what is the formula for finding orbital velocity?
26943
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785816
h = 4tanθ\nVh = dh/dt = (dh/dθ)(dθ/dt)\ndh/dθ = 4/cos^2θ\ncos^2θ = 4^2/(4^2 + 19^2)\nVh = (16 + 361)/4)(5/361)\nVh = (5/4)(16/361 + 1)\nVh = 1.3054 km/sec
338257
a) vcenter = (m1*v1+m2+m2)/(m1+m2)\n\nb) v1 rel = v1 - v center\nv2 rel = v2 - v center\n\nremember velocity is a vector so you must keep track of magnitude and direction
971222
It is not orbital velocity.\n\nIt is the escape velocity. In order for a black hole to exist the diameter of the mass should be less than the diameter at which the escape velocity is C.\n\n Now rewrite your question.
283082
An example is the rocket equation:\nV = Vex * log(m0/m1)\nwhere \nV is final velocity\nVex is the rocket exhaust velocity\nm0 is liftoff mass\nm1 is burnout mass\nThis equation says (e.g.) that you need a ratio of liftoff to burnout weights = e (2.718, base of natural logarithms) to achieve a V = Vex.\nNow suppose you know the final velocity you want and you know the mass of the object you want to impart the velocity to, which is a typical design situation.  Then you turn the equation around:\nM0=M1*e^(V/Vex)\n(The results can be surprising if you're unfamiliar with rocket propulsion.  With a state-of-the-art engine like the Shuttle main engines, Vex is about 15000 ft/sec. If you want 1000 pounds (empty rocket plus any payload) to reach 35000 ft/sec (~earth escape velocity) with an engine of equal efficiency, you'd need a liftoff weight of about 10000 lb.)
564410
The speed of an electron in a specific orbit is not a fixed quantity. Somewhat like the range of speeds of moleclules in a gas, the electron has a range of speeds.  When we talk of "THE speed" of an electron in an orbit, what is usually meant is the RMS speed.  RMS speed is the square ROOT of the MEAN SQUARED speed of the electron. \n\nSpeed of electrons in their orbits could be determined in various ways. One method is using the d'Broglie relationship together with the uncertainty relation, and calculate it quantum mechanically. For details, one could refer to modern physics/quantum mechanics books.\n\nFor the second part, taking for example the simplest case of H-atom, electrons that orbit in a circular orbit around the proton have a speed of v = SQRT{(ke^2)/(mr)}. Looking at this relationship one can easily tell that the speed of electrons depends on:\n1. Mass m of the atom. So, it should vary for different elements.\n2. Radius r of the orbit. So, it should vary with the different shells.
787829
SOHCAHTOA\nsinθ = O/H = a/c\ncosθ = A/H = b/c\ntanθ = O/A = a/b
1035497
(v - v0) / (t - t0)\n\nv - final velocity\nv0 - initial velocity\n\nt - final time\nt0 - initial time
587882
FORMULA: speed = frequency * wavelenght (sorry cant get symbols to work on yahoo)\n\n(1) 0.2= 1/0.5 * w   ...(as f=1/T)\n=> 0.2=2w\n=> w=0.1\n\n(2) s= 25*(10^8) * 0.12    ...(12cm= 0.12m)\n=> s= 3*10^8\n=> s= 300000000 m/s\n\n(3) 2= f * 0.5\n=> f= 2/0.5\n=> f= 8\n\n8 waves strike dock each second.\n\npretty easy questions.
605786
When a spacecraft is in orbit it needs no energy. That's the simple answer. However, you need to get it into orbit in the first place, and this means accelerating the object to the speed it needs to orbit the earth - around 18,000 mph. How much energy you need then depends on the mass of the object or spacecraft - this can be calculated from 1/2 mv^2. \n\nTo escape the earths gravity altogether the spacecraft needs to go faster, about 25,000 mph or 7 miles per second. So if you were going to Mars for example, once you achieve that speed you need no more energy.
748368
first get the time from the horizontal motion  t = .51 / 1.2 s (uniform motion, no acceleration here)\n\nthen use h = 1/2gt^2 for the vertical motion (with initial velocity zero)
666681
1/2 * mass * velocity^2.
842146
Well the earth orbit's the sun \n2AU * PI / year\n\nAn AU is the average distance of the Earth from the sun in it's orbit.  Although all orbit's are ellipses, the equation of a circle should approximate close enough.\n\n1AU =149.598×10^6 km.\n2AU = 299196000 km \n\nThe earth moves really fast so we should convert years into seconds.  The earth orbits the sun once about every 365.25 days (the .25 is why we have leap years).\n\n365.25 * 24 * 60 * 60 = 31557600 seconds.\n\nSo now all we must figure out to get the speed in km/sec we calculate \n(299196000 * PI) / 31557600\n=29.785280109497\n\nSo the Earth orbits the sun at a speed of 29.785280109497 km/sec.  That's pretty fast!\n\nWikipedia gives a max orbital speed of 29.783 km/sec and a min orbital speed of 29.291 km/sec, so my estimate was pretty darn close.
161857
Solve for the mass of the planet.  You are given a diameter and are told that it is sphereical, so solve for the solume.\nV = 4/3 pi*r^3\n\nOe you know the volume, using its density, solve for the mass,\nD = m/ v\n\nFind the orbital distance of the moon from the planet's center of mass.\nThis distance, r, would be the moons distance above the planet's surface, plus the planets radius.\n\nFind the grvitational acceleration caused by the planets mass on the moon using Newtons law of gravitation,\ng = G*m/r^2,\nwhere m is the mass of the planet, r is the distance from the center of mass, and G is the Universal gravitational constant (6.67 E-11).  Solving this will give you g, the gravitational accelerating the moon experiences towards the planet.\n\nWe are told the orbit is perfectly circular, so we can solve for the orbital speed of the moon by setting the centripetal acceleration eqaul to the gravitational acceleration.\ng = v^2/r\nwhere v is the orbital speed and r is thr radius.
772033
Draw a picture of it with arrows representing the vertical and horizontal velocities as 2 legs of a right triangle.  This triangle will show you that the angle with the vertical is arctan 3.5/5.5 = theta\n\nThen the flight path V will be Vv/cos(theta) or Vh/sin(theta)\n\nSince you're ruling out pythagoras, you really need to find the angle first, then the speed.......
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.
1102478
It varies between 30.287 km/s and 29.291 km/s with 29.783 km/s being the average\n\nThe reason for the varying speeds is that Earth's orbit is not perfectly circular.  At some times, it closer to the sun, causing the orbital velocity to become faster.  As it moves further away from the sun, the velocity gradually decreases to the minimum before beginning acceleration again.