## Chapter 13

### Check Your Understanding

The force of gravity on each object increases with the square of the inverse distance as they fall together, and hence so does the acceleration. For example, if the distance is halved, the force and acceleration are quadrupled. Our average is accurate only for a linearly increasing acceleration, whereas the acceleration actually increases at a greater rate. So our calculated speed is too small. From Newton’s third law (action-reaction forces), the force of gravity between any two objects must be the same. But the accelerations will not be if they have different masses.

The tallest buildings in the world are all less than 1 km. Since *g* is proportional to the distance squared from Earth’s center, a simple ratio shows that the change in *g* at 1 km above Earth’s surface is less than 0.0001%. There would be no need to consider this in structural design.

The value of *g* drops by about 10% over this change in height. So $\text{\Delta}U=mg({y}_{2}-{y}_{1})$ will give too large a value. If we use $g=9.80\phantom{\rule{0.2em}{0ex}}\text{m/s}$, then we get

$\text{\Delta}U=mg({y}_{2}-{y}_{1})=3.53\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{10}\phantom{\rule{0.2em}{0ex}}\text{J}$

which is about 6% greater than that found with the correct method.

The probe must overcome both the gravitational pull of Earth and the Sun. In the second calculation of our example, we found the speed necessary to escape the Sun from a distance of Earth’s orbit, not from Earth itself. The proper way to find this value is to start with the energy equation, Equation 13.5, in which you would include a potential energy term for both Earth and the Sun.

You change the direction of your velocity with a force that is perpendicular to the velocity at all points. In effect, you must constantly adjust the thrusters, creating a centripetal force until your momentum changes from tangential to radial. A simple momentum vector diagram shows that the net *change* in momentum is $\sqrt{2}$ times the magnitude of momentum itself. This turns out to be a very inefficient way to reach Mars. We discuss the most efficient way in Kepler’s Laws of Planetary Motion.

In Equation 13.7, the radius appears in the denominator inside the square root. So the radius must increase by a factor of 4, to decrease the orbital velocity by a factor of 2. The circumference of the orbit has also increased by this factor of 4, and so with half the orbital velocity, the period must be 8 times longer. That can also be seen directly from Equation 13.8.

The assumption is that orbiting object is much less massive than the body it is orbiting. This is not really justified in the case of the Moon and Earth. Both Earth and the Moon orbit about their common center of mass. We tackle this issue in the next example.

The stars on the “inside” of each galaxy will be closer to the other galaxy and hence will feel a greater gravitational force than those on the outside. Consequently, they will have a greater acceleration. Even without this force difference, the inside stars would be orbiting at a smaller radius, and, hence, there would develop an elongation or stretching of each galaxy. The force difference only increases this effect.

The semi-major axis for the highly elliptical orbit of Halley’s comet is 17.8 AU and is the average of the perihelion and aphelion. This lies between the 9.5 AU and 19 AU orbital radii for Saturn and Uranus, respectively. The radius for a circular orbit is the same as the semi-major axis, and since the period increases with an increase of the semi-major axis, the fact that Halley’s period is between the periods of Saturn and Uranus is expected.

Consider the last equation above. The values of ${r}_{1}$ and ${r}_{2}$ remain nearly the same, but the diameter of the Moon, $({r}_{2}^{}-{r}_{1}^{})$, is one-fourth that of Earth. So the tidal forces on the Moon are about one-fourth as great as on Earth.

Given the incredible density required to force an Earth-sized body to become a black hole, we do not expect to see such small black holes. Even a body with the mass of our Sun would have to be compressed by a factor of 80 beyond that of a neutron star. It is believed that stars of this size cannot become black holes. However, for stars with a few solar masses, it is believed that gravitational collapse at the end of a star’s life could form a black hole. As we will discuss later, it is now believed that black holes are common at the center of galaxies. These galactic black holes typically contain the mass of many millions of stars.

### Conceptual Questions

The ultimate truth is experimental verification. Field theory was developed to help explain how force is exerted without objects being in contact for both gravity and electromagnetic forces that act at the speed of light. It has only been since the twentieth century that we have been able to measure that the force is not conveyed immediately.

The centripetal acceleration is not directed along the gravitational force and therefore the correct line of the building (i.e., the plumb bob line) is not directed towards the center of Earth. But engineers use either a plumb bob or a transit, both of which respond to both the direction of gravity and acceleration. No special consideration for their location on Earth need be made.

As we move to larger orbits, the change in potential energy increases, whereas the orbital velocity decreases. Hence, the ratio is highest near Earth’s surface (technically infinite if we orbit at Earth’s surface with no elevation change), moving to zero as we reach infinitely far away.

The period of the orbit must be 24 hours. But in addition, the satellite must be located in an equatorial orbit and orbiting in the same direction as Earth’s rotation. All three criteria must be met for the satellite to remain in one position relative to Earth’s surface. At least three satellites are needed, as two on opposite sides of Earth cannot communicate with each other. (This is not technically true, as a wavelength could be chosen that provides sufficient diffraction. But it would be totally impractical.)

The speed is greatest where the satellite is closest to the large mass and least where farther away—at the periapsis and apoapsis, respectively. It is conservation of angular momentum that governs this relationship. But it can also be gleaned from conservation of energy, the kinetic energy must be greatest where the gravitational potential energy is the least (most negative). The force, and hence acceleration, is always directed towards *M* in the diagram, and the velocity is always tangent to the path at all points. The acceleration vector has a tangential component along the direction of the velocity at the upper location on the *y*-axis; hence, the satellite is speeding up. Just the opposite is true at the lower position.

The laser beam will hit the far wall at a lower elevation than it left, as the floor is accelerating upward. Relative to the lab, the laser beam “falls.” So we would expect this to happen in a gravitational field. The mass of light, or even an object with mass, is not relevant.

### Problems

$7.4\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{\mathrm{-8}}\phantom{\rule{0.2em}{0ex}}\text{N}$

a. $7.01\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{\mathrm{-7}}\phantom{\rule{0.2em}{0ex}}\text{N}$; b. The mass of Jupiter is

$\begin{array}{}\\ \\ {m}_{\text{J}}=1.90\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{27}\phantom{\rule{0.2em}{0ex}}\text{kg}\hfill \\ {F}_{\text{J}}=1.35\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{\mathrm{-6}}\phantom{\rule{0.2em}{0ex}}\text{N}\hfill \\ \frac{{F}_{\text{f}}}{{F}_{\text{J}}}=0.521\hfill \end{array}$

a. $9.25\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{\mathrm{-6}}\phantom{\rule{0.2em}{0ex}}\text{N}$; b. Not very, as the ISS is not even symmetrical, much less spherically symmetrical.

a. $1.41\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{\mathrm{-15}}{\text{m/s}}^{2}$; b. $1.69\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{\mathrm{-4}}\phantom{\rule{0.2em}{0ex}}{\text{m/s}}^{2}$

a. $1.62\phantom{\rule{0.2em}{0ex}}{\text{m/s}}^{2}$; b. $3.75\phantom{\rule{0.2em}{0ex}}{\text{m/s}}^{2}$

a. $5.85\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{10}\phantom{\rule{0.2em}{0ex}}\text{J}$; b. $\mathrm{-5.85}\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{10}\phantom{\rule{0.2em}{0ex}}\text{J}$; No. It assumes the kinetic energy is recoverable. This would not even be reasonable if we had an elevator between Earth and the Moon.

a. $5.08\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{3}\phantom{\rule{0.2em}{0ex}}\text{km}$; b. This less than the radius of Earth.

$1.89\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{27}\phantom{\rule{0.2em}{0ex}}\text{kg}$

a. $4.01\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{13}\phantom{\rule{0.2em}{0ex}}\text{kg}$; b. The satellite must be outside the radius of the asteroid, so it can’t be larger than this. If it were this size, then its density would be about $1200\phantom{\rule{0.2em}{0ex}}{\text{kg/m}}^{3}$. This is just above that of water, so this seems quite reasonable.

a. $1.66\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{\mathrm{-10}}\phantom{\rule{0.2em}{0ex}}{\text{m/s}}^{2}$; Yes, the centripetal acceleration is so small it supports the contention that a nearly inertial frame of reference can be located at the Sun. b. $2.17\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{5}\text{m}\text{/}\text{s}$

$1.98\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{30}\phantom{\rule{0.2em}{0ex}}\text{kg}$; The values are the same within 0.05%.

Compare Equation 13.8 and Equation 13.11 to see that they differ only in that the circular radius, *r*, is replaced by the semi-major axis, *a*. Therefore, the mean radius is one-half the sum of the aphelion and perihelion, the same as the semi-major axis.

The semi-major axis, 3.78 AU is found from the equation for the period. This is one-half the sum of the aphelion and perihelion, giving an aphelion distance of 4.95 AU.

### Additional Problems

a. $1.85\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{14}\phantom{\rule{0.2em}{0ex}}\text{N}$; b. Don’t do it!

The value of *g* for this planet is 3.8 m/s^{2}, which is about one-fourth that of Earth. So they are weak high jumpers.

a. The escape velocity is still 43.6 km/s. By launching from Earth in the direction of Earth’s tangential velocity, you need $43.4-29.8=13.8\phantom{\rule{0.2em}{0ex}}\text{km/s}$ relative to Earth. b. The total energy is zero and the trajectory is a parabola.

a. $1.3\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{7}\phantom{\rule{0.2em}{0ex}}\text{m}$; b. $1.56\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{10}\phantom{\rule{0.2em}{0ex}}\text{J}$; $\text{\u2212}3.12\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{10}\phantom{\rule{0.2em}{0ex}}\text{J}$; $\mathrm{-1.56}\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{10}\phantom{\rule{0.2em}{0ex}}\text{J}$

a. $6.24\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{3}\phantom{\rule{0.2em}{0ex}}\text{s}$ or about 1.8 hours. This was using the 520 km average diameter. b. Vesta is clearly not very spherical, so you would need to be above the largest dimension, nearly 580 km. More importantly, the nonspherical nature would disturb the orbit very quickly, so this calculation would not be very accurate even for one orbit.

a. 323 km/s; b. No, you need only the difference between the solar system’s orbital speed and escape speed, so about $323-228=95\phantom{\rule{0.2em}{0ex}}\text{km/s}$.

Setting $e=1$, we have $\frac{\alpha}{r}=1+\text{cos}\theta \to \alpha =r+r\text{cos}\theta =r+x$; hence, ${r}^{2}={x}^{2}+{y}^{2}={(\alpha -x)}^{2}$. Expand and collect to show $x=\frac{1}{\mathrm{-2}\alpha}\phantom{\rule{0.1em}{0ex}}{y}^{2}+\frac{\alpha}{2}$.

Substitute directly into the energy equation using $p{v}_{p}=q{v}_{q}$ from conservation of angular momentum, and solve for ${v}_{p}$.

### Challenge Problems

$g=\frac{4}{3}\phantom{\rule{0.1em}{0ex}}G\rho \pi r\to F=mg=\left[\frac{4}{3}\phantom{\rule{0.1em}{0ex}}Gm\rho \pi \right]\phantom{\rule{0.1em}{0ex}}r$, and from $F=m\phantom{\rule{0.1em}{0ex}}\frac{{d}^{2}r}{d{t}^{2}}$, we get $\frac{{d}^{2}r}{d{t}^{2}}=\left[\frac{4}{3}\phantom{\rule{0.1em}{0ex}}G\rho \pi \right]\phantom{\rule{0.1em}{0ex}}r$ where the first term is ${\omega}^{2}$. Then $T=\frac{2\pi}{\omega}=2\pi \sqrt{\frac{3}{4G\rho \pi}}$ and if we substitute $\rho =\frac{M}{4\text{/}3\pi {R}^{3}}$, we get the same expression as for the period of orbit *R*.

Using the mass of the Sun and Earth’s orbital radius, the equation gives $2.24\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{15}{\text{m}}^{2}\text{/s}$. The value of $\pi {R}_{\text{ES}}^{2}\text{/}(1\phantom{\rule{0.2em}{0ex}}\text{year})$ gives the same value.

$\text{\Delta}U={U}_{f}-{U}_{i}=-\frac{G{M}_{\text{E}}m}{{r}_{f}}+\frac{G{M}_{\text{E}}m}{{r}_{i}}=G{M}_{\text{E}}m\left(\frac{{r}_{f}-{r}_{i}}{{r}_{f}{r}_{i}}\right)$ where $h={r}_{f}-{r}_{i}$. If $h\text{<}\phantom{\rule{0.2em}{0ex}}\text{<}{\text{R}}_{\text{E}}$, then ${r}_{f}{r}_{i}\approx {R}_{\text{E}}^{2}$, and upon substitution, we have

$\text{\Delta}U=G{M}_{\text{E}}m\left(\frac{h}{{R}_{\text{E}}^{2}}\right)=m\left(\frac{G{M}_{\text{E}}}{{R}_{\text{E}}^{2}}\right)h$ where we recognize the expression with the parenthesis as the definition of *g*.

a. Find the difference in force,

${F}_{\text{tidal}}==\frac{2GMm}{{R}^{3}}\text{\Delta}r$;

b. For the case given, using the Schwarzschild radius from a previous problem, we have a tidal force of $9.5\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{\mathrm{-3}}\phantom{\rule{0.2em}{0ex}}\text{N}$. This won’t even be noticed!