The centripetal force acting on a satellite in orbit...?
155890
*is a center-directed force\n\n\nnot "*acts as an unbalanced force on the satellite" because, it is balanced by centrifugal force. \n\nnot "*changes the direction of the satellite" because direction is changed by only tangential forces and not the force acting towards center.
767660
A rocket is subjected to four forces in flight; weight, thrust, and the aerodynamic forces, lift and drag.\n\nat rest I will denote the launch pressure by the symbol ps. The pressure exerts an equal force on all sides of the pressure vessel, the cone of the engine. Because the rocket is free to move along the launch mechisims, the pressure exerts a vertical force given by: \n\nFp = (ps - po) * A \n\nwhere Fp is the pressure force, po is the atmospheric pressure, and A is the cross-sectional area of the rocket tube. The net force acting on the rocket is the difference of the vertical pressure force and the weight W of the rocket: \n\nFnet = (ps - po) * A - W \n\nTo simplify the analysis, we are going to assume that the length of the launch tube is small relative to the length of the rocket. As the rocket moves, the pressure remains fairly constant inside the tube. In reality, there is a slight decrease in pressure as the volume increase, but we are going to neglect that effect to simplify this analysis. With a constant pressure, and constant pressure force, we can use a simple algebraic form of Newton's second law to determine the acceleration a, velocity v and the distance y which the rocket moves along the launch tube: \n\nFnet = m * a \n\nm * a = (ps - po) * A - W \n\nwhere m is the mass of the rocket This equation can also be written as: \n\nW * a = g * ((ps - po) * A - W) \n\nwhere g is the gravitational acceleration, which is equal to 32.2 ft/sec^2 or 9.8 m/sec^2 on the surface of the Earth. The value of g is different on the Moon, and on Mars. The weight W is equal to the mass times the gravitational acceleration. Now divide both sides by the weight to determine the rocket acceleration along the launch tube: \n\na = g * ((ps - po) * A / W - 1) \n\nFor a constant acceleration, the velocity v and distance traveled y are given by: \n\nv = a * t \n\ny = .5 * a * t^2 \n\nwhere t is the time. Using the length of the launch tube L, we can now solve for the lift off time - TLO by using the second equation: \n\nTLO = sqrt (2 * L / a) \n\nand then using this time, we can use the first equation to solve for the velocity at the end of the tube: \n\nv = TLO * a \n\nv = sqrt ( 2 * L * a) \n\nAs the rocket clears the end of the launch tube (t = TLO+), the air inside the rocket, at pressure ps, will rush out the exit and the pressure will equalize with free stream inside the tube. In reality, this will provide an additional amount of thrust T given by: \n\nT = m dot * Ve \n\nwhere m dot is the air mass flow rate out of the tube, Ve is the exit velocity of the air. For an air rocket, with no nozzle and small volume, this additional thrust is small relative to the weight of the rocket, and we will ignore this effect. \n\nThe velocity v at the end of the launch tube is used as the initial velocity in the flight equations to determine the flight trajectory of a rocket.
208881
Any object in circular motin is acted on by a centripetal force.\n\nA cycle or a car or a train moving in a circular path. The frictional force provides the necessary centripetal corce.\n\nA stone tied to the end of a string and made to move in a circular path. The tension in the string provides the centripetal force.\n\nMoon rotates about the Earth. The gravitational force provides the necessary centripetal corce.
506306
the net force acting on the moon is directed towards earth.\n\nthe net force acting on earth is directed towards the moon.\n\nthe force on the earth is weak and the earth is very massive so it only causes the earth to wobble, however the force on the moon is large and it is less massive so it orbits the earth.
25445
The earth is APPROXIMATELY round due to gravitational forces.
612424
I haven't done any physics problems for ages, but I'll give it a try.\n\nThe centripetal acceleration = V^2/R \n\nBut, because the acceleration due to gravity is about 9.8m/sec^2 and this is important to the answer, you first want to express V in m/sec;\n\nV = 360km/hr = 360,000m/3,600sec = 100m/sec.\n\nSo the centripetal acc. = (100m/sec)^2 /500m = 10000/500 = 20m/sec^2, which is about 20/9.8 = 2.04 g-forces. This force is always directed towards the center of rotation. \n(Note: if you were on a spinning wheel space station where gravitational forces were negligible, and it had a radius of 500m and had a tangential V of 100m/sec, and you were walking around the inside of the station at the 100m extremity, you would have the apparent weight of 2.04 gs)   \n\nAs far as his "weight" is concerned: if the seat of the plane measured his weight at different points around the loop then it would vary. \n\nAt the top of the loop: his recorded weight would be 70kg x (20-9.8)/9.8 = 72.86kg. This is because gravity is reducing the  weight. \n\nAt the bottom: it would be 70x(20+9.8)/9.8 = 212.86 kg\nThis is because gravity is adding to the weight.\n\nAt the mid-points gravity and the centripetal acc. would be perpendicular to each other and the recorded weight would simply be 70x20/9.8 = 142.86kg\n\nI hope that helps.
724756
For a body in circular motion there will be a centripetal force\n= M V V /R.\n\nIn this case the tension of the string provides this force.\n\nMaxinmum tension = 50N.\n\nThat is the maximum centripetal force is 50 N\n \nM V V /R = 50\n\nV V = 50 x R / M = 50 x 1/ 2 =25 (m/s)^2.\n\n The maximum speed is V = 5 m/s.
990074
Picture a weight on a string that you spin overhead. The shorter the string, the faster you have to spin the weight to keep it horizontal. It's the same with an orbiting body. A geostationary satellite is placed at the proper orbital radius that its rotational velocity is the same as that of the Earth, so it remains placed over the same spot on the globe as it speeds forward in its orbit.
755334
Analyze the forces at the top of the loop:\n\nThe force of gravity pulling the ball downward,\nthe upward centrifigal force keeping the ball on the track,\nand the force of the track pushing the ball downward.\n\nThe answer to the question is when the centrifigal force is exactly equal to the weight, so the force of the track equals zero.\nmg=mw^2R\nw is the angular velocity\nw=sqrt(g/R)\n\nOnce you know the angular velocity, compute the equivalent linear velocity by mutlipying by R\n\nthen use mgh=1/2mv^2\nto compute h\n\nWhere is point Q?\n\nj
912301
Well, for centripetal force:\n\nForce = (mass) * (velocity squared) / radius\n\nand from there?  I'm lost.  I've tried plugging in all your information and it feels like we're missing a known variable.  My first assumption was to put in 7.18 as the force, but then we don' t the mass.   And if we plug in (7.17*mass), we're left with force as a variable.   \n\nOye, good luck.  It's been 10 years since I've taken physics.
64466
Yes! There is centripetal acceleration on the ball.  It is actually accelerating in towards the center, but other forces keep it on the edge.
672475
Since the proton is moving in a circular pattern, the force given is the centripetal force.  I know that it it worded weirdly, but I believe that is just to throw you off.  The equation for the centripetal force due to circular motion is F = m * (v^2) / r  \n\nwhere F is the force, m is the mass of the object, v is the velocity, and r is the radius of the circular path.\n\nSo we set up the following:\n\n8.00 * 10^-13 N = (1.67 * 10 ^-27 kg) * (v^2) / (0.8 m) \n\nI used 0.8m instead of 80cm because we need to be in standard SI units.  This will make sense later. Rearrange and solve for v.\n\nV = sqrt[ (8.00 * 10^-13 N) * (0.8 m) / (1.67 * 10^-27 kg) ]\n\nSince N = (kg*m)/(s^2)\n\nwe can re-write as:\n\nV = sqrt[ (8.00 * 10^-13 (kg*m)/(s^2)) * (0.8 m) / (1.67 * 10^-27 kg) ]\n\nthe kg cancel out:\n\nV = sqrt[ (8.00 * 10^-13 (m)/(s^2)) * (0.8 m) / (1.67 * 10^-27) ]\n\nSimplifying you get:\n\nV = sqrt(3.83 * 10^-14 m^2/s^2)\n\nSince they tell you that the sqareroot of 3.83 is 1.96, then we get our final answer to be:\n\nV = 1.96*10^7 m/s\n\nLet me know if you need furhter clarification on this problem.
256930
dba_gregg is correct.\n\nThe gravity pulls the front while the back is still supported by normal reaction force. Thus a torque is acting so as to rotate the bus with the front down. The angular momentum produced is conserved and the bus begins to rotate about the free axis passing through its centre of mass.
907263
my answer might be childish,but one can understand it easily\nThe gravitational force of earth acting at any point on its surface has a direction towards center of earth.Every matter on the surface is pulled towards the center of earth with this gravitational force.\ndo u know the value of this sticking force acting on every one kg of matter on earth--it is about 10N
93072
As we are not, the Earth is spinning at a pretty constant speed (1 rotation / day).  Based on the position we are on the Earth, we are traveling around at different speeds, the closer one is to the equater, the faster one is traveling.\nAt the same time, the force of gavity is pulling us toward the center of the Earth, this force acts as the centripetal force (NOT to be confused with the ficticious centrifugal force) that keeps us in circular motion.\n\nIf the Earth's rotation we to increase, the force needed to keep up down on the ground would need to increase too.  However, the force (gravity) is only so strong, it can only pull up down by some much.  So it one were to increase the spin of the Earth enough, it is possible for the centripetal acceleration to equal the gravitational acceleration....therefore we would be weightless.\n\nBut now the question is, what happens is we further increase the Earth's rotation past this point of weightlessness?\nIf you spin the Earth faster, our weight would not be enough to keep us in the circular motion we are currently in, so it would have the effect of seeminly lifting up off the ground.  HOWEVER, we would not fly up into space.  The moment we leave the ground, the centripetal acceleration needed is reduced (a = v^2 / r, when r increases, a decreases).  If we could maintain a constant speed once we leave the ground, we would eventually lift up enough to that we get to a height so that the centripetal acceleration equals the gravitational acceleration again and we would hover at that height.
833966
The question is easy enough if you think about it properly.  What you want is the speed at which the centrifugal force (given by mv^2/r) equals the gravitational force given by the bank, and that will be mg sin(2.5 degrees).  Note that m cancels out in this, and you are given r, so v is the only unknown and you can solve for it with a scientific calculator.
371595
approx 19 miles/second.\n\nAny force from centripetal acceleration you might feel would be more than compensated by the gravitational attractions from the Sun and Earth themselves.\n\nIt might be a measureable amount, but just barely.