Does anyone know how much thrust is required to overcome Earths gravitational forces?

Answer 1

There is no set amount of "thrust" which is required to escape Earth's gravity as this thrust (a force) could potentially be applied over a great or short distance, causing totally different values for the work done by the force.
And of course, the answer would depend on the mass of the object the thrust is accelerating.

There is something known as the "escape velocity", the initial velocity an object must have as it leaves the surface of the Earth (or the massive body in question) in order to forever escape its gravitational pull. In this case, the force or mass is not a factor in whether or not the object ever resturns to Earth.
The escape velocity on the surface of Earth is about 7 miles per second (25,000 miles per hour).

Answer 2

The thrust does not matter so long as it is enough to keep the object moving away from the earth and accelerating (gaining speed) For example, the Saturn V rocket with an Apollo spacecraft on it weighed 6.5million lbs at takeoff. It generated 7.5 million pounds of thrust in the first stage to overcome that 6.5million pounds. However, at 2minutes and 30 seconds into each flight, the first stage cut off and the second stage started up with only 1 million pounds of thrust, but that was plenty enough to take it approximately 100 miles up where the 3rd and final stage with 200,000lbs of trust brought it up to 17,500mph and placed it in a near circular parking orbit of 100miles.

After nearly 2 orbits, the 3rd stage would re-ignite at a specific point and the spacecraft would be hurled in the direction where the moon would be 3 days later at a speed of nearly 25,000mph.

Now we will find out why 25,000mph.

To leave the planet earth's gravitational field, one must do work from the surface of the earth until you reach infinity to leave (have a rocket with a lot of stages on it!). That said work is the closed integral of the F vector dotted into the displacement vector or:

W = Energy = § F dot ds.

In this cace F = -GmMe/r^2 Where m = mass of rocket and Me = mass of earth.

So you integrate; § (-GmMe/r^2) dr from Re to infinity

Re = radius of earth.

That gives you:

W = GmMe/r evaluated between Re and infinity

W = GmMe/Re.

Now what did all of that work turn into? Kinetic energy. So the kinetic energy of the rocket is 1/2mv^2

KE = 1/2mv^2 = GmMe/Re the m's will cancel and multiply both sides by 2 to get:

v^2 =2GMe/Re

or v = SQRT(2GMe/Re).

We give this a special name Ve = SQRT(2GMe/Re) and that is called the escape velocity and it depends on the mass and the radius of the planet you are leaving--nothing else!

If you go plug in the numbers and do the rest of the math, you will find that Ve is approximately 25,000 mph for the earth.

Now go look up the mass and radius for Jupiter and see what you get!!!

Answer 3

Trust me on this one too: The escape velocity of Earth is 9.8 meters per second per second squared (that's 9.8 m/s PER second^2). Every conventional rocket and space shuttle hoping to launch itself straight into the sky needs to have the correct proportional amount of hydrogen fuel relative to its total weight that will produce a force powerful enuff to overcome that magic number. Feel free to verify this with any physics text or professors if you wish. (i hope this helps).

Answer 4

The better question is how much velocity, since the thrust depends on the mass. I think there's a pretty good answer at:
http://www.physlink.com/Education/AskExperts/ae158.cfm

Answer 5

I don't think I would trust Goku-san on this one.
Escape velocity for earth is about 7 miles per second.

Read:
http://en.wikipedia.org/wiki/Escape_velocity

Answer 6

a CRAPLOAD. Earth is friggin huge xD
No but if i was to make an estimate, I'd say enough to move a standard space shuttle to near lightspeed, which btw, is pretty DAMN hard.

Answer 7

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