How do you find orbital velocity?
97099
Please take a look at:\n\nhttp://en.wikipedia.org/wiki/Orbital_velocity
1084130
The projectile motion is the motion of a projector starting from origin (x=0 & y=0) with initial velocity (v) & inclination (a) where (a) is the angle between X-axis & (v) so the only accelerations affecting the projector are:\nx''=0 & y''=-g --------> (1) & (2)\n\nTo calculate velocity (x' & y'):\nfrom (1) & (2) we get:       \nx''=dx'/dt=0  \ntherfore: x'=constant=v cos(a) [from initial condition] --------> (3)\n \nfrom (3) we get: y''=dy'/dt=-g\ntherfore: y'=-gt+constant\n\nfrom initial condition:\nat t=0 ----> y'=v sin(a) & constant=0\ntherfore: y'=v sin(a)-gt ------> (4)\n\nThe parametric equation:\nfrom (3): dx/dt=v cos(a)\ntherfore: x=tv cos(a)+constant\n\nfrom initial condition:\nconstant=0\ntherfore: x=tv cos(a) -----> (5)\n\nfrom (4): dy/dt=v sin(a)-gt\ntherfore: y=tv sin(a)-0.5g(t^2)+constant\n\nfrom initial condition:\nt=0 therfore y=0 & constant=0\ntherfore y=tv sin(a)-0.5g(t^2) -----> (6)\n\nfrom (5): t=x/[v cos(a)]          sub in (6)\ntherfore\ny=[x tan(a)]-{ [g(x^2)]/[2(v^2)] }*[1+ tan^2 (a)] \nthis equation is called The Cartizian Equation\n\nTo find The Range ------> sub y=0 in The Cartizian Equation\ntherfor:\nR(range)=[2v sin(a) cos(a)]/g
899353
1) you basically need to draw out a triangle.  cot-1(x) returns the angle whose cotangent is equal to x, so let theta=cot-1(5/12).  this means cot(theta)=5/12, or tan(theta)=12/5.  the triangle associated with this problem is a 5-12-13 right triangle, so the sine of theta is just 12/13.\n2) break the velocities down into x-components and y-components, using the formula vx=vcos(theta) and vy=vsin(theta):\nvx=v1cos(theta1) +v2cos(theta2)\n  =580cos(332) +60cos(45)\n  =512+42\n  =555 mph\nvy=v1sin(theta1) +v2sin(theta2)\n  =-272+42\n  =-230 mph\nnow, use the pythagorean theorem to find the magnitude of the plane's velocity:\nv=sqrt(vx^2+vy^2)\n  =sqrt(307510+52838)\n  =sqrt(360349)\n  =600 mph\nnow, use the formula tan(theta)=vy/vx to find the direction:\ntheta=arctan(555/-230)\n  =-22.5 degrees\n  =337 degrees
790665
starting from the very beginning\nv ≡ ds/dt\na ≡ d²s/dt² = dv/dt\ndv = adt\nif a is constant,\n∫dv = a∫dt\nv = at + C\ncall C v0, or initial velocity:\nv = v0 + at\nv = 0 +(9.80662 m/sec^2)(2 sec)\nv = 19.61324 m/s
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.......
790378
h=-16t^2+48t+50\nfor first part put t=3\nh=-16*9+48*3+50\nsolve and subtract 50 to find height above projection\nfor second part diff the eq. to get velocity\ndh/dt=-32t+48\nat max height this velocity is zero\n0=-32t+48\nt=3/2\nput this in given eq. to find height
693415
You take the avaerage of the instantaneous velocity over an interval of time.\n\ntake say five instantaneous velocity measurements over about 40 seconds. then take the average of these five measurements to get a linear velocity. to make your answer more accurate just take more than 5 mearsurements and average them in the same matter.
723707
s'(t) =(t-3)(-2t^{-3}) + t^{-2}\n=( -2t+6 +t)/ t^{-3}\n=(6-t) / t^{-3}\nso the velocity is 0 when the top =0\ni.e 6-t-0\nt=6\ns
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.
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
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.
106990
Period is proportional to R^3/2.\nThis is an answer I posted to bombay2pac2000 under Astronomy & Space 8 hours previously, and it applies here as well.\nKepler's 3rd law says: Orbital period T = 2*pi*R^1.5/SQRT(GM)\nwhere R is the radius (in the case of a circular orbit), G is the universal gravitational constant and M is the mass of the central body.  In orbit, centrifugal acceleration CA balances gravitational acceleration GA.  CA=w^2R (where w is omega, the orbital angular rate which = 2pi/T). So CA = (2pi/T)^2*R. From Newton we know GA = GM/R^2. Put 'em together, \nCA = GA\nR*(2pi)^2/T^2 = GM/R^2 \nT^2 = R^3*(2pi)^2/GM\nT = 2*pi*(R^3/2)/SQRT(GM)
1090272
11038.509198867 kph.\n\nThe math: a circular geosynchronous equatorial orbit has a radius of approximately 42,164 km (from the center of the Earth).  To get the diameter of the orbit,  pi * 2 * 42164, or 264924.2207728 km diameter for the entire orbit.  Divide by 24 to get the distance traveled in one hour, 11038.509198867 kph.
680534
Use the equation from the link below.  You will find that \nv/c = 0.14365, or\nv = 0.14365 c\nor\nv = 43,100 km/s\n\nThe object is moving away from the observer at this rate.\n\nBy the way, you probably won't get credit unless you can work out the math steps I left out.
872212
Use the certical component to find the time in the air and since the horizontal velocity does not change; Dx = vx * time\n\nDx = v^2 / g\n\nheight = v^2 / (2g)\n\nHope this helps without actually solving it for you.
200570
the velocity of the ambient
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).