Physics needed: Kepler's first and third laws.

   The Appendix gives a brief derivation of Kepler's first and third laws, starting from constant energy and constant angular momentum.

The problem: Calculate, using Kepler's third law, the time needed for a spacecraft to travel from Earth to Mars, stay on Mars a while, and then return to Earth. Assume circular orbits for both Earth and Mars and a spacecraft orbit that just touches the two planet orbits. Given: the radii of the planet's orbits are 1 and 1.6 A.U. respectively. Round off all periods to simple fractions of a year. Sketch the planetary positions at the time of launch and at the time of arrival at Mars, together with the spacecraft orbit from Earth to Mars, and again for the return trip.

The setting: In 1969 and for a few years thereafter, humans visited the Moon. The next major goal for human travel is Mars, perhaps by the year 2020. Even now, Mars is being explored intensively, by computer-directed spacecraft, because of indirect (and debatable) evidence that some form of primitive life may have occurred on Mars long ago.

   To send humans to Mars is technically much more complicated. A massive spacecraft will be needed to assure human survival. Its launch into an orbit to Mars will need the most powerful rockets available. Given likely rocket power, even twenty years from now, an orbit from Earth to Mars will have to be chosen to minimize the required energy. With the approximation that the orbits of Earth and Mars are circles, which is adequate for our purposes, the minimum-energy orbit is an ellipse (practically a circle) that just touches the orbits of Earth and Mars. To visit and stop at Mars, the orbit will consist of two half ellipses. The spacecraft will be launched from Earth, perform half an orbit in traveling from Earth to Mars, land on Mars, later be launched from Mars, perform half an orbit, and land on Earth. It must stay on Mars long enough so that the Earth will be in the right position for the space craft to meet Earth. The question is: how long will the astronauts be absent from Earth?

Physics needed: Energy equation for orbits around the Sun. The Appendix gives a short derivation starting from constant angular momentum and constant total energy.

The problem: Using the energy equation for a Keplerian orbit around the Sun, evaluate v² in (m/s) ² for the Earth (assuming a circular orbit at r = 1 A.U., and for the minimum-energy orbit from Earth to Mars when the spacecraft is at Earth's orbit (but free of Earth's gravity) r = 1 A.U.. Evaluate the difference in v². Compare to the value of v² needed to escape the Earth's gravity. Which takes less energy, leaving the Earth's gravity or changing from circular to elliptical orbit around the Sun?

   Also evaluate the difference in v² between the space probe in its minimum-energy orbit arriving at Mars (r = 1.6 A.U.) and in its circular orbit with Mars (but free of Mars' gravity) at r = 1.6 A.U. Compare to the v² needed to escape Mars' gravity (mass of Mars = 0.11 Earth's mass, radius of Mars = 0.53 Earth's radius).

The setting: The space craft taking humans to Mars and back to Earth on a minimum-energy orbit (tangent to the orbits of Earth and Mars, both assumed circular) must use rocket propulsion to escape Earth's gravity, then to accelerate to acquire the minimum-energy orbit to Mars, later to accelerate to match the speed of Mars, and finally to reduce the speed of falling onto Mars; still later, rocket propulsion is again needed to escape Mars' gravity, to slow down to acquire the minimum-energy orbit to Earth, to slow down to match the speed of Earth, and finally to reduce the speed of falling to Earth. This problem shows that all of these maneuvers require significant energies. They are compared to the energy needed to escape Earth's gravity because most students have seen pictures of major rocket launchings and have at least an intuitive feeling that this energy is quite large.

The solution: The relevant formula for the velocity of a planet or smaller object in an orbit around the Sun, with semi-major axis a and distance r from the Sun, is v² = GMo[(2/r) - (1/a)]. For Earth orbit, v²= 8.9x10⁸ (m/s) ², for the spacecraft at Earth orbit, r = 1 and a = 1.3, v² is 1.23 times larger, so the extra v²that must be imparted is v² = 2.0 x10⁸ (m/s) ². For comparison, the value of v²needed to escape Earth's gravity, is v²= 2GM_E/r_E = 1.25x10⁸ (m/s) ². Conclusion: The energy needed for the space craft to escape the Earth's gravity is less than the energy needed, after escaping from Earth's gravity, to place the craft into the correct orbit around the Sun.

   At Mars, the energy required to change from the elliptical orbit to the circular orbit moving with Mars is given by v²= (5.5-4.2)x10⁸ (m/s) ² = 1.3x10⁸ (m/s) ², once again a substantial energy. The energy to escape from Mars is v²= 0.25x10⁸ (m/s) ².

Interpretation: The energies needed to change orbit are quite comparable to the energies needed to escape Earth's gravity. The energies are near the limits of technical capability if the spacecraft is heavy enough to carry humans safely to Mars and back. The use of v² has in effect focused on the energies needed by the unfueled space craft occupied by humans. Additional energy is needed for fuel, which is reduced every time the rockets are fired, and possibly discardable rocket stages.

   For space probes without humans, it is no longer necessary to use the minimum-energy orbit. For instance, the spacecraft Ulysses reached Jupiter on a much shorter orbit. It then gained energy while flying past Jupiter (see problem 2.2), enough so as to fly over the south pole and then over the north pole of the Sun.

Didactics:

1. The energy equation is an appropriate topic for lecture. However, then the students can solve the problem in groups, with different students serving as recorders and reporters than for problem 1.1. If one group reports some part of the problem incorrectly, let other student groups recognize the mistake. The professor need only make sure that the incorrect group figures out what they did not understand correctly.

2. The energy equation simply means: kinetic energy plus gravitational energy equals a negative constant, the total energy. The derivation in the Appendix merely identifies the constant. Check the constant: It yields the correct result for a circular orbit, r = a.

3. Students may be tempted to evaluate v(final)-v(initial) during the time the rocket operates and then square that difference. They will get different answers because they are implicitly placing themselves into different frames of reference. Energy is frame-dependent. All the energies of the spacecraft and planet orbits are evaluated in one frame, stationary with respect to some distant star.

4. Check that your class realizes: no rocket power is needed while in orbit to Mars. Rocket power is used briefly to provide the necessary energy, but then the spacecraft moves without effort along its new Keplerian orbit, until the rockets must again be used briefly upon reaching Mars.

5. Check that your class understands the minimum-energy orbit qualitatively: Once free of Earth's gravity, the spacecraft is in a circular orbit around the Sun. The spacecraft must be accelerated so that it "overshoots" Earth's orbit to reach further out. Once at Mars, it must again be accelerated to achieve a circular orbit at the distance of Mars. If it were not accelerated, it would continue on its elliptical orbit and begin to fall back toward r = 1 A.U.. You can check that the class understands this : Ask for equivalent arguments about the decelerations on the way back from Mars to Earth. (At Mars, the spacecraft must be decelerated in order to "undershoot" Mars' orbit and begin to fall toward r = 1 A.U.; at Earth, the spacecraft must again be decelerated because otherwise the craft would start traveling outwards again.)

6. Once again, students should work out this problem using no more than two significant digits. The differences in v²computed here are very sensitive to the approximations made, especially the approximation that Mars moves in a circular orbit.

Physics needed: Two-body problem, for circular orbits. Doppler shift only for interpretation.

The problem: A star is observed to change its radial velocity, v^*(rad), sinusoidally with time, with an amplitude v^* and period P. Assume that the star moves because it is pulled by a planet orbiting in a circle around that star. Assume that our line of sight to the star is parallel to the plane of the planet orbit. The observed star velocity is relative to the center of mass, and the mass of the star, M^*, is known. Derive an equation for the mass of the planet, M_p, and its distance from the star, r, in terms of v^*, M^*, and P. Then express M_p in units of Jupiter's mass (M_J / Mo = 10^-3), r in units of A.U., M^* in units of Mo, and P in years. (The velocity v^* remains in m/s; Mo, G, A.U., and the year are among the units given in the Introduction.)

   Check that your equations work for Jupiter: its orbital speed is v_J = 12 km/s.

   Evaluate M_p and r, in these units, for the star 47 UMa : M^* = 1.05 Mo, v^* = 44 m/s, and P = 3 years; and for 51 Peg: M^* = 1.0, v^* = 58 m/s, and P = 4.2 days.

The setting: The Sun is a fairly normal star. Therefore, most astronomers expect that there are other stars with planets orbiting around them. But it has been difficult to detect planets. They cannot be photographed because they are too faint compared to the immediately adjacent star. The contrast is smaller in the infrared. With improving infrared technology, perhaps planets can be photographed in the infrared within a few years.

   Since 1995, orbiting planets can be detected by the corresponding motion of the star (action and reaction). In the case of the Sun, Jupiter's gravitational pull on the Sun causes the Sun to perform a small orbit around the Jupiter-Sun center of mass, with a velocity of 12 m/s = 43 km/hour. (Given momentum conservation in the center-of-mass frame, this velocity is Jupiter's orbital velocity of v_p= 12 km/s multiplied by the mass ratio M_J/Mo = 10^-3.) After long technical development, periodically changing Doppler shifts as small as 10 m/s can now be measured for a few dozen reasonably bright stars.

The solution: We assume that the planet's orbital plane is in our line of sight. Then the amplitude of the observed v^*(rad) equals the actual orbital velocity of the star. Since the observed v^*(rad) varies sinusoidally in time, the orbit is circular. The planet, moving around its star, accelerates the star. The three fundamental equations are v^*/v_p = M_p/M^* (which represents momentum-conservation in the center-of-mass frame), 2r/P = v_p, and v_p²/r = GM^*/r² (force balance for the planet). Elimination of the unknown v_pgives two equations for r and M_p involving the observed quantities v^* and P, and the known M^*. One gets (M_p / M^*)³= v^*3P/(2GM^*) and r³=P²GM^*/4² . The latter should look familiar! For changing units, set M_p' = M_p/M_J and then M_p = M_J (M_p / M_J) = 10^-3MoM_p', use the value of Mo given in the Introduction. Similarly for other quantities. On introducing the asked-for units, one obtains M_p³ = 4.4x10^-5v^*3PM^*2 and r³ = P²M^*, where the primes have been omitted.

   For Jupiter, v^* = v_J (M_p / Mo) = 12 m/s, P = 12, M^* = 1, and the result is indeed M_p=1, r = 5.2, correct. Given the data in the problem, for 47UMa, M_p= 2.3, r = 2.1; for 51 Peg, M_p = 0.46, r = 0.05.

Interpretation: Given that v^* < 3m/s cannot now be measured, we can not yet detect planets with masses similar to Earth, only planets with masses similar to Jupiter. Nevertheless, the elation is great: We have finally detected planets around other stars! However, two possible objections have to be kept in mind:

   First, if the orbit is inclined to the line of sight, the actual orbital velocity is larger than the measured radial velocity and the properly deduced mass of the planet is larger. Thus the above procedure yields only a minimum planet mass, M_p > 2.3 for 47UMa, M_p > 0.46 for 51 Peg. One argues that, if the inclination angles are random, then most deduced minimum masses will be "close" (by better than a factor two) to the actual masses.

   Second, star spots and motions in the stellar surface might cause an apparent Doppler shift. For the stars observed so far no such confusion is expected.

   Surprise! No one expected the variety of orbits that have been detected so far.

1. Some planets such as the planet of 47 UMa are indeed similar to Jupiter in minimum mass and orbital radius.

2. Several planets with nearly circular orbits are at a surprisingly small distance from the star, with orbital periods of merely a few days. Examples are the planets of 51 Peg and of Boo, M_p > 3.87M_J, and r = 0.046 A.U. With their Jupiter-like masses, one expects these planets to be gaseous (mostly hydrogen, like Jupiter). How can gaseous atmospheres so close to the star survive the intense heating by the star? Have these planets perhaps formed only recently or perhaps recently arrived from a larger orbit, or both? All the central stars are (selected to be) fairly similar to the Sun. Clearly we do not yet understand the formation of planetary systems such as our solar system.

3. Some derived orbits are very eccentric orbits. Examples: 70VirB, with M_p > 8.1 M_J, P = 117 days, has an orbit reaching from 0.3 to 0.7 A.U. from the star; and 16CygB, with P = 2.2 years, has an orbit reaching from 0.6 to 2.8 A.U. Many astronomers are convinced that planets must form out of a rotating disk of gas and dust. Such planets must form with a nearly circular orbit. Therefore, these astronomers argue, any object with a very elliptical orbit must actually be a star which formed together with the central star in a common cloud of gas. But not everyone agrees.

   There are no planets detected with P around 10 years or longer, in part because such planets would cause v^* to be unobservably small, in part because accurate data have been taken and analyzed for only a few years, so that the oscillation in v^*(rad) cannot yet have become obvious.

Didactics:

1. The conversion of units is essential to recover a sense of the physical quantities such as planetary masses and years that we are familiar with.

2. The mass of the planet is important in relating v^* to v_p, but it has been neglected wherever it appears together with M^*, in particular, in Kepler's third law.

Physics needed: Two body problem; formula for Doppler shift.

Didactic purpose: An error analysis typical for frontier science.

The problem: Two spectra are given for the spectroscopic binary Mizar A ((UMa ). At the time of spectrum A, the spectrum lines are single. At the time of spectrum B, two days later, the lines are double. Their separation informs us of the radial velocity of the stars relative to each other.

   Given the wavelength calibration on the spectrum, measure the relative velocity in km/s using the Doppler formula. Estimate the uncertainty in your value : If your best measurement is x km/s, what is the largest y so that x+y or x-y might be correct values, given this spectrum? Your estimate of the uncertainty will be compared to the range of velocities measured by others. According to Kepler's third law (extended for two objects of similar masses M₁and M₂), M₁+M₂depends on the cube of the velocity difference. What is the uncertainty in M₁+M₂?

   The two lines in spectrum B are shifted nearly equally to the right and left of the single line in spectrum A. That suggests the two stars have nearly equal masses . In the center-of-mass frame, M₁v₁= M₂v₂. Estimate the uncertainty in the ratio of masses, given your estimated uncertainty in radial velocity.

Didactic purposes: A stretch of the imagination. Interpretation of uncertain data.

Physics purpose: Demonstration of the existence of black holes.

The problem: Deduce the mass of the stellar black hole called Cyg X-1, given that its visible orbiting companion has a mass of M₁= 33 Mo with an uncertainty plus or minus 9Mo, an orbit with a period of 5.6 days and a velocity amplitude of 76 km/s. Assume a circular orbit.

1. Assume M₁ = 24Mo, the orbital plane is parallel to our line of sight (sini=1). This yields the minimum mass for the invisible object.

2. Assume a more probable M₁= 33Mo and orbital inclination given by sini= 0.5. To do this, first correct the modified Kepler's third law so that it is valid for an orbit of an unknown inclination, and then eliminate one of the velocities in terms of the other (remembering action and reaction).

Physics background: Stars create their own gravity. If a star becomes smaller, the speed needed to escape from its surface increases. Eventually, when the star becomes sufficiently small, the escape speed exceeds the speed of light, c, and nothing can escape from the surface, not even radiation. Such an object is referred to as a black hole. A black hole occurs once the object shrinks so that its surface is within the "Schwarzschild radius", r = 2GM/c², where M is the mass of the object . If M = Mo, then r = 3 km. [As a mnemonic, this radius follows from the "classical" relation for escape speed c: 1/2 mc²= GMm/r . Actually the Schwarzschild radius was derived from general relativity by Karl Schwarzschild in 1916.]

Astrophysical background: The first good candidate as a black hole was the brightest x-ray source in the star constellation Cygnus, called Cyg X-1. In 1971, radio astronomers, detecting radio bursts from the same direction of the sky, located the x-ray emitting object much more precisely. They showed that it is in the same position as a hot (surface temperature about 3x10^4oK) and very large (radius about 20 solar radii) star which orbits about an invisible companion. The theory for the lives of stars (section V) says that the companion might be invisible because it is an extremely small star called a neutron star, with a radius of only 10 km, or because it is truly a black hole. Theory also says: Neutron stars can have a mass no greater than 5Mo, for otherwise they collapse to become a black hole. Therefore, if we can show that the mass of the invisible object exceeds 5Mo, then the invisible companion must be a black hole. (Several later problems deal with neutron stars, starting problem 3.1.)

The solution: We start with the modified form of Kepler's third law, M₁+M₂= (v₁+v₂)³ (P/2G). Here the v are the actual orbital velocities, but v₂is not observed. a) From momentum conservation in the center-of-mass frame, we know v ₂= v₁M₁/M₂. Kepler's third law becomes v₁³ (P/2G) = M₂³/( M₁+M₂)². With P = 5.6 days and v₁= 76 km/s, the left side becomes 0.25 Mo. With M₁ = 24 Mo, one gets M₂= 6 Mo . b) Now we need to realize that the actual orbital velocities are larger than the observed radial components, v = v_r/sini . Kepler's third law now becomes v_r1³ (P/2G) = (M₂sini) ³/( M₁+M₂)² . The left side is still 0.25 Mo. For M₁= 33 Mo and sini=0.5, one gets M₂= 15 Mo.

Interpretation: The mass derived for the invisible object is high enough for the object to be a black hole, even in the improbable case a). In fact, sini =1 is not possible because the x-ray source attached to the invisible object is not eclipsed by the star and, also, the computed separation of the two objects would be no larger than the sugergiant's radius, which would imply a huge flow of gas from the supergiant to the black hole. There is some mass flow, but much more modest. To explain some observed light variations of the visible star with the same period as the orbit, sini=0.5 has been deduced. The mass of the visible supergiant is uncertain in part because the spectrum of such a rare star is normally difficult to translate to a mass, in part because the visible star cannot be quite normal, situated so close to a sink for its gas and a powerful x-ray source.

   To summarize, the evidence for this black hole is very good, though by itself it does not convince all astronomers. There are now other stellar objects with very good evidence for a black hole. The existence of stellar-mass black holes is no longer in doubt.

   How can we say that x-rays and radio waves are associated with stellar black holes when black holes do not radiate? The answer is that gas flows from the surface of the visible star, falls toward and begins to swirl around the black hole. Indeed there is optical evidence for this gas. As it swirls ever closer to the black hole, it reaches temperatures on the order of 10⁷K, emitting x-rays. Only then does it fall into the black hole and disappear. Some of the x-rays, flickering on time scales less than one second, tell us that the gas falls into the black holes in chunks.

Didactics:

1. Most students readily accept the notion of black holes. It is worth pointing out that astronomers are a conservative and skeptical crowd (as is necessary to avoid mistakes in this frontier science). Black holes with stellar masses were known theoretically for decades, but many astronomers doubted seriously that they exist in nature. The general acceptance of the reality of stellar-mass black holes had to wait until the necessary x-ray technology came along. Similarly, the reality of long-predicted neutron stars had to await the development of radio technology and the discovery of pulsars (see problem 3.4).

2. Since the derivation of Kepler's third law in the Appendix is rather cumbersome, it may be useful to provide to the class (or have them derive) the same law for circular orbits. The two objects have a common period P, so v₁= 2 r₁/P and v₂ = 2r₂/P. Action, reaction, and the force balance yields M₁v₁²/r₁ = M₂v₂²/r₂ = GM₁M₂/(r₁+r₂)². If the velocity amplitudes are both measured, the ratio of the masses derives from M₁v₁= M₂v₂. Using a = r₁+r₂= r₁ (1+M₁/M₂), and v = v₁+v₂ = v₁ (1+M₁/M₂), one obtains most directly M₁+ M₂= av²/G, from which the usual forms follow by using 2a = vP.

Relevant physics: Gravity within a spherical distribution of mass, at a distance r from the center, is given by g(r)=GM(r)/r², as if all the mass M(r) out to a sphere of radius r were placed at the center. An outline of the proof appears in the Appendix. The relation is analogous to that found for some electrostatic problems.

Physics implication: Used for our Galaxy, g(r) leads to the inference of much invisible, "dark" matter in our Galaxy.

The astronomical setting: The stars that we see in the Milky Way constitute part of a large rotating disk of stars and gas that we call the Milky Way Galaxy or usually simply the Galaxy (written with a capital G). In the part of the sky containing the star constellation Andromeda, one can see (with binoculars) a spiral galaxy, called the Andromeda galaxy or M31, that probably looks similar to our Galaxy seen from a far. (M31 is 2x10⁶light years from us.) If the Sun were placed into the picture of the Andromeda galaxy, it would be near the outer part of the disk of stars. Most of the stars are a few light years apart. The Sun and the stars near us rotate about the center of the Galaxy with a speed of about 220 km/s, at a distance from the center of about 2.7x10⁴light years. (Newer data might change that distance by 10%.) Judging from the brightness of the center of the Andromeda galaxy, most of the stars of that galaxy are relatively near the center and distributed roughly spherically about the exact center. Although the center of our Galaxy is largely obscured from our vision by interstellar dust, we can tell that in our Galaxy also there is a roughly spherical distribution of stars around its center and distributed roughly spherically about the center. If we take this observation literally, then the Sun is outside most of the mass distribution and we can treat all the mass of the Galaxy as being at the center.

Didactic purpose: A further stretch of the imagination. Some actual space data.

Physics needed: Kepler's third law, thin-triangle relation d = r.

The problem: Calculate the mass of the black hole at the center of the galaxy M84 if gas observed 0.1 seconds of arc from the black hole circles the black hole with a speed of 400 km/s. Assume a distance of 5x10⁷light years.

The solution: Since 1 radian = 2x10⁵ seconds of arc, the gas is observed at a distance (5x10⁷light years)/2x10⁶= 25 light years from the center. Using Kepler's third law and the units used for the solar system, P = 2r/v = 4x10¹²s = 1.3x10⁵years, r = 1.6x10⁶A.U., M = r³/P² = 3x10⁸Mo.

The setting: In 1963, quasars (short for quasi-stellar objects) were discovered to be among the most distant objects in the Universe, giving off hundreds of times as much radiation as do large galaxies, and yet, judging from their light variations, as small as light weeks. These strange properties led to the suggestion that the power must come from gravitational energy, specifically from material falling into a very massive and very tiny object, a black hole. A black-hole mass of 10⁸Mo was suggested, but this value seemed quite incomprehensible at that time. Now we know that many galaxies contain very powerfully radiating centers, involving stellar masses up to 10⁹Mo. But are these centers really black holes? They might be merely dense accumulations of stars.

   In 1997, the Hubble Space Telescope (2.4 meter diameter, in orbit around Earth) observed the center of a galaxy (not a quasar) named M84. This galaxy is a member of a "nearby" cluster of galaxies, located merely some 5x10⁷light years from us. The galaxy is so nearby that the telescope could resolve light from merely 25 light years away from the center of M84 and measure the orbital velocity of the radiating gas. The deduced mass of 3x10⁸Mo is not particularly unusual, but the resolved small distance from the center is. If this mass consists of 3x10⁸stars distributed through a sphere of radius only 25 light years, then the stars are so densely packed (5x10³per cubic light year) that they would have collided by now, and the result would be a black hole.

Interpretation: The evidence for a very massive black hole at the center of the galaxy M84 is very good. Telescopic evidence for most other black holes is not as good. But the consensus now says that many galaxies indeed contain a black hole at the center.

   The range of black-hole masses is large. The galaxy M87 has a black hole of 2x10⁹Mo! One of the two small companion galaxies to the Andromeda galaxy (picture in problem 1.6) probably has a black hole of about 3x10⁶Mo. Our Milky Way Galaxy, although one of the larger galaxies, probably contains only a modest black hole with about 3x10⁶Mo (see didactic 3, below). Perhaps the smallness of our black hole is fortunate for our existence on Earth. Quasars are still believed to contain massive black holes, but they are too far away for good telescopic evidence.

   Why do bright galaxy centers and quasars with massive black holes shine so powerfully? The energy can be derived from roughly one star per year (or as much mass in clumps of gas) spiraling toward the black hole. A star is broken up by gravitational tidal forces, and its gases heat up and shine powerfully just before crossing the Schwarzschild radius. At this rate, a black-hole mass of 10⁸Mo is accumulated in merely 10⁸years. Many quasars are in galaxies that are colliding with other galaxies. The complicated gravity within the colliding galaxies sends stars falling toward the black hole. After some time, no more stars (or gases) are available from the surroundings, and the powerful radiation stops. On the scale of the age of the Universe (roughly 13x10⁹years), these are transitory phenomena. The data: The photograph represents the spectrum obtained by the Hubble Space Telescope. The diagram is intended to help interpret the spectrum. When the light from the galaxy falls onto the image plane of the telescope, where a photo of the galaxy might be taken, a thin rectangular slit is placed on that plane which transmits only the light from a slit-shaped region that includes the exact center of the galaxy. The length of the slit is about 600 light years at the galaxy. In the photo of M31 in Problem 1.6, this length would extend over roughly 1% of the size of that picture.