![]() If velocity is (-) (down or to the left) and x (motion) is speeding up, then acceleration will have a (-) sign in front of it. To figure out the acceleration, use the chart and go by velocity and motion. So if you assign a (+) to velocity when it goes up or to the right, then a (-) would be assigned to down and the left. Basically, I didn't quite get it 100% until I started using this. The relative motion of two objects that are acted upon by a central force can be described by Newton's 2nd Law as if they were a single mass with a value called the "reduced mass".Here is something that may help you all discern between when to use a (-) or (+) sign in front of acceleration. If you can obtain the individual orbit radii r 1 and r 2 then you can use the center of mass condition with the measured mass sum to obtain the individual masses These relationships are important in the study of visual binaries. If we use the convenient astronomical unitsĪ is the radius, and to elliptical orbitsĪnd from just the period and orbit radius you can obtain the sum of the masses m 1 + m 2. Since the period T of the orbit is given by then the motion equation can be written which reduces to This leads to Kepler's 3rd law (the Law of Periods) which is useful for the analysis of the orbits of moons and binary stars. If you are riding on one of the masses, the relative motion equation has the same form if you substitute the reduced mass Hole through center of Earthįrom the gravity force and the necessary centripetal force: ![]() If the Earth were of uniform density (which it is not!), the acceleration of gravity would decrease linearly to half the surface value of g at half the radius of the Earth and approach zero as you approached the center of the Earth. If you drilled a hole through the center of the Earth, the acceleration of gravity would decrease with the radius on the way to the center of the Earth. Please note that the above calculation gives the correct value for the acceleration of gravity only for positive values of h, i.e., for points outside the Earth. The value of g at any given height, say the height of an orbit, can be calculated from the above expression.Ībove the earth's surface at a height of h = m = x 10 6 m, which corresponds to a radius r = x earth radius, the acceleration of gravity is g = m/s 2 = x g on the earth's surface. The weight of an object is given by W=mg, the force of gravity, which comes from the law of gravity at the surface of the Earth in the inverse square law form:Īt standard sea level, the acceleration of gravity has the value g = 9.8 m/s 2, but that value diminishes according to the inverse square law at greater distances from the earth. HyperPhysics***** Mechanics ***** Rotation Since it acts always perpendicular to the motion, gravity does not do work on the orbiting object if it is in a circular orbit. The force of gravity in keeping an object in circular motion is an example of centripetal force. The orbit can be expressed in terms of the acceleration of gravity at the orbit. Setting the gravity force from the universal law of gravity equal to the required centripetal force yields the description of the orbit. The circular orbit is a special case since orbits are generally ellipses, or hyperbolas in the case of objects which are merely deflected by the planet's gravity but not captured. ![]() Gravity supplies the necessary centripetal force to hold a satellite in orbit about the earth.
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