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Astronomy 2010
Problems in Planetary
Astronomy
Fall_2016
Day-6
ANNOUNCEMENTS
•Homework Set 5: Chapter 4 #31, 32, 33, 34, 37, 38,
41, 42, 44
•Project 1 will be due October 17. In addition to a
written paper, you need to prepare a short oral
presentation. Make your choice ASAP
NEWTON’S 1ST LAW OF MOTION
The Law of Inertia
An object in straight line uniform motion will
continue that motion unchanged unless some
external force acts on it
This was law based entirely on the work
of Galileo
NEWTON’S SECOND LAW
The Force Law:
F = ma
The acceleration a body experiences is directly
proportional to the net force acting on it and inversely
proportional to its mass


 v d v
a

t dt
Acceleration and force
are actually vectors


F  ma
VECTORS HAVE BOTH MAGNITUDE
AND DIRECTION
A vector can be broken
into horizontal and vertical
components using a few
simple trigonometric
relationships. These
relationships come from
the Pythagorean Theorem
relating the sides and
angles of a right triangle.
NEWTON’S THIRD LAW
The Action-Reaction Law
For every force there is an equal and opposite
reaction force
USING NEWTON’S LAWS
A 215 kg sailboat experiences a force from the wind
of 75.0 N at 30° north of east while the current pushes
it with a force of 50.0 N due south. What is the
magnitude and direction of the acceleration of the
boat?
SOLUTION
North-South Components
Wind: (75.0N)sin30° = 37.5N
Current: (50.0N)sin270° = -50.0N
Net = 37.5N – 50.0N = -12.5N
East-West Components
Wind: (75.0N)cos30° = 64.45N
Current: (50.0N)cos270° = 0.0N
Net = 64.45N + 0.0N = 64.45N
Fnet 
 64.45 N    12.5 N 
2
2
 66.14 N
 12.5 N 
  10.48  349.5
 64.45 N 
  tan 1 
Fnet  ma  a 
Fnet
a  66.14 N
 0.308 m s2
215kg
m
The acceleration is 0.31 m/s2 at 10° south of east
IN ADDITION TO HIS THREE LAWS
OF MOTION, NEWTON
FORMULATED THE UNIVERSAL
LAW OF GRAVITY
FGrav
m1m2
 G 2 rˆ
r1 2
G is the universal gravitational constant (6.67 x 10-11
),
m1 and m2 are the masses of the two objects, r1-2 is the
distance between them and r̂ is the direction of the line
connecting the two masses.
N  m2
kg 2
EASY EXAMPLE
Using the information in the Appendix, determine
the gravitational force of the Earth on the Moon.
Which way does it act?
What is the gravitational force of the Moon on the
Earth? Which way does it act?
SOLUTION
ME = 5.97 x 1024 kg
MM = 7.35 x 1022 kg
rE-M = 3.844 x 105 km

mE mM
F G 2
rˆE  M  6.67 1011
rE  M
 5.97 10 kg  7.35 10

 3.844 10 m 
24
N m
kg 2
2
8
22
kg

2
 1.98 1020 N
Gravity is always an attractive force so the Earth pulls
on the Moon: the direction is from the Moon to the Earth.
The force of the Moon on the Earth is exactly the same
but in the opposite direction (Newton’s 3rd Law)
1.98 x 1020 N directed from the Earth to the Moon
NEWTON DERIVED KEPLER’S
3RD LAW USING UNIVERSAL
GRAVITATION AND HIS LAWS
OF MOTION
P
4

3
r
GM *
2
2
Rearranges to give
4
r
M* 
 2
G P
2
3
Where M* is the total mass of the system. For a planet
orbiting a star, this is effectively the mass of the star. For
a moon or satellite orbiting a planet this is effectively the
mass of the planet unless the moon is large.
NEWTON’S FORM OF KEPLER’S 3RD
EXAMPLE
The two moons of Mars, Phobos and Deimos, orbit
the planet at 9,380 km and 23,460 km respectively. If
their periods are 0.32 days (Phobos) and 1.26 days
(Deimos), what is the mass of Mars?
The equation to use has “G” in it which has units of
(Nm2/kg2) so the 1st thing to do is unit conversions:
km to m and days to seconds
9380 km 1000 m km  9.38 106 m
23,460 km 1000 m km  2.346 107 m
0.32 days  86400 s day  2.7648 10 4 s
1.26 days  86400 s days  1.0886 105 s
NEWTON’S FORM OF KEPLER’S
3RD EXAMPLE CONTINUED
P
4
4 r

 M 
3
r
GM
GP2
2
M
2

2 3
4 9.38 10 m
2
6

3
 6.39 1023 kg
2.7648 10 s 
4 2.346 10 m 
M
 6.45 10
1.0886 10 s 
6.67 10
6.67 10
11 Nm 2
kg 2
2
11 Nm 2
kg 2
4
7
2
3
23
5
Within round-off error of being the same.
The textbook value is 6.42x1023 kg
kg
NEWTON’S FORM OF KEPLER’S 3RD
PROBLEMS
The first extra-solar planet discovered orbits the
star 51 Pegasi. If the semimajor axis is 0.052 AU
and the orbital period is 4.23 days, what is the
mass of 51 Pegasi?
Triton, the largest moon of Neptune, orbits the
planet at a distance of 3.548x105 km every 5.88
days. Use this information to determine the mass
of Neptune.
EXAMPLE 1 EASY SOLUTION
Using units of AU and years, the ratio of the cube
of the semimajor axis to the square of the orbital
period will give the mass in solar masses. So, we
only need to convert the period from days to years.
4.23days  365.25 days
year  0.01158 years
Now use the ratio of A3 to P2 to find the mass in
solar masses
0.052 AU 

A3
 M solar masses 
 1.0484M Sun
2
2
P
 0.01158 years 
3
1.0484M Sun 1.98911030 kg M Sun  2.085 1030 kg
EXAMPLE 1 HARD SOLUTION
Since G has units of (Nm2/kg2), distances must be in
meters and periods in seconds so do unit conversions
first
0.052 AU 1.49598 1011 m AU  7.779 109 m
4.23days  86400 s day  365472s
Next, chose the equation to use and
algebraically solve it for the mass.
2
2 3
4

4

r
2
3
P 
r  M 
GM
GP 2
Finally, plug in numbers to solve
4  7.779 10 m 
4 r
30
M


2.09

10
kg
2
2
11 Nm2
GP
6.673 10
365472s 
kg 2 
2
2 3

9

3
EXAMPLE SOLUTION 2
Do unit conversions first
3.548 105 km 1000 m km  3.548 108 m
5.88days  86400 s day  508032s
Use the same equation as the previous example
4  3.548 10 m 
4 r
26
M


1.02

10
kg
2
2
11 Nm2
GP
6.673 10
508032s 
kg 2 
2
2 3

8

3
THE
BLACKBODY
SPECTRUM
Stefan’s Law relates
the temperature to
the total energy
emitted per square
meter
F  T 4
  5.67  10 8
W
m2 K 4
Wein’s Law relates the
temperature to the wavelength of
maximum intensity
0.0029
max (in meters) 
T
BLACKBODY EXAMPLE
Determine the wavelength of maximum intensity of the
Earth and the total energy emitted by it. Assume it is a
perfect blackbody with a temperature of 288 K and a
radius of 6.378x106 m
For wavelength of maximum intensity, use Wein’s law
max
0.0029 K  m 0.0029K  m


 1.01 105 m  10.1 m
T
288K
Next, find total energy emitted: flux times surface area
Eemitted   T 4  4 R 2 

 4 6.378 10 m
 1.99 1017 W
6
 5.67 10
2
8
W
m2 K 4
  288K 
4