How do figure the escape velocity of various objects such as neutron stars, planets or asteroids for example?
924128
quark's  formula is slight incorrect\n\nVescape = sqrt(2GM/r)\n\nwhere M is mass of the body being escaped from (i.e the planet or star or asteroid)\n\nand r is the distance between the object and the body being\nescaped from\n\nand G is the gravitational constant.\n\nTo get this formula is a engery equation\n\ninitial kinetic energy required is the potential energy needed\nto bring the potential energy from current state to state\nat inifinite distance (i.e to zero)\n\nIn a gravitational field the potentail engery = - GMm/r\nfor object of mass m.\n\nSo to get this to zero you need a kinetic energy of\n\n1/2mv^2 = GMm/r\n\nsolve for v and you get sqrt(2GM/r).
750537
acceleration = v^2/r or v*v/r (this is m/s^2 so divide by 9.81 to get g's)\n\nwhere v is the velocity\n\nr is the radius\n\nBe careful, think about the one g the pilot feels from the pull of the earth. ;)
515286
I once wrote a simple program to simulate orbital motion. It went something like this:\n\nX=0\nY=7000\nVX=7.5\nVY=0\nLABEL:\nR=SQRT(X*X+Y*Y)\nAX=-398600*X/R/R/R\nAY=-398600*Y/R/R/R\nVX=VX+AX\nVY=VY+AY\nX=X+VX\nY=Y+VY\nPLOT X,Y\nGOTO LABEL\n\nIt was originally in BASIC, but the above is pseudo code. The number 398600 is the Earth's gravity parameter, which is basically the mass of the Earth multiplied by the universal gravitational constant, so this program simulates Earth's gravity. The object starts 7000 km from the center of the Earth traveling at 7.5 km/s. Gravity varies as the inverse square of distance. The X component is just negative X divided by R cubed and the Y component is negative Y divided by R cubed. Use these to calculate the X and Y acceleration, use the X and Y acceleration to calculate the X and Y velocity, use the X and Y velocity to calculate the new position in the X,Y plane. Then use the new X and Y position to start over and keep doing this to plot the path of the object.
1082685
Simply equate gravity function with centripetal acceleration.\n\nGMe/(r+h)^2 = V^2/(r+h)....where r is earth radius and h is height\n\nsolve for v
844533
from what I remember conservation of momentum states that: \nm1v1=m2v2\nwhere in this case, m1= mass of the astronaut\nm2=mass of the sat.\nv1= unknown\nv2=speed of the sat.\n\nyou can find the speed by solving the above equation for v1:\nv1=m2v2/m1\n\nFrom there, once you have the speed you can calculate the distance by multiplying the speed (v1) times the time (7.5 s) given in the problem.
605585
It sounds like you need three things\n\nA starting point for each star.\n\nA vector for each star (that takes care of direction and speed)\n\nGravity (The more dense a star is in relation too its mass the greater its gravitational pull)\n\nOnce you have worked out the specific gravity for each object (star) the only things that vary are the vector of each star and thus its new position.\n\nGood luck!! I know any gravitational calculation with three or more object gets very complicated very quickly.
553694
You are assuming there is an absolute reference to measure their speeds against. All speed has to be measured against some standard. But, for the sake of your discussion, let's assume you have two large asteroids in space that are motionless relative to each other. You launch super SciFi spaceships from each, aiming near to the other so they will pass. Each spaceship accelerates to 60% of C (speed of light). You are asking if they won't see the other zoom by at 1.2C.\n\nYou are forgetting that as they move to 60% of C, time is also affected. (google for the twins experiment). Time slows to an extent that they will never see anything approaching at C or in excess of C.\n\nHope this helps.\n\nChuck Taylor\nhttp://groups.yahoo.com/group/lunar-observing/
787635
At Launch = 16.21 km/s (36,300 mph)\nConverts to:  58,356 Km/hr or 3.41 AU/Year \n\nAfter Jupiter Gravity assist = 21 km/s (47,000 mph)\nConverts to:  75,600 Km/hr or 4.41 AU/year\n\nPluto Flyby = 13.78 km/s (30,850 mph)\nConverts to:  49,608 Km/hr or 2.90 AU/year
238117
If by "calculate" you mean process the results of experimental meaurements of speed, force, etc., then you measure the initial total energy (potential, kinetic) of the test object, measure all known energy input that occurs along the path (integral of applied external force times velocity dot product), and subtract from the total of those two the final total energy. The result is the energy lost to dissipative processes like friction and air resistance.\nIf instead you use calculate in the sense of predicting the dissipative losses, you need a model of the object (its drag coefficient and the coefficient of friction). If the path is constant-slope and speed and air density are constant, you can calculate the end state analytically because drag force is constant and so is normal force and thus friction. Otherwise you have nonlinearities which require a time-stepped simulation to solve.
1069779
Draw a free-body diagram of the ball and examine it's accelerations - if you assume wind and air resistance are negligible and that the ball is accelerated straight up, then you're left with its initial acceleration (68.57 m/s^2) and gravity (9.8 m/s^2 if I remember correctly) \n\nThe ball will decelerate by 9.8 m/s every second, you can divide to find the amount of time the ball takes to reach its apogee (point of zero acceleration), then integrate the acceleration curve to find the velocity curve, which can be integrated again to find the distance traveled.  I'm not sure it can be done without the calculus.
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
241909
The escape velocity is the minimum velocity an object needs to obtain in order to leave the surface of the Earth and never return, it will break free of the Earth's gravitational field forever.\nIt is not necesary to obtain escape velocity just to go into space.  While orbiting the Earth, you are most definatly still trapped within the Earth's gravitational field and you will eventually return to the surface.\n\nIf an object launched from the ground with an initial velocity lower than the escape velocity, the object will eventually (it may be take a very long time) return to the Earth and fall back down.
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!
910104
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 not even a light can escape. 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.).
535160
what do you want to calculate? gravitation? emission? something else?
946723
To escape the Earth, you have to go over 7 miles per second.  That's the escape velocity you need to escape the gravitational pull of the Earth.  The gravitational pull of a black hole (inside the event horizon) exceeds the escape velocity of light.  In other words, light would have to move faster than 186,000 miles per second in order to escape.  Since it can't, light can't escape the gravitational pull.  No sucking required.
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."