Force to escape gravity?

How would i calculate the force needed to escape earths gravity, if i have the x mass of an object? I think i would need two vectors, one for force and one for gravity. maybe one for acceleration too?

thanks

I am usually a bit more meticulaous that I am going to be for this post. I am in a hurry. Sorry in advance for typo's and errors in my hurried thought processes. Use qhat I say as a starting point for further thought.

Your question is a bit more complicated that you think. You have to apply a force for some amount of time to achieve a velocity.

To escape the Earth's gravitational field, you need to achieve escape velocity, which varies with how far from the Earth you happen to be. A NASA site can probably provide a table showing escape veloctiy versus ditance from the Earth.

I think escape velocity is about 25,000 Miles per hour at the earth's surface. I also think that escape velocity refers to movement parallel to a radius vector.

The farther you get from the Earth, the less the required escape velocity. I think that NASA has a web site with a lot of data on this subject.

To do the calculations, you have to work with the equation Force = Mass * Acceleration for the space ship or projectile. You also have to subtract Force = G*EarthMass*Mass / D^2 to account for gravity pulling you back. You use Velocity = Acceleration * Time to determine current velocity.

Using the above formulae, you can work out the velocity at various distances from the Earth. When the velocity equals escape velocity for that distance, you have your answer.

Unfortunately the problem is even more complicated. In practice rocket fuel weighes a lot. As you burn it, the space ship gets lighter. This is significant enough that it must be considered. Next, rockets are usually launched in the direction of the Earth's rotational motion to gain some velocity from that motion. There are probably other complications.

Escape velocity at the earth's surface is:

1.12*10^4 m/sec = 6.95 miles/sec

thanks, i had no idea that it was that complex. I am doing vectors in physics now, and i knew about gravity, that the accel due to gravity is 9.8 m/s, and we just learned the equ f=ma. thanks for the information you provided. i think that if i do a little more research, i should be able to figure it out.

Quote:

---

Unfortunately the problem is even more complicated. In practice rocket fuel weighes a lot. As you burn it, the space ship gets lighter. This is significant enough that it must be considered. Next, rockets are usually launched in the direction of the Earth's rotational motion to gain some velocity from that motion. There are probably other complications.

---

I am not particularly concerned about this, becuase i am thinking about it more theoretically. say there was a 1cm^3 lead cube, and it had "Magic powers" that would push it up, how much force would it take.

I have another question. Is escape velocity dependent on mass? i wouldn't think that it would be, because gravity pulls equally on all objects, but i am not sure.

Anyway, thanks for the help Guv, and Jim. I will try and work on it more and let you know if i figure out the answer. :D

Escape velocity is dependent on mass of the planet or star you are trying to escape from and the current distance from the center of the planet or star.

I am pretty sure it is also dependent on the direction of the motion. I think the velocity component parallel to a radius vector must be equal to escape velocity. I am not certain of this, but it seems reasonable.

The mass of the escaping object affects the force or energy required to achieve escape velocity, but the escape velocity is not dependent on this mass.

An instantaneous force F > gravity opposite to the gravity would bring two objects apart. Think of it as vector sum between two vectors in opposite directions.
gravity = G * m1 * m2 / r^2
G being the gravity constant: 6.67E-11 Nm^2/kg^2
m1 and m2 being the mass of the objects and r the distance between their mass centers.

to escape earth's gravity field you'd need a force of approximately your mass*9.8m/s^2

thanks. I am starting to understand this more now. I am new to physics, and i am wondering, what is the "radius vector" that you guys are talking about?

radius vector, I think Guv refers to the vector difference between the objects mass centrums.

Assuming an escape from a spherical planet or star, imagine a line from the center of the sphere to a point on the surface and continuing outward.

Such a line defines a direction relative to the sphere. This is the direction I was trying to describe when I used the term radius vector. This is a piece of jargon I made up for this thread. I do not think it is a technical term for this subject matter.

People in Physics prefer working with axes that are perpendicular to each other. The radius vector is from the center of earth (a sphere) outward. Any motion perpendicular to this is tangential and a waste of energy. The minimum about of energy to escape a planet's surface will be when all the motion is along a radius.

Speaking of force to escape a planet's surface is ackward because as kedaman wrote, the force of gravity is changing with distance from the planet's center plus if you are in a rocket, the mass of the rocket is changing with time, (fuel is burning). You also need to know the exhaust speed of the hot gases that is pushing you up into space. Mathematically, this is solvable but doing it manually is not pleasant. I recommend setting the equations up and letting Maple or Mathematica solve the bugger. You could then plot force vs time or rate of fuel burning or any other way that appeals to you.

For this reason most people ask "what is the escape velocity?" and use Conservation of Energy to find the answer. As Guv said, the escape velocity is independent of the mass being hurled in to space. When this speed gets up to the speed of light, we have a black hole.

Hope this helps.

ackward or not is for the other issue, I think Sail wanted explanation for a simpler issue, the instantaneous force. The acceleration of an object could be described as
a=SF/m
The acceleration vector a is the sum of all forces SF on the mass m.
In this case, the force needed + gravity (opposite to each other) sums to a acceleration vector opposing gravity, that is >0 will separate the objects (instaneously)

ok, based on my figuring, and what you guys have said so far, this is what i have:

- I can't use the opposite of acceleration due to gravity (g = -9.8 m/s^2) because, when increasing altitude, the gravity decreases, so it must be a squared function?

- The mass will not affect the acceleration due to gravity.

- kedaman (sorry, as i have said, i am new to physics) is g = 6.67E-11 Nm^2/kg^2 the constant that i would use against the force pushing against the earth? so would i subtract g from the force of the rocket engine?

- Another question. I may be missing something, but wouldn't this equation, in it's MOST BASIC sense basically be subtracting two vectors. the vector of the force of the rocket subtracted from the force of gravity pulling on it?

Sorry if i have been unclear. And thanks again for all the help! :D

Sail, as Kedaman said earlier the force of gravity was a function of the distance from the center of the Earth F = G * M * m / R^2. G is Newton's universal gravitational constant. G is not a vector. The entire gravitational force equation is a vector pointing towards the center of earth, F down. m * a = sum of all the forces = F up - F down.

When you want to start adding or subtracting things, make certain everything has/have the same unit(s). So no, you can not subtract G from the force of the rocket. But as you said in the most basic sense, you do want to take the force of the rocket and subtract the force of gravity to find the net force. Just make certain the forces are along the same axis.

>- The mass will not affect the acceleration due to gravity.

The acceleration will be reverse proportional to the mass; F=ma

>- I can't use the opposite of acceleration due to gravity (g = -9.8 m/s^2) because, when increasing altitude, the gravity decreases, so it must be a squared function?

it's a differential equation, it doesn't mean it's useless, but it suggests that you could have use of another formula. This is what the others have been suggesting, but based on your original question about instaneous force, I think you want explanation for something simpler, Newton's a=SF/m

>- kedaman (sorry, as i have said, i am new to physics) is g = 6.67E-11 Nm^2/kg^2 the constant that i would use against the force pushing against the earth? so would i subtract g from the force of the rocket engine?

G, the gravity contant is not a force, force has the unit newton=kilogram meter per second^2, G is the gravitational contant that tells you how much force is due to gravity, it's a natural constant that you have to calculate with for all forces due to gravity.

>- Another question. I may be missing something, but wouldn't this equation, in it's MOST BASIC sense basically be subtracting two vectors. the vector of the force of the rocket subtracted from the force of gravity pulling on it?

In MOST BASIC sense, the most fundamental equation you need to know is the force balance equation. SF=0
from which you can derive a resultant force equation a=SF/m on the mass m.
Note that I use bold text for vectors. The greek sigma letter stands for "sum of", and the expression sigma F means the sum of all force vectors (not difference) on an object, also known as resultant force. The resultant force will determine in which direction the mass will accelerate, because F=ma. Looking at that equation you see that the mass will accelerate proportional to the force, but reverse proportional to the mass, that is if an object is heavy, it's harder to set in motion/stop/generally accelerate

Since F is a vector unit, two opposite vectors will take out each other, the sum of two opposite vectors will be 0, thinking about differences will confuse you at times, so it is easier to think of F as a vector and resultant vector of as a sum of them.

thanks a lot kedaman.

well, after thinking about it for a while, i basically get the ideas behind it, but, i don't understand how to put it all into an equation.

my physics teacher will be helping me with it a few chapters from now. thanks for all the help everyone! :D:D:D

Basically this is a problem which might require numerical approximation methods instead of being amenable to analytical solution.

Assuming that you know the rate at which the rocket fuel burns and the amount of force per pound of fuel consumed, you can work with what happens in short periods of time.

You know the initial height, velocity, and weight of the rocket. Initial height and velocity are probably zero, although distance from the center of the planet or star might be easier to work with than using zero as height at the surface. The gravitational constant (G), mass of the earth, and approximate Earth radius are known.

With the above information and the Force equations, you can calculate the velocity and height a short time after the start of rocket firing. You now have a new rocket weight (some fuel has burned), a new height, and a new velocity. You can use the equations on the new data and figure out where you will be in the next time increment. The equations have been posted earlier. I think the pertinent ones are the following.

NetForce = RocketMass * Acceleration, or Acceleration = NetForce / Rocketmass

FuelForce = Function(FuelParameters, Time) This I have no idea about, but it must be known to NASA for a given rocket fuel and an assumed rate of burning.

GravityForce = G * RocketMass * EarthMass / Distance^2, where distance is distance from the center of the earth.

NetForce = Fuelforce - GravityForce

NewVelocity = OldVelocity + Acceleration * TimeIncrement

NewDistance = OldDistance + OldVelocity * TimeIncrement

NewRocketMass = OldRocketMass - MassConsumedFuel

At each step in the calculations, check calculated height versus a table (or formula) giving escape velocity as a function of distance from the center of the Earth.

thank you very much Guv. that helped clear things up. :)

Sorry to jump in...but I was just wondering...im in Grade 10 so I guess i dont know anything compared to all of you, but i was wondering whether gravity can be defined as a constant. Like can you say that the gravity for Earth is <a certain number> and the sun is some other number? Or is it far more complex than that?

gravity depends on the distance to the mass. If you stand at earth surface you will weigh more than if you stand in an airplane 5 km above the surface. The gravity is inverse proportional to the quadratic distance, which means you weigh 4 times less at a 2 times distance from earth centrum.
Furthermore gravity depends on the masses, if you stand on jupiter you will weigh a lot more, that is proportional to the mass of jupiter. Similar if you have mass of 80kg you will weigh more than your brother with 60kg mass.
Therefore this formula:

gravity = G * m1 * m2 / r^2

well, so according to that, would it be true, that austronauts are not really "weightless" when orbiting the earth?

There's no point in universe where earth gravity is 0

WOW, i never knew that. that is pretty incredible since at time the distance between anything can be billions of light years!

the force required to break free of earth's gravity varies with the distance to the core and the shape of the rocket. The downward acceleration to eartion is 9.8m/s approximately, there is a specific formula, but is has no meaning for small mass. The upward component of the force must be greater than the attraction to erth.

let:
FAv = vertical component of force applied
Ff = Force of friction = K v^2, where K is a constant depending on the shape of the object and v is the velocity
M = mass
g = gravitational constant

for lift-off to occur:
FAv - Ff > M * g

most simple calculations assume Ff = 0, for a perfect ball it is really close to 0 any way. To be really accurate (for large bodies) you should find the proper formula for determining gravitational attraction and recompute several times (if in a loop or sometime0 to be more exact.


PS:

ah-ha! there is the formula:
gravity = G * m1 * m2 / r^2 <from kedaman>
byut does anyone remember the value of G to get m/(s^2) ?

for really high speeds you should recalculate M for the gain in mass of an object as it approaches lightspeed <see Theory of Relativity>

why would it not work for small mass?

The gravitational attraction formula works for all masses, but the difference is so minimal on small masses that the correction is beyond your your percision when measuring factors (eg. mass, distance, etc.). Eg. the attraction/acceleration maybe actually 9.800000123 m/s^2 instead of 9.8. The difference becomes important when the mass approaches the mass of a small planet or moon and larger. Same thing goes with gain of mass due to acceleration; its is only important as an object appraoches light speed.

Ahhhhhhh....i see. thanks a lot for the info.

Previously posted by somebody: Earth’s gravity is never zero. This is in theory. In practice: At great distances it becomes so small that it cannot be measured (see below).

If you are falling, you are weightless. This is difficult to explain, but imagine standing on a scale bolted to the floor of an elevator which is moving down. As the elevator moves faster, the scale will register less and less weight. If the elevator moves fast enough, your feet will no longer be in contact with the scale, at which point the scale will register your weight as zero.

An astronaut in orbit is weightless for a similar reason. A passenger in an airplane is not weightless, because his motion is not exactly synchronized with gravity. The passenger in the airplane will seem to weight less than he does on the surface due to being farther away from the Earth. Assuming the radius of the Earth is 4000 miles and the airplane is 5 miles above the surface, a 200 pound man will weight about 199.5 pounds. This calculation is based on the ratio between 4000^2 and 4005^2.

A billion miles from Earth, a 200 man would weigh .000 000 003 20 pounds, assuming that the Earth is the only object exerting a gravitational force on him. A billion miles is only about 1.5 light hours, which is not even outside our solar system (it is well inside the orbit of Pluto). The gravitational force diminishes by a factor of 4 every time you double the distance. At millions or billions of light years, it is damn small.

1/x^2 tend to go to 0 when x goes towards infinity. But it is false to assume that it is 0 at a certain distance x, it goes closer and closer but never hits 0. In practice at great distance that is, you might dismiss it as a term in a calculation if the effect on the result is unnoticable.

The Earth moves about 67,000 miles per hour in its orbit around the sun. At the Equator the surface is moving about 1000 miles per hour due to rotating once per day.

NASA has to take these velocities into account when it sends a space craft to mars, the asteroid belt, or elsewhere beyond Earth's gravitaitonal field.

For a Mars mission I imagine they have a computer doing constant calculations of the positions of Earth and Mars for hours or even days before launch time. Even while the space craft is stationary on Earth, it is moving relative to Mars.

Astronauts in training, practise weightlessness in aeroplanes, I read. The plane describes a precise parabolic arc, and at a certain point, the passengers are in free fall in a free fall plane, hence the weightlessness. I would only do this with a pilot I trusted fairly well, ie I'd get a PI to make sure he wasn't bonking my Mrs and wanted me to die from fright.

But to answer your question, I remember vaguely fighting this in Applied Maths I in 1974...

I found an equation for Ve (escape velocity) in an encyclopeadia, but I do remember the maths to get to it were complicated. All posters above are right, with the equations, but it's the combination of them all tht gets to your answer. I can't remember how to do that, but, nonetheless, heres an eqn for Ve...

Ve= sqrt(2GM/r)

from which it appears that Ve depends only on the planet's mass, not the escapee's.