Gravitaional and Centripetal Forces ```Name: Paul Status: educator Grade: 9-12 Location: MA Country: N/A Date: 6/21/2005 ``` Question: I am confused by something I read in the archives here (see below) saying that centripetal force slightly reduces the effects of gravity. How is that the case? It seems to me that the direction of the centripetal force (from a spinning earth) is perpendicular to the vector of gravity and should not affect it. *http://www.newton.dep.anl.gov/askasci/phy99/phy99x82.htm The centripetal force due to rotation, whether on the Earth, or rotation with respect to another frame of reference, is actually directed TOWARD the axis of rotation. So in effect we are always falling "toward" the center of the Earth by an amount that keeps us on the surface of the Earth. If you could somehow "turn off gravity" we would continue to move in a direction tangential to the surface of the Earth where we are standing. This would only be non-zero if we were standing at one of the poles. Vince Calder Paul, You must first realize that centripetal force is NOT a force. Centripetal force is an effect of forces directed toward a central point. IF all true forces add up to a large enough net force, directed at the center of a circular path, then these forces are able to KEEP the object on that circular path. The formula for centripetal force is a calculation of how much force is needed. When you rotate with the Earth, some of the gravitational force is needed just to keep you on the Earth's surface. If this is all there were, you would feel weightless. This is why orbiting in space results in "weightlessness". The ship is moving fast enough to use up all of gravity just to keep from shooting off into space. Without forces, everything moves in a straight line, not in a circular orbit. Because you do not need ALL of gravity to just keep you on the Earth's surface. It is too much, so it would pull you inward if the ground were not there. The ground provides a normal force to oppose that portion of gravity that you do not need. It is this normal force, how hard your feet and the ground push against each other, that bathroom scales actually measure as your weight. Dr. Ken Mellendorf Physics Instructor Illinois Central College Dear Paul, The archives you reference were perhaps not carefully enough written. The rotation of the earth has NO effect on the gravitational attraction between the earth and an object on its surface or any other object for that matter. As Newton stated in his law of gravitation, F = GMm/r^2, where F is the force between two masses M and m separated by a distance r. G is the gravitational constant 6.6726E-11 N m^2 kg^-2. So two 10 kg masses (about 22 lb each) placed 5 m apart (about 16.4 ft) will exert an attractive gravitational force (gravity is always attractive) on each other of about 2.67E-10 N (about 6E-11 lb). Note that I use a notation where E-2 = 0.01 = 10^-2. So E-11 = 0.00000000001 and 6E-11 lb=0.00000000006 lb. In principle this applies only to two point masses of zero size (or at least of negligible size compared to their separation.) However, as Newton showed (he had to invent integral calculus to show this) a uniform spherical mass distribution produces a gravitational force on a small mass outside that sphere as if all the mass in the sphere were concentrated at the center of the sphere. So a 100 kg person on the surface of the earth (radius = 6.37E6 m and mass 5.98E24 kg) is attracted toward the center of the earth with a force of 983.4 N (about 220 lb). So he is accelerated toward the center of the earth with an acceleration given by a = F/m = 9.83 m/s^2. If he is at the equator, part of this acceleration (a = v^2/r = 0.034 m/s^2, where v = 463 m/s = 1035 mi/hr) is needed to keep the person equally distant from the center of the earth and so in a circular orbit about the center of the earth. Looked at slightly differently, a force F = ma = 3.4 N =0.76 lb would just keep a 100 kg person (220 lb) on the surface of the earth at the equator. Notice also that if a day were only 1.5 hours long, the full weight of a person would just keep him on the earth at the equator. This explains while all satellites in low earth orbit have a period of about 1.5 hr. Notice finally that a person at the equator will have an acceleration towards the center of the earth reducing his distance from the center of the earth of 9.80 = 9.83 - 0.03 m/s^2 where the 0.03 m/s^2 is the centripetal acceleration needed to keep the person in a circular orbit about the center of the earth and so always at the same distance from the center of the earth. I also have to point out that the centripetal acceleration is towards the center of the earth and so is exactly parallel to the gravitational force and the acceleration due to gravity. Best, Dick Plano, Professor of Physics emeritus, Rutgers University No, centripetal force is a vector pointing from an object toward the point about which it is revolving, so it acts in the same direction as gravity. In fact, in this case, centripetal force is exerted *by* gravity. Tim Mooney Hi Paul- I better excerpt what they said in "phy99x82.htm": If the Earth were not rotating...its gravitational field would be unchanged... [but] the acceleration "due to gravity" would be very slightly more than 9.8 m/s^2 anywhere but at the poles because there would no longer be any centrifugal force reducing the effect of gravity. Grayce If anything, the spin of the earth lessens the effect of gravity... Larry Krengle ...9.8 m/s/s is a measured value, and it includes the effect of the Earth's spin.... If the Earth were not spinning, the measured value would be .034 m/s/s larger. Tim Mooney Paul, I am not sure which way you are thinking the centripetal-force-vector points. What I think: - gravity vector g^ points "down", towards center of spherical Earth, making a vector normal to the local plane of earth's surface, parallel but opposite the vector r^. * - "centripetal force" vector cp^ points "outwards" from the N-S spin axis of Earth, always perpendicular to it. Imagine a cylindrical frame of reference: r^ = {R, theta, Z}. (Z being parallel to N-S axis, R extending radial & perpendicular to that axis, theta being an angle, the geographic longitude). Vector cp^ is parallel to direction of distance R. - anywhere on the equator, "outwards" = "up": parallel and opposite to "down". There, centripetal force subtracts directly from gravity. (about 0.33%) - anywhere else but at the poles, cp^ is not quite parallel to, but still subtracts from gravity g^ to a lesser extent, proportional to the cosine of the latitude: (1.0..0.0). Now, the Earth's "spin vector" or "angular momentum vector", as used in vector physics, is pointed along the N-S axis. But the local centripetal force is a different vector from that, and it has a different direction or magnitude at every location r^. I am not sure how to clearly derive it from the spin vector using vector operations. Something like the time_derivative of (the Cross_Product of (spin^ & r^) ).* That cross-product gives the local velocity due to spin, and is parallel to earth's surface at each location. An end-to-end set of velocity vectors makes a circle around the earth, tracing out a latitude line. But centripetal force derives from _acceleration_ due to spin, which is the time derivative of velocity. The time derivative of a circular motion is 90 degrees rotated at each point around this circle, pointing towards the center of the circle. This makes it vertical at the equator and roughly vertical elsewhere. They (cp^ and g^) are not perpendicular. * (r^ being the local 3-D vector position with respect to earth's center) I hope that helps Jim Swenson Click here to return to the Physics Archives

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