Download Astronomy 100 Homework #2 Solutions 1, Problem 3‐27 (2 pts

Astronomy
100
Homework
#2
Solutions
1,
Problem
3‐27
(2
pts).
According
to
Kepler’s
third
law,
(b)
Jupiter
orbits
the
Sun
more
quickly
than
Saturn.
Because
the
semimajor
axis
is
smaller,
the
orbital
period
P
=
a3/2
will
be
smaller.
2.
Problem
3‐40
(2
pts).
The
recently
discovered
object
Sedna
orbits
our
sun
beyond
Pluto,
at
an
average
distance
of
509
AU.
What
is
its
orbital
period?
a=509
AU,
P2
=
a3.
P
=
a3/2
=
5093/2
=
11,480
years
3.
Problem
3‐41
(3
pts).
The
recently
discovered
Eris,
which
is
slightly
larger
than
Pluto,
orbits
the
Sun
every
560
years.
What
is
its
average
distance
(semimajor
axis)
from
the
Sun?
How
does
its
average
distance
compare
to
that
of
Pluto?
a
=
P2/3
=
5602/3
=
67.9
AU.
This
is
further
from
the
Sun
than
Pluto,
which
has
a
semimajor
axis
of
~40AU.
4.
Problem
4‐28
(2
pts).
When
visiting
another
planet,
(b)
your
mass
would
be
the
same
as
on
Earth,
but
your
weight
would
be
different.
Weight
depends
on
the
strength
of
gravity.
5.
Problem
4‐34
(2
pts).
If
Earth
were
twice
as
far
from
the
Sun,
the
force
of
gravity
attracting
Earth
to
the
sun
would
be
(c)
one‐quarter
as
strong.
Gravity
follows
an
inverse
square
law
in
proportion
to
distance.
6.
Problem
4‐36
(2
pts).
If
the
Moon
were
closer
to
the
Earth,
high
tides
would
be
(a)
higher
than
they
are
now.
As
the
Moon
gets
closer,
the
gravitational
force
of
attraction
increases.
The
difference
in
the
force
of
attraction
between
different
parts
of
Earth
and
the
Moon
will
also
increase.
7.
Problem
4‐44
(3
pts).
Imagine
another
solar
system,
with
a
star
of
the
same
mass
as
the
Sun.
Suppose
there
is
a
planet
in
that
solar
system
with
a
mass
twice
that
of
Earth
orbiting
at
a
distance
of
1
AU,
what
is
the
orbital
period
of
this
planet?
Explain.
Making
the
approximation
MPlanet
<<
MStar,
Newton’s
version
of
Kepler’s
3rd
Law
tells
us
PPlanet2=
aPlanet3
/
MStar
(solar
masses).
Since
the
mass
of
the
star
hasn’t
changed
in
this
problem,
M=1
and
P
=
a3/2
=
(1
AU)
3/2
=
1
year.
8.
Problem
4‐45
(4
pts).
Suppose
a
solar
system
has
a
star
that
is
four
times
as
massive
as
our
Sun.
If
that
solar
system
has
a
planet
the
same
size
as
Earth
orbiting
at
a
distance
of
1
AU,
what
is
the
orbital
period
of
the
planet?
Explain.
Again,
Newton’s
version
of
Kepler’s
3rd
Law
is
PEarth2
=
aEarth3
/
M(solar
masses).
Here,
M=4.
So
P2
=
a3
/
4
=
(1
AU)
3
/
4
=
0.25.
P=
√0.25
=
0.5
years.
Extra
Credit:
Problem
4‐42
(+2
pts).
Use
the
data
in
Table
4.1
to
answer
each
of
the
following
questions.
a)
Compare
the
energy
of
a
1‐megaton
H‐bomb
to
the
energy
released
by
a
major
earthquake.
The
energy
released
by
an
earthquake
is
2.5*1016
J
/
4*1015
J
=6.25
times
larger
b)
If
the
United
States
obtained
all
its
energy
from
oil,
how
much
oil
would
be
needed
each
year?
Annual
US
energy
consumption
/
Energy
released
by
burning
1
liter
of
oil
=
1020
J
/
1.2*107
J
=
8.3*1012
liters
of
oil.
c)
Compare
the
Sun’s
annual
energy
output
to
the
energy
released
by
a
supernova.
The
energy
released
by
a
supernova
is
~1044
J
/
1034
J
=
1010
times
larger.