♥◘•○gRav!ty○•◘♥

Gravity
There is one important question which we have avoided discussing until now. Why do objects fall towards the surface of the Earth? The ancient Greeks had a very simple answer to this question. According to Aristotle, all objects have a natural tendency to fall towards the centre of the Universe. Since the centre of the Earth coincides with the centre of the Universe, all objects also tend to fall towards the Earth's surface. So, an ancient Greek might ask, why do the planets not fall towards the Earth? Well, according to Aristotle, the planets are embedded in crystal spheres which rotate with them whilst holding them in place in the firmament. Unfortunately, Ptolemy seriously undermined this explanation by shifting the Earth slightly from the centre of the Universe. However, the coup de grace was delivered by Copernicus, who converted the Earth into just another planet orbiting the Sun.

So, why do objects fall towards the surface of the Earth? The first person, after Aristotle, to seriously consider this question was Sir Isaac Newton. Since the Earth is not located in a special place in the Universe, Newton reasoned, objects must be attracted toward the Earth itself. Moreover, since the Earth is just another planet, objects must be attracted towards other planets as well. In fact, all objects must exert a force of attraction on all other objects in the Universe. What intrinsic property of objects causes them to exert this attractive force--which Newton termed gravity--on other objects? Newton decided that the crucial property was mass. After much thought, he was eventually able to formulate his famous law of universal gravitation:

Every particle in the Universe attracts every other particle with a force directly proportional to the product of their masses and inversely proportional to the square of the distance between them. The direction of the force is along the line joining the particles.

Incidentally, Newton adopted an inverse square law because he knew that this was the only type of force law which was consistent with Kepler's third law of planetary motion.

Consider two point objects of masses $m_1$ and $m_2$, separated by a distance $r$. As illustrated in Fig. 104, the magnitude of the force of attraction between these objects is
\begin{displaymath} f = G \frac{m_1 m_2}{r^2}. \end{displaymath} (541)

The direction of the force is along the line joining the two objects.

Figure 104: Newton's law of gravity. \begin{figure} \epsfysize =2in \centerline{\epsffile{newton.eps}} \end{figure}

Let ${\bf r}_1$ and ${\bf r}_2$ be the vector positions of the two objects, respectively. The vector gravitational force exerted by object 2 on object 1 can be written
\begin{displaymath} {\bf f}_{12} = G \frac{{\bf r}_2-{\bf r}_1}{\vert{\bf r}_2-{\bf r}_1\vert^3}. \end{displaymath} (542)

Likewise, the vector gravitational force exerted by object 1 on object 2 takes the form
\begin{displaymath} {\bf f}_{21} = G \frac{{\bf r}_1-{\bf r}_2}{\vert{\bf r}_1-{\bf r}_2\vert^3} = - {\bf f}_{21}. \end{displaymath} (543)

The constant of proportionality, $G$, appearing in the above formulae is called the gravitational constant. Newton could only estimate the value of this quantity, which was first directly measured by Henry Cavendish in 1798. The modern value of $G$ is
\begin{displaymath} G = 6.6726\times 10^{-11} {\rm N m^2/ kg^2}. \end{displaymath} (544)

Note that the gravitational constant is numerically extremely small. This implies that gravity is an intrinsically weak force. In fact, gravity usually only becomes significant if at least one of the masses involved is of astronomical dimensions (e.g., it is a planet, or a star).

Let us use Newton's law of gravity to account for the Earth's surface gravity. Consider an object of mass $m$ close to the surface of the Earth, whose mass and radius are $M_\oplus =5.97\times 10^{24} {\rm kg}$ and $R_\oplus = 6.378\times 10^6 {\rm m}$, respectively. Newton proved, after considerable effort, that the gravitational force exerted by a spherical body (outside that body) is the same as that exerted by an equivalent point mass located at the body's centre. Hence, the gravitational force exerted by the Earth on the object in question is of magnitude
\begin{displaymath} f = G \frac{m M_\oplus}{R_\oplus^{ 2}}, \end{displaymath} (545)

and is directed towards the centre of the Earth. It follows that the equation of motion of the object can be written
\begin{displaymath} m \ddot{\bf r} = - G \frac{m M_\oplus}{R_\oplus^{ 2}} \hat{\bf z}, \end{displaymath} (546)

where $\hat{\bf z}$ is a unit vector pointing straight upwards (i.e., away from the Earth's centre). Canceling the factor $m$ on either side of the above equation, we obtain
\begin{displaymath} \ddot{\bf r} = - g_\oplus \hat{\bf z}, \end{displaymath} (547)

where
\begin{displaymath} g_\oplus = \frac{G M_\oplus}{R_\oplus^{ 2}} = \frac{ (6.6... ....97\times 10^{24})}{(6.378\times 10^6)^2} = 9.79 {\rm m/s^2}. \end{displaymath} (548)

Thus, we conclude that all objects on the Earth's surface, irrespective of their mass, accelerate straight down (i.e., towards the Earth's centre) with a constant acceleration of $9.79 {\rm m/s^2}$. This estimate for the acceleration due to gravity is slightly off the conventional value of $9.81 {\rm m/s^2}$ because the Earth is actually not quite spherical.

Since Newton's law of gravitation is universal, we immediately conclude that any spherical body of mass $M$ and radius $R$ possesses a surface gravity $g$ given by the following formula:
\begin{displaymath} \frac{g}{g_\oplus} = \frac{M/M_\oplus}{(R/R_\oplus)^2}. \end{displaymath} (549)

Table 6 shows the surface gravity of various bodies in the Solar System, estimated using the above expression. It can be seen that the surface gravity of the Moon is only about one fifth of that of the Earth. No wonder Apollo astronauts were able to jump so far on the Moon's surface! Prospective Mars colonists should note that they will only weigh about a third of their terrestrial weight on Mars.


Table 6: The mass, $M$, radius, $R$, and surface gravity, $g$, of various bodies in the Solar System. All quantities are expressed as fractions of the corresponding terrestrial quantity.
Body $M/M_\oplus$ $R/R_\oplus$ $g/g_\oplus$
Sun $3.33\times 10^5$ 109.0 28.1
Moon 0.0123 0.273 0.17
Mercury 0.0553 0.383 0.38
Venus 0.816 0.949 0.91
Earth 1.000 1.000 1.000
Mars 0.108 0.533 0.38
Jupiter 318.3 11.21 2.5
Saturn 95.14 9.45 1.07
Gravitational potential energy
We saw earlier, in Sect. 5.5, that gravity is a conservative force, and, therefore, has an associated potential energy. Let us obtain a general formula for this energy. Consider a point object of mass $m$, which is a radial distance $r$ from another point object of mass $M$. The gravitational force acting on the first mass is of magnitude $f = G M/r^2$, and is directed towards the second mass. Imagine that the first mass moves radially away from the second mass, until it reaches infinity. What is the change in the potential energy of the first mass associated with this shift? According to Eq. (155),
\begin{displaymath} U(\infty) - U(r) = -\int_{r}^\infty [-f(r)] dr. \end{displaymath} (550)

There is a minus sign in front of $f$ because this force is oppositely directed to the motion. The above expression can be integrated to give
\begin{displaymath} U(r) = -\frac{G M m}{r}. \end{displaymath} (551)

Here, we have adopted the convenient normalization that the potential energy at infinity is zero. According to the above formula, the gravitational potential energy of a mass $m$ located a distance $r$ from a mass $M$ is simply $-G M m/r$.

Consider an object of mass $m$ moving close to the Earth's surface. The potential energy of such an object can be written
\begin{displaymath} U = -\frac{G M_\oplus m}{R_\oplus +z}, \end{displaymath} (552)

where $M_\oplus$ and ${R_\oplus}$ are the mass and radius of the Earth, respectively, and $z$ is the vertical height of the object above the Earth's surface. In the limit that $z\ll R_\oplus$, the above expression can be expanded using the binomial theorem to give
\begin{displaymath} U \simeq -\frac{G M_\oplus m}{R_\oplus} + \frac{G M_\oplus m}{R_\oplus^{ 2}} z, \end{displaymath} (553)

Since potential energy is undetermined to an arbitrary additive constant, we could just as well write
\begin{displaymath} U \simeq m g z, \end{displaymath} (554)

where $g = G M_\oplus/R_\oplus^{ 2}$ is the acceleration due to gravity at the Earth's surface [see Eq. (548)]. Of course, the above formula is equivalent to the formula (125) derived earlier on in this course.

For an object of mass $m$ and speed $v$, moving in the gravitational field of a fixed object of mass $M$, we expect the total energy,
\begin{displaymath} E = K + U, \end{displaymath} (555)

to be a constant of the motion. Here, the kinetic energy is written $K= (1/2) m v^2$, whereas the potential energy takes the form $U= - G M m/r$. Of course, $r$ is the distance between the two objects. Suppose that the fixed object is a sphere of radius $R$. Suppose, further, that the second object is launched from the surface of this sphere with some velocity $v_{\rm esc}$ which is such that it only just escapes the sphere's gravitational influence. After the object has escaped, it is a long way away from the sphere, and hence $U=0$. Moreover, if the object only just escaped, then we also expect $K=0$, since the object will have expended all of its initial kinetic energy escaping from the sphere's gravitational well. We conclude that our object possesses zero net energy: i.e., $E = K+U = 0$. Since $E$ is a constant of the motion, it follows that at the launch point
\begin{displaymath} E = \frac{1}{2} m v_{\rm esc}^{ 2} - \frac{G M m}{R} = 0. \end{displaymath} (556)

This expression can be rearranged to give
\begin{displaymath} v_{\rm esc} = \sqrt{\frac{2 G M}{R}}. \end{displaymath} (557)

The quantity $v_{\rm esc}$ is known as the escape velocity. Objects launched from the surface of the sphere with velocities exceeding this value will eventually escape from the sphere's gravitational influence. Otherwise, the objects will remain in orbit around the sphere, and may eventually strike its surface. Note that the escape velocity is independent of the object's mass and launch direction (assuming that it is not straight into the sphere).

The escape velocity for the Earth is
\begin{displaymath} v_{\rm esc} = \sqrt{\frac{2 G M_\oplus}{R_\oplus}} = \sqr... ...mes(5.97\times 10^{24})}{6.378\times 10^6}}= 11.2 {\rm km/s}. \end{displaymath} (558)

Clearly, NASA must launch deep space probes from the surface of the Earth with velocities which exceed this value if they are to have any hope of eventually reaching their targets.