Astronomy 3 - Solution Set 8

Astronomy 3 -Solution Set 8

  1. Describe the location of the equinoxes and solstices in the Uranian Sky. What are the seasons like on Uranus?

    At the summer equinox, the sun will be near the celestial pole of Uranus. As the Uranian year progresses, the sun will circle around the celestial pole and move further and further away, until it approaches the celestial equator in the Uranian Fall. This is the equinox. The sun will soon drop below the horizon for the duration of the winter. When the sun sets, you will be in continuous night for the next decade or two until the planet finally approaches Spring and the sun comes back to the celestial equator.

  2. Why is the belt-zone circulation difficult to detect on Uranus?

    Uranus is very cold, so cloud layers form at a low level in the atmosphere. The upper part of the atmosphere is cloudless hydrogen gas and it obscures the cloud layers. The hydrogen gas participates in any belt-zone circulation, but its motion is difficult to detect because of the lack of clouds.

  3. What is the escape velocity from the surface of an icy moon with a diameter of 20 km? (HINTS: The density of ice is 1 g/cm3. The volume of a sphere is (4/3)pr3

    The given information is the moon's diameter, so the radius is 10 km = 10,000 m. Convert the density to kg/m3:

    d = ([1 g/(cm3)]) ([1 kg/1000 g])([100 cm/1 m])([100 cm/1 m])([100 cm/1 m]) = 1000 kg/m3

    Find the mass: M = (4/3)pr3 d = (4/3) p(104 m)3 (103 kg/m3) = 4.19x1015 kg

    Now use the familiar formula for the escape velocity:

    Ve = [2GM/R]m1/2 = Ö{[(2(6.67x10-11 m3/kg  s2)(4.19x1015 kg))/(104 m)]} = 7.2 m/s

  4. Given the size of Triton's orbit (R = 355,000 km) and its orbital period (P = 5.877 days), calculate the mass of Neptune.

    First you convert the given period to seconds:

    P = (5.877 days) ([24 hours/1 day])([60 min/1 hour])([60 sec/1 min]) = 5.08x105 sec

    Then convert the distance from Neptune to Triton to meters:

    R = (355,000 km) ([1000 m/1 km]) = 3.55x108 m

    Finally, the formula you need is:

    M = [(4p2 R3)/(G P2)] = [(4p2 (3.55x108 m)3)/((6.67x10-11 m3/kg s2) (5.08x105 s)2)] = 1.025x1026 kg = 17 Earth masses

  5. How can Titan keep an atmosphere when it is smaller than airless Ganymede?

    Titan has a lower escape velocity than Ganymede. This makes it harder for Titan to keep an atmosphere, since gas molecules don't have to travel as fast to escape.

    However, Titan is farther from the sun and hence colder. This means that the gas molecules are traveling more slowly. In fact, they're traveling so slowly that Titan is able to hold on to an atmosphere.

  6. If you piloted a spacecraft to visit Saturn's moons and wanted to land on a geologically old surface, what features would you look for? What features would you avoid?

    Old surfaces are heavily scarred by craters. Young surfaces are smooth and have little scarring.

  7. Why do we expect small satellites to be irregular in shape? Give examples and exceptions.

    Small satellites have very weak gravitational fields, which makes them unable to pull themselves into a spherical shape.

    The fact that small satellites are sometimes captured asteroids is less important.

    Examples include the moons of Mars, Phobos and Deimos.

    To find a counterexample, you need to find a moon that has very little structural integrity, so that it's easy for the gravity to deform the moon into a sphere. You might look for small, icy moons.

    Another way to find a counterexample is to find a moon that will be intensely heated, so that the rock will melt and form a sphere in spite of the moon's small size. Look for small satellites near large planets, where tidal heating is intense. Saturn's moon Pan is an example.