Transcript Day-11

Astronomy 2010
Problems in Planetary
Astronomy
Fall_2016
Day-11
ANNOUNCEMENTS
•The last exam and second project presentations will be
during the final exam period: Tuesday December 6 @
10:30am
•Homework: Chapter 9 # 39,40,41, 44, 45
TIDAL STRESS AND ORBITAL
RESONANCE
WHAT WOULD THE GRAVITATIONAL
FORCE BE INSIDE THE EARTH?
Newton showed that, as long
as the body is spherically
symmetrical, the only thing
that matters is the mass of the
sphere underneath you. The
spherical shell above you does
not contribute to the
gravitational force.

M m
Fgrav  rˆG
2
R  d 
M’ is the mass of the sphere
with radius (R-d) and d is the
distance below the surface
GRAVITATIONAL ACCELERATION
DECREASES AS YOU MOVE INWARD
r
This assumes a uniform density. For a real planet, the
density increases as you go down so the acceleration
increases at first then starts to decrease.
EXAMPLE
Assuming a uniform density, what
is the acceleration due to gravity
one fifth of the way to the center
of the Earth?
First, find out how much mass is below:
mass
3
density 
 M   43  R  d  
volume
Use (R-d) = 4/5 Rearth = 0.8 x 6.378x106 m and  = 5,515 kg/m3

M    5.102 10 m
4
3
6
 3.069 1024 kg

3
kg
5515

m3 
EXAMPLE SOLUTION
Now find “g” from the gravitation formula

M m
M

Fgrav  rˆG
 mg  divide by m  g  G
2
2
R  d 
R  d 
Use M’= 3.069x1024 kg and (R-d) = 5.102x106 m

g  6.67 1011
 7.86 m s2
N m
kg 2
2
 5.102 10 m
3.069 1024 kg

6

2
TIDES ARE THE RESULT OF A
DIFFERENCE IN GRAVITY
MATHEMATICAL TREATMENT OF
TIDAL FORCE
Start with the gravitational force between two point
masses a distance r apart

Mm
Fgrav  rˆG 2
r
Now make M a sphere with radius Dr so that a point on
the surface of M is R ± Dr from m (ignore the directions
for now) thus
Mm
Fgrav  G
2
R  Dr 
MATHEMATICAL TREATMENT OF
TIDAL FORCE
Pull out an R2 term in the denominator
Mm
Mm
1
Fgrav  G
G 2
2
R 1  Dr
R  Dr 


R
2
Now expand this in a Maclaurin series expansion
Mm
1
Fgrav  G 2
R 1  Dr


R
2
Mm
2Mm Dr
 G 2 G 2
 ...
R
R R
The first term is just the normal gravitational force
between the two masses. The second term (and
other terms) is the tidal force
SO THE TIDAL FORCE GOES AS
ONE OVER R3
2MmDr
Ftidal  G
3
R
Note that Dr is the radius of the body the
tides are acting on and R is the center-tocenter distance between the two bodies.
EXAMPLE
Calculate the tidal force of the Earth acting on a
1kg mass on the surface of the Moon
For this problem m is the mass of the Earth
(5.9736 x 1024 kg) , M is the one kilogram mass, Dr is
the radius of the Moon (1.7374 x 106 m) and R is the
Earth-Moon distance (3.844 x 108 m)
Ftidal  G

2MmDr
R3
 6.67 10
11 N m
kg 2
2



3.844 10 m 
2 1.0kg  5.9736 1024 kg 1.7374 106 m
 2.437 105 N  24 N
8
3

THE ROCHE LIMIT
When the tidal force on a body
exceeds the binding force
between the particles the body
is composed of, it is at the Roche
Limit. If the binding force
holding a moon together is
simply its self gravity, the Roche
1
limit is
 M
d Roche  R 2
 m



3
R is the radius of the primary body,
M is the density of the primary and
m is the density of the moon
ORBITAL RESONANCE
When the orbital
period of a small body
is a small integer ratio
of the orbital period of
a larger body, orbital
resonance occurs
SOMETIMES ORBITAL RESONANCE
LEADS TO STABILITY
LAGRANGE POINTS ARE BALANCE
POINTS
Lagrange points are points where the gravitational
force between two objects is balanced. Some are
stable (L4 and L5) and others are not (L1, L2 and L3)
ANGULAR MOMENTUM
Angular momentum is a basic property of anything
than moves along a non-straight path. For the most
part, we will deal with simple circular motion.
ANGULAR MOMENTUM IS AN
IMPORTANT QUESTION IN THE
FORMATION OF THE SOLAR
SYSTEM
Angular momentum is conserved in
the collapse from a large cloud to a
flattened disk with a central bulge
ORBITAL ANGULAR MOMENTUM
If the orbital radius is
much greater than the
radius of the planet, it
will act like a point mass.
We usually use the
symbol L for angular
momentum.

 
L  mr  v
ANGULAR MOMENTUM IS A VECTOR
The angular momentum vector is
perpendicular to the plane of the orbit. Use the
right-hand-rule to find the direction
ROTATIONAL
ANGULAR
MOMENTUM
DEPENDS ON THE
OBJECT’S SHAPE
AND MASS
DISTRIBUTION

 
L   r  v dV
 is the mass distribution
(density function), r is the
location of each point
mass and v is its velocity
FOR UNIFORM BODIES THE
INTEGRAL SIMPLIFIES TO


L  Iw
I is the moment of inertia of the
body and w is it’s rotational
velocity. I can be calculated
based on the shape of the
object. The chart at left shows
the value of I for a variety of
shapes and rotation axis.
EXAMPLE
Using the information in the textbook appendix,
find the total angular momentum of Venus.
Rvenus = 6,052 km = 6.052x106m
Rorbit = 108.2x106 km =1.082x1011m
Prot = 243.02 days
Porbit = 0.6152 years
mVenus = 4.87x1024 kg
Lrot = Iw
For a sphere I = 2/5mR2
Lrot = 4/5mR2/T
 45
4.87 10

w = 2/T


2
2
kg 6.052 106 m
 2.135 1031 kg m s
243.02days  86400 sec days
24
Lorbit = mvRorbit = m(2Rorbit/T)Rorbit


1.082 10 m
11
2
24
40 kg m
=2mR2orbit/T  2 4.87 10 kg 0.6152 years  3.156 107 sec year  1.845 10
2
s
EXAMPLE OF CONSERVATION OF
ANGULAR MOMENTUM
In a supernova the central iron core of a massive
star collapses from an Earth-sized object (radius
equals 6.378x106m) to a neutron star with a
diameter of 20 km. If the star’s core had a
rotational period of 2.4 days before the collapse,
what is the rotational period after the collapse?
EXAMPLE SOLUTION
Rotational angular momentum of a solid rotating sphere
L = Iw = (2/5mR2)(2/T) = 4/5mR2/T
So conservation of angular momentum is
4
2

mR
5
before
Tbefore

4
2

mR
5
after
Tafter

2
Rbefore
Tbefore

2
Rafter
Tafter
Solve for Tafter and plug in numbers
Tafter 
2
Rafter
2
before
R

10 10 m 
2.4days  86400

6.378 10 m
2
3
Tbefore
6
2
sec
day
  0.5097 s