what id the trigger force that make the ejected object return to earth after it has lost energy?
365397
OK, relativity DOES NOT come into play here, because the velocities involved are not remotely high enough.\n\nYou are talking, it seems, about an object being projected up into the air and falling back to earth.  This must occur below escape velocity or it would leave the earth.  Relativistic speeds are far, far, far, far higher than that.  So, your question is in fact not related to relativity in any way, as much as you seem to think it is. SO I will answer your question 'normally'.\n\nWhen the object is projected upwards, it possesses a certain amount of kinetic energy whcih, as it rises, it converted into potential energy which increases with height.  When the KE finally runs out, the object will peak and as gravity, which exists by the way, will pull it down the PE will be converted back into KE and thus motion, in a downward direction in this case.\n\nAll I can say after this is to stand under it and maybe after it hits, you'll understand a little better.
393875
The way I see it.... u calculate its impulse with the ground.... mass1*velocity1 (pre-shock)=m2*v2 (after shock) then caculate it's potential energy by newton's law of gravity using v2....(god I'm being a GEEK!)
1045984
No one knows the answer to that. However, matter seems to cause a "dent" in space. The greater the mass of this matter, the greater the "dent" it creates. Other matter (with less mass) gets influenced by matter with larger dents. This causes small matter to orbit and free fall toward big matter.\n\nHowever, is this "dent" magnetic or electric? No one knows. However it is a fascinating thing to ponder!
641452
opposite force being applied
728806
flying faster than the gravitational pull that is pulling you back
442513
The gravitational force holds it into a spherical shape.
1183764
But...blasting matter in the direction of gravity's pull IS propulsion! :)\n\nYou can't get any "work" without expending energy of some sort.  There's no way to move anything without using some sort of energy.  Perhaps one day that will involve anti-gravity devices, but they'll have to use anti-gravitons in some way to work, which will involve imparting energy to the anti-gravitrons, which will be propulsion (just a different kind).\n\nThe problem with gravity is that it's pretty weak compared to the other fundamental forces.  The weak nuclear force is millions of times stronger that gravity, which means its anti-force is also milltions of times stronger.  Gravity works well for pulling large masses together into planets and stars, but on a small local level it's not that usable...there will probably be better ways to getting about than anti-gravity.
200339
Conservation of momentum.  Newton tells us that for every action, there is an equal and opposite reaction, such that in a collision, the net momentum remains the same.  When you realese gas backwards, the momentum imparted on the gas away from the space craft needs to be ofset by imparting momentum upon the astronaut in the direction of the space craft.\n\nWhen you fire a gun, the momentum of the bullet forward is equal to the momentum of he gun backward.  Since the gun is much bigger than the bullet, it doen't move as fast, and the force is spread over a larger asrea (your hands and shoulder).  But the kick backwards is the same sort of principle: the equal and opposite reaction.
512623
I will state this here, in yet another question: THERE IS NO SUCH THING AS CENTRFUGAL FORCE!! Look it up in a basic physics textbook like Giancoli or Serway. Or even do a google search. You have to understand what reference frame you're talking about. The fact that your body tends to want to move to the outside of the curve when your car goes around in a circle has nothing to do with anything other than the fact that your body wants to keep going along in a straight line (conservation of momentum), but the car gets in front of it at every turn (semi pun intended). What is actually happening is that the car is an accelerating reference frame.\n\nAs for the moon/earth question... imagine a man 6' tall throws a tennis ball horizonatally with respect to the earth's surface. The ball will go forward a certain distance (because of it's velocity tangential to the earth's surface) as it falls the ~6' to the ground (because gravity pulls it down). The faster he throws the ball (horizontally), the further it will go before it hits the ground. If he throws it fast enough then the amount that the ball travels forward and the amount that it drops towards the earth's surface can match the curvature of the earth, and the ball will orbit the planet (neglecting air resistance, as we physicists always do) and come around and bonk the thrower in the back of the head. The moon is just a much larger tennis ball, thrown by a much larger man.\n\nYour eqn. is correct, but it only describes the force acting upon the moon. It is not the equation of motion, which detirmines what actually happens to the moon. The equation of motion is:\n\nF = m * a  (F = force, m = mass of moon, a = acceleration of moon around earth)\n\nIn the case of the circular motion of the moon around the earth, "a" is defined by:\n\na = (v^2)/r (where v is the magnitude of the velocity of the moon and r is the center-to-center distance from the earth to the moon)\n\nPlugging one into the other you get:\n\nG * M * m/r^2 = m * (v^2)/r (where M is the earth's mass)\n--> G * M/r = v^2\n\nSo, as long as the tangential velocity, v = sqrt(G * M/r), the moon falls just enough towards the earth as it travels just enough tangentially second by second, that it continues to orbit around the earth. (Note: there is NO CENTRIFUGAL FORCE anywhere in any of these eqns.!!)
709931
kaksi_guy: I think charlesellis753 is correct.  If we interpret the question as a fill in the blank, mass appears to be the best answer.\n\nIf the ______ of an object doubles, then the force required to accelerate it at the same rate will double.\n\nThat sentence is true is the blank is filled with 'mass'.
1002624
Well actually on thing first: the reason why on a spacecraft or in a spacestation air is sucked out into space, is because gas as being air rises and fills as much room as it can expand, but as space is empty it will try to consume that space. This gets vacuum mixed up with air/gas escaping.\nSo nebulas, dust, gas and other remnants can stay in one place. Although the gas expands and converts dark matter to a larger mass (process=0.000023% conversion). As also the mass expands, the same way as galaxies expand the universe like the particles of the nova gas.\nIf a spaceship or spacestation was to explode it would be a flash of light expanding on the particles that were ejected by the explosive expansion, feeding on the explosion power. Basically the explosion would push out the mass rather than the mass expanding on itself. The term of explosion in space is not yet literally confirmed but some people name it a "nova". Although it means "explosion of gas from gas cloud" to our sense. There may not be air but the dark matter would provide a vibe cushion. Although a smaller sound it will still be a "sound" that probably only certain ears can hear as well as the energy production would create a certain spectrum of sound.\nOnly man-made satellites have been broken and fallen to Earth, but I don't know of any explosions, actually they're 50% explosion proof I think.\nAlso think as if you're in a spaceship and you are travelling at 100 mph all heavier loose matter will be forced down by gravity and weight as liquid and gas and air as heavy loose matter to be heavy and travelling will cause these heavier loose matter to be forced into your face as you are gaining speed this is a vacuum of force "forced matter vacuum". With the door wide open while travelling will cause a pump or "matter pump vacuum". Making loose heavy matter to be forced to the bottom in gravity and backwards by force vacuum "also 'gravitional vacuum' ". This is called the "circulation vacuum". Like liquid you would see heavier elements go to the base and the lightest at the top this is the same physic as that of forcing heavier loose matter back in a spaceship as "circulation vacuum". This force is relative to gravity "relativity".
94632
Assuming (1) the Earth isn't torn apart by the impact of such a meteor and (2) the Earth wasn't molten rock in the center, you could safely expect this ball to go through the hole toward the center of the Earth, pass through the center and start moving towards the opposite end of the Earth.\n\nBecause the ball won't have gained enough kinetic energy to then break away from Earth's gravity (in the same way a pendulum always comes back and finally, to rest), the ball will then slow to a stop and return back towards the center.  Just as a pendulum will, the ball will oscillate back and forth for a while until its potential energy evens out to zero and it stops, motionless, at the exact center of the Earth.\n\nGive it a try and let us know how it all works out, okay?
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).
368139
If the object has onboard power, then the speed is not really relevant to whether it will leave the atmosphere, just when. If you just fire an object upwards and it has no self-propulsion, however, then you need to calculate the change in gravitational potential energy (find out how far the atmosphere extends) and add the energy lost to air friction (this depends on velocity). Equate this to the object's initial kinetic energy and solve for initial velocity v. Your initial velocity will have to be larger than this.
103247
The kind of force? This is normal force - the deformation incurred along a plate boundary (be it convergent or strike-slip) is caused by the repulsive force of particles of baryonic matter. It's the exact same force that is exerted when you push your hand against a wall.\n\nWhat drives plate motion is convection of the underlying magma, which is heated by nuclear decay deep in the earth.\n\nEarthquakes are the result of stored energy from continued strain on a region. They aren't really a force themselves, they are simply the result of a release of stored energy when rock surfaces slip suddenly and release P and S waves. (Along with a whole bunch of other waves, Love and whatnot, I can't remember everything)
1067489
I take it you mean the mass at the end of the pendulum?\n\nThe pendulum swings back and forth because gravity attracts the mass at the end of the pendulum.  It speeds up on its way down and has built up a velocity by time it reaches the bottom.  Gravity then slows the object by the exact same rate it sped it up when it was on its downswing.  The result is the mass swings back up to the exact same height it started out at (assuming your pendulum has no friction and that there is no air resistance, of course).\n\nAs to why the amount of mass doesn't matter:  forget about up or down, you just have two objects existing in space.  The Earth's gravity pulls the object towards the Earth and the strength of this pull depends on the combined mass of the Earth and the object.  The object has some natural resistance to being pulled from its location (i.e. - it has inertia).  The mass of the object is its natural resistance to being moved.  In other words, the mass is contributing equally to the strength of the force pulling it and to its abilty to resist being moved.\n\nMathematically, gravitational force is:  F=GMm/(r^2) where the force of gravity is F, the universal gravitational constant is G, M is the mass of the Earth, r is the distance from the center of the Earth, and m is the mass of the object.\n\nThe effect of this force is expressed by:  F=ma where F is the force of gravity, a is how fast the object will accelerate, and m is the mass of the object.\n\nSubstituting, you have GMm/r^2 = ma.  You have a 'm' on both sides of the equation, meaning mass cancels itself out when you're only concerned with the motion resulting from gravity, not the actual force, itself.
981333
http://www.gsfc.nasa.gov/scienceques2002/20030516.htm