Sanchez
Astronomy Karen 104
Professor Wilcots
DIS 404
21 November 2010
Homework 4
1.
a) Uranus rings within Roche limit: DR= 2.44 (pplanet/pmoon)1/3R planet
Uranus density: 1.270
Uranus radius: 25,559 km
Uranus average moon density: 1.477 (taken from pg. A-12)
Work shown:
The Roche Limit of Uranus is 59,302.74 km and since Uranus farthest rings are 51,00
km away from the planets center (as shown in the book), all of Uranus rings reside
within the Uranus Roche limit.
b) Mars Roche Limit: The density of the moon of Mars is 1.75 realtive to water because it
is the average of Mars two moons- Phobos and Deimos who have densities of 2.0 and
1.5, respectively. Mars Roche Limit is 10,857 km
Work shown:
Phobos lies within the Red Planets Roche limit. Since it has a semi-major axis of 9,380
km but Deimos does not due to its semi-major axis of 23,500 km. If their semi-major
axes get smaller by 20%, Phobos would have a semi-major axis of 7504 km (9380 * .2=
1876, 9380-18760= 7504) which is still within the Roche limit. However, Demois would
then have a semi-major axis of 18,800 km which is still outside the Roche limit. In 5
Myr, if the moons of Mars decay by up to 20%, both Phobos and Deimos would reside
inside and outside of the Roche limit, respectively.
2.
a) Gravitational force exerted on a 1kilogram rock by the Moon when the rock is located
at:
i) Center of Earth-
ii) The side of the Earth facing the Moon-
iii) The side of the Earth that is opposite the moonWe experience tides because the moons tidal stresses stretch Earth into an
elongated shape- tidal bulge. Due to friction, Earths rotation drags its tidal bulge around,
out of perfect alignment with the Moon and ocean tides arise and fall as the rotation of
the Earth carries us through the oceans tidal bulges. In this instance, the tidal force
exerted on a 1kilogram rock by the Moon when the rock was located on the side of the
Earth facing the Moon was greater than the side of the Earth that is opposite the moon
because of the elongating and stretching of Earth from the forces, thus this causes tides to
rise and fall each day. When it was located on the side of the Earth facing the Moon, the
force was greater, thus there was a stretch.
b) The gravitational force exerted by the Sun and the Moon on a 1 kg rock on opposite
sides of the Earth during a full moon and during a new moon are:
F tidal-Sun:
F tidal- Full Moon:
F tidal- New Moon:
During a new moon and a full moon, the gravitational pull of the moon and sun are
combined (spring high tide). When the Earth, Sun and Moon are in a line, the
gravitational forces of the Moon and the Sun both contribute to the tides. . This reaffirms
the previous problem with the
c) If the moon were closer to the Earth such that it had an orbital period of 24 hours, the
tides would act in the same manner that they do because it will be over the same part of
the Earth. The elongating would still occur because the moon is permanently deformed
and tidally locked to Earth.
3.
a) The summer triangle crosses the meridian at around 2-3 p.m. now that it is midNovember. In September, I found that the summer triangle crossed the meridian at around
7 p.m. Given the changes time in from September to November, one can guess that the
summer triangle will cross the Meridian at around 12 p.m. on Christmas.
b)
4.
a) Ios surface are and volume are 41396452.24 km2 and 2.504 x 1010 km3 with a radius of
1,815 km.
Volume of sphere: 4/3 r3
Surface area of sphere= 4 r2
Work Shown:
b) The volume of volcanic material deposited on Io surface each year is 75134.56 km3
2.504x1010(0.000003 km)= 75134.56 km3
c) It would take 333,270 years for volcanism to perform the equivalent of depositing Ios
entire volume on its surface.
Work Shown:
d) Assuming the age of the Solar System is 4.5 billion years old, Io might have turned
inside out 135025.47 times.
Work Shown:
5. Closer spectroscopic observations of a comet making its closest approach to the Sun
would reveal that it is has gases it releases the closer it is to the Sun due to a heating up of
gases and an increased velocity due to angular momentum. Presumably, the opposite goes
for the spectroscopic observations of a coment when it is at the distance of Uranusvelocity and energy relative to the velocity of the comet are lower because it does not
have the Sun to heat up the nucleus and create energy to create a tail of released gases.
The difference for the observations is due to the Sun being closer at the closer
observation; the Sun gives massive amounts of energy and heat to the comet and it speeds
up as it is closer, creating a cloud of gas and dust (coma). If I were to measure the
brightness of the two spectra at the two different points, I could measure its Temperature
using its ratio.
6. Asteroids dont have tails because unlike comets, they dont contain ice and gas.
Instead, asteroids are composed of metals and rocky material whereas comets are made
up of ice, dust and rocky material. Asteroids are thought to have formed closer to the Sun
where ices melted due to the heat. Comets lose material with each orbit when they
approach the sun because their ice melts and vaporizes to form a tail. Additionally,
asteroids have more circular orbits and group together whereas comets tend to have more
elongated orbits that go out further from the Sun. Both of these factors prevent asteroids
from having a tail.
7.
a) Keplers Third Law: p2=a3
Eris orbital period (P): 557 sidereal years
Eris mean distance from Sun: 67.668 AU
b) Pluto should be brighter than Eris by around 4 times because it is around 2 times as
close. Brightness is a square function of distance, therefore, 22= 4. The three things in
determining how bright a planet or dwarf planet appears to be as seen from Earth are 1.
Temperature 2. Size 3. Distance
c) The biggest inclination of degrees relative to the ecliptic is Eris with 44.187 degrees.
Multiplied by two to find the entire amount of degrees on both sides of the ecliptic, one
gets 88.374 degrees. Since this only surveys a portion of the sky, we must multiply that
by 360 degrees to find the total amount of sky to observe. With a total of 31,815 degrees,
in order to detect the dwarf planets, we need 31,815/10 (degrees per night) in order to
figure out the amount of days we need to observe the dwarf planets. The total is 3182
nights to survey the entire area.