwards on the rocket. Technically, this means any acceleration could place it at any altitude; i.e., flicking it would put it in orbit.
895548
Minimum speed for what?\nIf you want to leave earth orbit then you need to be over the earth escape velocity.
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!
804445
First you want to know the acceleration due to gravity of the planet Glob. At the surface the gravitational acceleration (g) can be determined by equating the weight of an object with the gravitational force law\n\nm*g = G*M*m/R^2\n\nwhere R is the radius of Glob, M is Glob's mass, G is the universal gravitational constant, and m is the mass of the object. Solving for g yields\n\ng = G*M/R^2\n\nNow, throwing an object into the air, and neglecting air resistance, we can consider conservation of energy. The kinetic energy of the object when it is released by the astronaut is completely converted into potential energy when the object reaches the top of its travel.\n\n0.5*m*V^2 = m*g*H\n\nwhere V is the initial speed of the object and H is the height to which it traveled, and g is the gravitational acceleration solved for above. Solving for V we find\n\nV = sqrt(2*g*H).\n\nThe satellite traveling in a circular orbit of radius D above the planet has its centrifugal force balanced by the gravitational force of the planet.\n\nFc = m*v^2/D = G*M*m/D^2 = Fg\n\nSolving for the speed of the satellite (v) needed in order to maintain orbit at radius D yields\n\nv = sqrt(G*M/D).
647242
A rocket doesn't have to fly a certain speed to leave the earth.  A rocket has an engine which means that it needs no minimum speed to get out of the earth's gravitational field.  Escape velocity is based on an object which will not be experiencing powered flight.  A rock thrown upward from the ground will need an initial velocity of 11.2km/s upwards to leave the gravitational field of earth.  A rocket needs to go just fast enough to continue moving (assuming a VERY large amount of fuel to continue engine burn for this duration).
540447
Let's just talk about Earth. The earth's diameter is 12756 km. Its circumference is pi x 12756 = 40074 km. The earth spins 1 round in 24 hours. That's 1700 km/h, or 472 m/s. If it rotates in an OPPOSITE direction, it's like the Earth is rotating at 944 km/s and suddenly stops. What happens? You fly through the air at 944 km/s. All the houses fly through the air at 944 km/s. Whole forests fly at 944 km/s. Even whole mountain ranges and mountain tips fly at 944 km/s. The ocean's waters fly backward at 944 km/s. Talk about a uber tsunami. I don't know what the friction of air against these object would do to them. Can anyone solve that? Even the Earth's crust might become detached with its outer mantle, I think. Lava breaks out everywhere, global earthquakes... In short: All hell breaks loose. Man, that's even worse than the Death Star. Superman actually does that to Earth in one of his movies. He's a major nut!