where all the internal torques cancel out (being equal and opposite pairs), resulting in only the net external torque, , appearing on the right‐hand side. The left‐hand side of Eq. (2.60) is derived as follows, once again by the virtue of Eq. (2.56) and the fact that is the centre of mass:
(2.61)
A substitution of Eq. (2.61) into Eq. (2.60) yields the following equation of rotational kinetics of the body:
(2.62)
where
(2.63)
is the angular momentum of the body about its centre of mass, o. Thus a net external torque about the centre of mass of a body equals the time derivative of its angular momentum about the centre of mass.
If the body is rigid, then the distance between any two of its particles is a constant. Hence, the velocity of the elemental mass relative to is given by
(2.64)
where because Here is the angular velocity of a local reference frame, oxyz, rigidly attached to the body at , with unit vectors along , , and , respectively (Fig. 2.3), and is measured relative to the inertial frame, (OXYZ). Such a reference frame, oxyz, is termed a body‐fixed frame.
2.7 Gravity Field of a Body
The principles derived for the ‐body problem earlier in this chapter can be extended to the determination of the gravitational acceleration caused by a body whose mass, , is distributed over a large number of elemental masses, . In the following discussion, it is assumed that the other point masses are densely clustered together to form a body, away from the test mass, ; hence the test mass has the following equation of motion:
(2.65)
where
(2.66)
is the total potential energy of the system,
(2.67)
is the mass of the body consisting of the last particles,
(2.68)
with being the location of the test mass, , in an inertial reference frame, (OXYZ), and being the location of the centre of mass of the attracting body consisting of the remaining particles, which are located at . If it is further assumed that the test mass is negligible in comparison with the combined mass of the remaining particles constituting the body, that is, , then the test mass, , causes a negligible acceleration on the body. Consequently, the body can be assumed to be at rest, and the origin of the inertial reference frame, OXYZ, is moved to the centre of mass of the body, i.e., , , and . Hence, the equation of motion of the test mass becomes the following:
(2.69)
or, since the partial derivative on the right‐hand side yields only the terms for which either or equals 1, we have
(2.70)
In terms of the gravitational potential of the body at the location of the test mass, which is given by
(2.71)
the acceleration of the test mass is expressed by dividing the right‐hand side of Eq. (2.70) by as follows:
(2.72)
