What must be the velocity of an object of mass 270 kilograms in circular orbit about a planet of mass 8 *10^ 24 kilograms at a radius of 20000 kilometers? How long will it therefore take the object to orbit the planet?
The gravitational force is obtained by substituting the masses of the planet and the object, and the distance ( 20000 km)(1000 m/km) = 2E+07 meters for m1, m2 and r into the expression F = G m1 m2 / r ^ 2.
- gravitational force = 360.18 Newtons.
- gravitational acceleration = ( 360.18 Newtons)/ ( 270 kg) = 1.33399 meters/second ^ 2.
- v = 5165.2 meters/second.
- T = `ds / v = 1.25664E+08 meters/( 5165.2 meters/second) = 24328 seconds.
Note that the planet and the object in fact orbit their common center of mass. However, since the object is much less massive than the planet, the center of mass is essentially the same as the center of the planet. This assumes that the planet is spherical and that its mass is distributed symmetrically about its center. These assumptions would be appropriate to a communications satellite orbiting the Earth but would not be completely accurate for the Moon (the center of mass of the Earth-Moon system is located at a point near the surface of the Earth, not at its center). Nor would this model be particularly good for a spacecraft orbiting the Moon, since the Moon is denser on the side facing the Earth and is therefore not spherically symmetric.
This problem illustrates the general principle used in calculating the periodand velocity of a circular orbit. The general principle is that the centripetal force holding the object in its circular path is provided by the gravitational force on the object.
If we let v, r, m and M stand for the velocity of the object, the radius of the orbit (from the center of the planet to the small satellite), the mass of the satellite and the mass of the planet, respectively, we see that the centripetal force Fcent = m v ^ 2 / r on the object, whose mass is m, is equal to the gravitationalforce G m M / r ^ 2 between the object and the planet, whose mass is M.
The equality between centripetal force and gravitational force is expressed in the equation
This equation can be simplified by dividing both sides by m, which yields
The left-hand side is just the centripetal accelerationof an object moving with velocity v in a circle of radius r, and the right-hand side is the gravitational field strength, in m/s ^ 2, at distance r from the center of theplanet. This form of the equation shows that the speed of an object orbiting a planet is independent of the mass of the object.
This equation could be solved for v to obtain
or it could be solved for r to obtain
In the current problem we need to find v, so we solve the original equation for v and substitute the mass M ofthe planet and the radius r of the orbit.
To find the period T of the orbit (i.e., the time required to complete an orbit), we note that the distance traveled in the orbit is `ds = 2 `pi r, so
T = `ds / v = (2 `pi r) / (`sqrt(G M / r) = 2 `pi `sqrt(r^3 / (G M)).
It is not recommended tomemorize either of the equations derived above for v or that for r, nor that for T. Simply know that thecentripetal acceleration is equal to the gravitational force, and you havethe equation m v ^ 2 / r = G m M / r ^ 2, which can be easily solved for whatever variable you require. And know the relationship between distance, speed and required time.
The figure below shows the centripetal acceleration v^2 / r directed toward the center of the circle, which is the center of the Earth, and the gravitational force F directed toward the center of Earth.
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