Riemann surface
matlab

Question 2.3
The active gravitational mass $\left(m^A\right)$ of a particle is an attribute that enables it to establish a gravitational field in space, whereas the passive gravitational mass $\left(m^{\mathrm{P}}\right)$ is an atribute that enables the particle to respond to this field.
(a) Write Newton’s law of universal gravitation in terms of the relevant active and passive gravitational masses.
(b) Show that the third law of motion makes it unnecessary to distinguish between active and passive gravitational mass.

Solution
(a) The gravitational force $\mathbf{F}{12}$ that particle 1 exerts on particle 2 is proportional to the product of the active gravitational mass $m_1^A$ of particle 1 and the passive gravitational mass $m_2^{\mathrm{P}}$ of particle 2. Thus, the inverse-square law of gravitation is $$\mathbf{F}{12}=-G \frac{m_1^{\mathrm{A}} m_2^{\mathrm{P}}}{r^2} \hat{\mathbf{r}}$$
where $G$ is the universal constant of gravitation, $r$ is the distance between the particles and $\hat{\mathbf{r}}$ is a unit vector directed from particle 1 to particle 2. By the same token, the force $\mathbf{F}{21}$ which particle 2 exerts on particle 1 is $$\mathbf{F}{21}=G \frac{m_2^{\mathrm{A}} m_1^{\mathrm{P}}}{r^2} \hat{\mathbf{r}} .$$
(b) According to Newton’s third law, $\mathbf{F}{12}=-\mathbf{F}{21}$. It therefore follows from (1) and (2) that
$$\frac{m_2^{\mathrm{A}}}{m_2^{\mathrm{P}}}=\frac{m_1^{\mathrm{A}}}{m_1^{\mathrm{P}}} .$$
We conclude from (3) that the ratio of the active to the passive gravitational mass of a particle is a universal constant. Furthermore, this constant can be

14 Solved Problems in Classical Mechanics
incorporated in the universal constant $G$, which is already present in (1) and (2). That is, we can set $m^{\mathrm{P}}=m^{\mathrm{A}}$. There is no need to distinguish between active and passive gravitational masses; it is sufficient to work with just gravitational mass $m^{\mathrm{G}}$ and to write (1) as
$$\mathbf{F}{12}=-G \frac{m_1^G m_2^G}{r^2} \hat{\mathbf{r}}$$ Comment Evidently, the same reasoning applies to the notions of active and passive electric charge. Thus, if one were to write Coulomb’s law for the electrostatic force between two charges in vacuum as $$\mathbf{F}{12}=k \frac{q_1^{\mathrm{A}} q_2^{\mathrm{P}}}{r^2} \hat{\mathbf{r}}$$
where $k$ is a universal constant, a discussion similar to the above would lead to
$$\frac{q_2^{\mathrm{A}}}{q_2^{\mathrm{P}}}=\frac{q_1^{\mathrm{A}}}{q_1^{\mathrm{P}}}$$
Consequently, the ratio of active to passive charge is a universal constant that can be included in $k$ in (5); it is sufficient to consider just electric charge $q$.

Question 2.4
The inertial mass of a particle is, by definition, the mass that appears in Newton’s second law. Consider free fall of a particle with gravitational mass $m^{\mathrm{G}}$ and inertial mass $m^{\mathrm{I}}$ near the surface of a homogeneous planet having gravitational mass $M^{\mathrm{G}}$ and radius $R$. Express the gravitational acceleration $a$ of the particle in terms of these quantities. (Neglect any frictional forces.)

Solution
The equation of motion is
$$m^{\mathrm{r}} a=F$$
where $F$ is the gravitational force exerted by the planet
$$F=G \frac{M^{\mathrm{G}} m^{\mathrm{G}}}{R^2}$$
(see Question 11.17). Thus
$$a=\frac{m^{\mathrm{G}}}{m^1} \frac{G M^{\mathrm{G}}}{R^2}$$

E-mail: help-assignment@gmail.com  微信:shuxuejun

help-assignment™是一个服务全球中国留学生的专业代写公司