1.  [5] After 6 half-lives (4.2 billion / 0.7 billion = 6), the number of U-235 atoms is reduced by 2⁶ = 64. So, divide 512 bazillion by 64 to get the answer: 8 bazillion.

2. [5] the moon rises on the eastern horizon.  Therefore, if the moon is rising at midnight it would be directly overhead (i.e., crossing your meridian) 6 hours later, at about 6 a.m.  This is 3rd quarter moon.

3. [5] under present day conditions, high tide occurs shortly after the moon crosses your meridian, i.e., when it is highest in your local sky. Since the moon is at it's highest point in the sky at noon when it is a New moon and at 6 p.m. when it is at 1st quarter moon, the moon must be in between New and 1st quarter moons if high tide occurs at 3 p.m., and must be a little bit past halfway from New to 1st quarter moon because of the slight delay between the moon going overhead and the tidal response.  Of course, we get this same condition if the moon is exactly on the other side of the Earth as well, so the moon could also be a  little bit past halfway from Full to 3rd quarter moon. And there is no way to decide between these two choices.

       if answer  = "new" or "1st" or "full" or "3rd", -4 points

4. [4] a) 3.9 BY  b) 4.3 BY  c) 4.4 BY  d) 4.562 BY

    if wrong dates, but right order of magnitude (several BY yrs), -3 pts

5. [3] 900 times weaker since the force of gravity between Earth and Moon is inversely proportional to the square of the distance to the moon

     if "weaker" without the right numerical value, -1 pt

6. [5] The tides induced by the moon would be at least 30 times weaker than at present.  Since the tides induced by the Sun are 40% the strength of those produced by the moon today, the tides would be dominantly controlled by the Sun. Therefore, low tides would occur shortly after 6 am and 6 pm.

        if answer "same as today", - 4 pts (this is correct if lunar tides are dominant)

7. [5] Assuming C/Halley and Uranus both have 80 year orbital periods around the Sun, then we know from Kepler's third law that both must have the same semi-major axis, since the only thing that determines the orbital period around the Sun is the size of the orbit as measured by the semi-major axis.  We are told that a = 19.2 for Uranus; therefore, a = 19.2 for C/Halley.  We are also told that C/Hally has a closest approach distance to the Sun of 0.6 AU.  Since the major axis = 2a = 38.4 AU and the major axis is the sum of the closest approach and most distant approach (aphelion) distances, C/Halley's aphelion distance must be 37.8 (i.e., 37.8 AU + 0.6 AU = 38.4 AU). From the information in the table, it is evident that C/Halley's orbital path crosses the orbits of  Venus, Earth, Mars, Jupiter, Saturn, Uranus and Neptune (7 planets).

        -1 for each extra (Mercury) or missing (e.g., Neptune) planet
           note: Pluto can be included or excluded.  If you treat it as a round orbit, then the answer is "no."  If you remember that it is in an elliptical orbit that brings it as close to the Sun as Neptune, then the answer is "yes."  Either is acceptable.

8. [3] radioactive decay (or Uranium, Thorium and Potassium)

9. [3] A solar day is 24 hours.

10. [3] A solar day is longer.  The difference is best described as a few minutes longer (4 minutes) than a sidereal day.

11. [3] Full moon to full moon (a synodic month) is longer (29.5 days) than the moon's orbital period (27.3 days).

12. [4] a week is one fourth of a month, and is easily marked by the change from new to 1st quarter, 1st quarter to full, full to 3rd quarter, and 3rd quarter to new moon.

13. [6] A seasonal year and the orbital period of the Earth would be identical if the Earth didn't precess.  But because of the 26,000 year precession cycle, the Earth's polar axis slowly changes position relative to the Sun.  If we imagine that at a given point in space, the northern hemisphere has summer today, in 13,000 years, the northern hemisphere will experience winter at this same point in space.  Thus, the seasons "migrate" around the earth's orbit with a 26,000 year period.  Thus, the seasonal year and orbital period differ by about 365.25 days divided by 26,000, or about 20 minutes.

14. [6] The seasons occur primarily because the rotation axis of the Earth is tilted (by 23.5 degrees) relative to the plane of the Earth's orbit around the Sun (the ecliptic plane).

15. [6]  A solar eclipse can only occur if 1) the moon is at new moon and 2) new moon occurs when the moon is in the ecliptic plane (i.e., the moon is directly along the line between the Sun and Earth, not above or below this line, or the ecliptic plane).  We don't have solar (or lunar) eclipses every month because the moon does not orbit in the ecliptic plane; instead, the lunar orbit is tilted, or inclined, to the ecliptic, crossing the ecliptic at only two points each orbit.

16. [6] The tides have 1) slowed down the moon's rotation period so that it now rotates exactly once per orbit, 2) slowed down the Earth's rotation period from about 18 hours, 1 billion years ago, to 24 hours today, and 3) caused the moon to spiral outwards from the Earth, thereby increasing the distance from the Earth to the Moon and increasing the length of the Month.

      notes: answers should be clear: "the Earth slows down" is not a very meaningful answer.  What aspect of the Earth slows down?

17.[4] These three effects occur for any planet-moon system (or star-planet system, like the Sun and Mercury) and has been most profound with the Pluto-Charon system, in which the planet (Pluto) and moon (Charon) both rotate at the same speed (6.2 days) and the moon's orbital period (or their orbital period around each other) is also 6.2 days.  In the Earth-Moon system, only two of these three periods are identical, as the Earth still rotates fairly fast.

      Many other good answers are possible (e.g., the moon systems of the giant planets, the Mars-Phobos interaction; the Sun-Mercury interaction), if they are well explained.

18. [6] Carbon-10 would have only 4 neutrons (as all carbon atoms must have 6 protons).  With only 4 neutrons, the repulsive (electro-magnetic) force between the positively charged protons would be stronger than the attractive (strong nuclear) force between all 10 of these particles.  The nucleus needs two more neutrons (minimum of 6), to provide a bit more of the attractive force, in order to overcome the proton-proton repulsion so that  the nucelus can hold itself together.  Thus, nature can't build carbon-10 atoms that don't fall apart.

      The question asks "why." Simply saying "unstable" or "the number of protons must be at least equal to the number of neutrons" is not a good answer.

19.  [6] Impact craters: can be very large, generally low elevation (below surrounding topography), circular, surrounding rings of mountains, central peaks (for big ones), circular "eject blanket," radial lines of ejecta out to large distances, secondary craters

       Volcanic craters: most are fairly small, generally high elevation (at tops of mountains), irregular shaped,

20. [6]  From crater counting, we know that the lunar highlands have up to ten times more craters than the lunar maria, yet the highlands are not ten times older than the maria; they are only a little older 4.2-4.4 BY vs. 3.3 - 3.8 BY).  We also infer from the younger regions (the maria) that the cratering rate on the moon has been fairly constant for the last 3 BY.  Therefore, we conclude that at a time in the past from 3.5 - 4.5 BY ago, the moon endured a much greater rate of impact bombardment than from 3.5 BY ago to the present.  We call this early period the "heavy" or "early" or "late stage" bombardment.

        Other answers are possible also, if they are well explained.

21. [6]  In 3 billion years (3 x 10⁹yrs), the moon's orbit has expanded by

Using Kepler's 3rd Law, we can write that