Gravitation

Gravity or gravitation is a natural phenomenon by which all things with energy are brought toward (or gravitate toward) one another, including stars, planets, galaxies, and even light and sub-atomic particles. Gravity is responsible for many of the structures in the universe, by creating spheres of hydrogen—where hydrogen fuses under pressure to form stars—and grouping them into galaxies. On Earth, it gives weight to physical objects and causes the tides. It has an infinite range, although its effects become increasingly weaker on objects farther away.

Gravity is accurately described by the general theory of relativity which describes gravity not as a force but as a consequence of the curvature of spacetime, caused by the uneven distribution of mass/energy, and resulting in gravitational time dilation, where time lapses more slowly in lower (stronger) gravitational potential.

However, for most applications, gravity is well approximated by Newton's law of universal gravitation, which postulates that gravity is a force where two bodies of mass are directly drawn or attracted to each other according to a mathematical relationship, where the attractive force is proportional to the product of their masses and inversely proportional to the square of the distance between them.

Mass is a fundamental property of all particles, and in conventional theories of physics, from Newtonian mechanics to string theory, it is positive or zero. In addition to setting the scale of particle acceleration in response to forces and setting the speed limit for particles, mass gives rise to gravitational force. It should be noted that only zero mass particles, the photon and the gluon, travel at the speed of light.

Everyday activities like dragging something across the floor or throwing a medicine ball give us insight into what is called inertial mass, \(m_\text{inertial}\), which is the resistance of an object to acceleration under a net force \(m_\text{inertial} = \dfrac{F_\text{net}}{a}.\) The bigger something is, the more inertial mass it tends to have and the harder it is to move the object.

We know from our experiences that it is harder to carry a box of books up the stairs than a tea kettle, and there is also a common aphorism, "the bigger they are, the harder they fall." All these show that gravity is a mass-dependent force: the bigger something is, the more gravitational mass it has and the harder it is to move the object. That is, \(F_\text{gravity} \sim f(m_\text{gravitational})\), where \(f\) is some monotonically increasing function of \(m_\text{gravitational}\).

However, the relationship between these two masses is not clear, and neither is the form of \(f\). Happily, there is a simple experiment to demonstrate what's known as the equivalence principle.

Equivalence Principle

After centuries of controversy as to whether or not Galileo actually dropped anything from the Leaning Tower of Pisa, TV physics personality Brian Cox finally solved the problem of relating \(m_\text{inertial}\) to \(m_\text{gravitational}\). Using an enormous vacuum chamber, a feather and a bowling ball were dropped from a tall crane. The ball and feather were found to hit the ground at precisely the same time, which showed that their accelerations were identical: \(a_\text{feather}=a_\text{bowling ball}\). This result implies that the acceleration of a mass in a gravitational field is independent of its mass.

If we take this seriously, it shows that

\[ \frac{F_\text{gravity}(m_\text{gravitational})}{m_\text{inertial}} =\text{ constant}, \]

which we find from \(F_\text{net} = F_\text{gravity} = m_\text{inertial}a\).

All else being equal, the ratio of the force felt by an object due to its gravitational mass to its inertial mass is a constant. The logic of this condition demonstrates two things:

• The force of gravity must be in direct proportion to the gravitational mass: \(F_\text{gravity}\propto m_\text{gravitational}\).
• Gravitational mass and inertial mass are identical: \(m_\text{gravitational} = m_\text{inertial}\).

This makes it clear why all objects accelerate uniformly in the gravitational field of the Earth.

If we call the gravitational field \(\Gamma\), then \(F_\text{gravity}=m_\text{inertial}\Gamma\), so \(m_\text{inertial}\Gamma = m_\text{inertial}a\) implies that \(\Gamma = a\), i.e. the force of gravity on an object is given by the mass of the object, multiplied by the gravitational field strength, which has units of acceleration.

This gives rise to a question: Is there any fundamental distinction between acceleration and a gravitational field?

Equivalence Principle

The inertial and gravitational masses of an object are indistinguishable:

\[m_\text{inertial} = m_\text{gravitational}. \ _\square\]

We know how particles respond to gravitational fields, but what gives rise to gravitational fields? As it turns out, gravitational fields interact with particles through their masses, and gravitational fields arise from particles via their masses, as is the case with electric fields and charged particles. In other words, gravity is the force of interaction between massive objects. This is a revealing point.

Gravitational field's dependence on mass of source particle:

In the previous example, we found the force of gravity on a particle with mass \(m_a\) in a gravitational field was \(m_a\Gamma\). If we consider \(\Gamma\) to be the field of another particle of mass \(m_b\), located a fixed distance away, then we can say the force on particle \(a\) is \(m_a\Gamma_b\).

However, we can view this interaction the other way around, as the force felt by particle \(b\) due to the field from particle \(a\). From this perspective, the force can be written as \(m_b\Gamma_a\), yielding

\[\begin{align} m_b\Gamma_a &= m_a\Gamma_b \\ \frac{m_b}{m_a} &= \frac{\Gamma_b}{\Gamma_a} = \text{ constant}, \end{align}\]

which implies that \(\Gamma_i \propto m_i\). Thus, the force of gravity between two massive objects is proportional to the product of their masses \(F_\text{ab} \propto m_am_b\). Because particle mass is non-negative, the gravitational interaction between massive objects can only be attractive.

Now we nearly have a complete description of Newtonian gravity, and all that is left is the force's dependence on the distance between particles. Here, there are two ways of going. One way is to point out that experimental measurements of the force of attraction between massive objects show that the gravitational force decreases as the inverse-square of the distance between objects, \(r^{-2}\). However, we'll consider a more interesting approach.

Gravitational field's dependence on distance from source:

Let's envision the field as a bundle of lines emanating in all directions. The number of lines emerging from the particle is in direct proportion to the mass of the particle \((\Gamma_S \propto m_s),\) and the strength of the interaction with another particle \(m_r\) is proportional to the number of field lines from \(m_s\) which intersect \(m_r\).

If we consider the field of a point particle \(m_s\), it clearly must have spherical symmetry. Now, if we surround \(m_s\) with an imaginary sphere, it is clear that no matter how large or how small the sphere, the same number of field lines from \(m_s\) will penetrate the surface of the sphere. However, the density of the field lines which penetrate the surface will decrease as the sphere grows larger, and so will the number of field lines that penetrate a distant particle.

Concretely, the number of field lines is a constant, but the surface area of the sphere grows as \(4\pi r^2\), so the density of field lines shrinks as \(\Gamma_S \sim \frac{1}{4\pi}\frac{m_s}{r^2}\). We can state this in the form of a flux equation. If we call the flux of field lines from \(m_s\) through a surrounding surface \(\Phi_G\), then we have \(\Phi_G \sim \frac{m_s}{r^2}\). Now, since field lines point inward toward the source particle, the flux through the surface will be inward and thus have a negative sign.

Now we have the complete form for Newton's law of gravitation, and the gravitational field due to a particle \(m_s\) at a distance \(r\) is given by

\[\Gamma_s \sim \frac{m_s}{r^2}.\]

The \(\sim\) implies that there is a numerical prefactor we've left out. We'll call this number \(G\), the gravitational constant. Its measurement was no small historical achievement.

Newton's Law of Gravitation

The strength of the gravitational interaction between two particles of masses \(m_a\) and \(m_b\), separated by a distance \(r\), is given by

\[F_{ab} = G\frac{m_am_b}{r^2}. \ _\square\]

The above equation can also be expressed in vector form:

\[\begin{align} \vec{F}=G\dfrac{m_1m_2}{r^2}(-\hat{r})&=-G\dfrac{m_1m_2}{r^2} \hat{r}\\\\ \vec{F}&=-G\dfrac{m_1m_2}{|r|^2} \hat{r}. \end{align}\]

Submit your answer

Two identical rockets are in circular trajectories around Earth under the control of their engines. Rocket A is twice as far from the center of Earth as rocket B, yet they have the same centripetal acceleration. What is the ratio of the velocity of rocket B to that of rocket A, \(\dfrac{v_B}{v_A}?\)

We mentioned that the distance dependence in Newton's law can be obtained by considering the total flux of gravitational field lines through an imaginary sphere. A surprising corollary to the flux relation, which we will now prove, is that a particle located inside of a spherically symmetric shell of mass will feel no gravitational attraction to the shell.

Show that a particle \(m_r,\) located inside a spherical shell of total mass \(M_s,\) feels no gravitational attraction to the shell.

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We focus on the diagram of the shell that's intersected by the spherical sections of radii \(r_L\) and \(r_R\). The two sections intersect the same small solid angle \(\Delta\Omega\) (we exaggerate the size of the sections for clarity, but they are taken to be very small), and thus the entire shell can be calculated using opposing pairs of equal solid angles. The entire shell has mass \(M_s\) and thus has mass density \(\sigma = \frac{1}{4\pi}\frac{M_S}{R^2},\) where \(R\) is the radius of the shell. Particle \(m_r\) is attracted by both sections in opposite directions.

The section of radius \(r_L\) has a total mass of \(M_L = \sigma \Omega r_L^2\), and the entire section is located a distance of approximately \(r_L\) from the particle. Thus, the gravitational field strength at \(m_r\) due to the section is \[\sim\Gamma_L = G\frac{M_L}{r_L^2} = G\frac{\sigma\Delta\Omega r_L^2}{r_L^2} = G\sigma\Delta\Omega\] and points along the gold line toward the left.

You may have noticed that the previous calculation results in an \(r_L\)-independent expression. The calculation for the gravitational field due to the \(r_R\) section is similarly independent of \(r_R\) and is equal to \(G\sigma\Delta\Omega\), pointing along the gold line toward the right. I.e. the net force on \(m_r\) from the two sections of the shell is \(G\sigma\Delta\Omega-G\sigma\Delta\Omega=0\).

Because the shell can be deconstructed into a collection of such pairs, there is no net force on the particle, because the attraction of every patch of surface cancels exactly with its partner. Note that this result is general for any particle inside the spherical shell, and does not depend on the particular location inside of the shell. \(_\square\)

Challenge yourself

Using a similar line of argument, one can show that outside of a spherical shell of mass \(M_s\), the field strength is given by \(\Gamma_s = G\frac{M_s}{r^2},\) where \(r\) is the distance from the center of the sphere. Can you do it?

\[\sqrt{2}\frac{Gm_b^2}{d^2}\] \[2\frac{Gm_b^2}{d^2}\] \[2\frac{Gm_b^2}{d^2} + \frac{Gm_bM_s}{R^2}\] \[\sqrt{2}\frac{Gm_b^2}{d^2} + \frac{Gm_bM_s}{R^2}\]

A green sphere of mass \(m_b\) is placed inside a spherical shell of radius \(R\) and mass \(M_s\). Two spheres identical in mass to the green sphere are placed a distance \(d\) to the right and to the bottom, respectively, of it. What is the magnitude of the gravitational force on the green sphere?

Mechanics problems often approximate the potential energy change of an object near the surface of Earth to be \(\Delta U = mg\Delta h\), the mass of the object times the change in height times a field of constant strength \(g = 9.8\text{ m/s}^2\). However, we know that the gravitational field of Earth falls off as the square of the height above the Earth's surface \(\sim r^{-2} = \left(R_\text{Earth}+h\right)^{-2}\), where \(R_\text{Earth}\) is the radius of Earth, i.e. non-constant. The question stands, "How do we reconcile the approximation with the inverse square law of gravity?"

Potential Energy Approximation

Let's calculate the work required to lift an object a distance \(\Delta h\) in the gravitational field of Earth. Since the particle starts and ends with zero velocity, the work done is stored in the form of potential energy, \(W = \Delta U\). We have

\[\begin{align}W &= \int F(r)dr\\ &= \int_{R_\text{Earth}}^{R_\text{Earth}+\Delta h} G\frac{m_rM_\text{Earth}}{r^2} dr. \end{align}\]

The integral is simple and gives us

\[\begin{align} W &= -Gm_rM_\text{Earth} \left(\frac{1}{R_\text{Earth}+\Delta h} - \frac{1}{R_\text{Earth}} \right) \\ &= Gm_rM_\text{Earth} \frac{\Delta h}{\left(R_\text{Earth}+\Delta h\right)R_\text{Earth}}. \end{align}\]

Now, as long as the change in height \(\Delta h\) is small compared to the radius of Earth, we can say \(\left(R_\text{Earth}+\Delta h\right)R_\text{Earth} \approx R_\text{Earth}^2\). We therefore have

\[\Delta U = m_r\frac{GM_\text{Earth}}{R_\text{Earth}^2}\Delta h.\]

Now things are starting to look familiar. In the usual approximation we have \(\Delta U = m_r g \Delta h\) and here we have \(\Delta U = m_r \frac{GM_\text{Earth}}{r_\text{Earth}^2}\Delta h\). For the two schemes to be in agreement, it should be so that \(g = \dfrac{GM_\text{Earth}}{r_\text{Earth}^2}\).

Let's check

\[\begin{align} \frac{GM_\text{Earth}}{r_\text{Earth}^2} &\approx \frac{6.7\times10^{-11}\frac{\text{Nm}^2}{\text{kg}^2}\times 6\times10^{24}\text{ kg} }{ \left(6.4 \times 10^6 \text{m}\right)^2 } \\ &\approx \frac{6.7\times 6 \times 10}{6.4^2}\text{ m/s}^2\\ &\approx \frac{40\times10}{41}\text{ m/s}^2 \\ &\approx 9.8 \text{ m/s}^2. \end{align}\]

Thus, we have derived the connection between the usual \(g\) and the expression obtained from the approximation that underlies it:

\[\boxed{\displaystyle g = \frac{GM_\text{Earth}}{r_\text{Earth}^2}}.\]