Newton’s Law of Universal Gravitation
Every object in the universe is attracted to
every other object.
F= Gm1m2
r2
G = 6.67 X 10-11 N-m2/kg2
m1 = mass of one object
m2 = mass of second object
r = distance from center of objects
Cavendish proves the law in
1798
Everything in the solar system pulls on everything
else.
Sun pulls on Earth
All the other planets
also pull on the
Earth
a. Calculate the force of gravity between two
60.0 kg (132 lbs) people who standing 2.00 m
apart. (6.00 X 10-8 N)
b. Calculate the force of gravity between a 60 kg
person and the earth? Assume the earth has a
mass of 5.98 X 1024 kg and a radius of
6,400,000 m (4,000 miles). (584 N)
A planet has a mass four times that of the earth, but
the acceleration of gravity is the same as on the
earth’s surface. Relate the planet’s radius to that
of the earth (2Re).
A 2000-kg satellite orbits the earth at an altitude of
6380 km (the radius of the earth)above the earth’s
surface. What is the force of gravity on the
satellite?
F= Gm1m2 = (6.67 X 10-11 N m2/kg2)(2000kg)(5.98 X 1024 kg)
r2
(6,380,00 m + 6,380,00 m)2
F = 4900 N
A simpler solution:
At the earth’s surface:
FG = mg
At twice that distance:
FG = (1/2)2mg = ¼ mg
FG = (¼ )(2000 kg)(9.80 m/s2)
FG = 4900 N
What is the net force on the moon when it is at a
right angle with the sun and the earth?
Relevant Data:
MM = 7.35 X 1022 kg
ME = 5.98 X 1024 m
MS = 1.99 X 1030 m
rMS = 1.50 X 1011 m
rME = 3.84 X 108 m
Calculate each force separately:
FME
Earth
1020 N
FME = 1.99 X
FMS = 4.34 X 1020 N
FR2 = FME2 + FMS2
q
FMS
FR = 4.77 X 1020 N
tan q = opp = FME
adj
FMS
q = 24.6o
Sun
FR
Consider an object (any object)
Weight = mg (Force due to gravity)
mg = GmmE
rE2
g = GmE
rE2
mE =g rE2 = (9.8 m/s2)(6.38 X 106 m)2 = 5.98 X 1024 kg
G
6.67 X 10-11 N-m2/kg2
Calculating “g”
g = GmE
rE 2
g = (6.67 X 10-11 N-m2/kg2)(5.98 X 1024 kg)
(6.38 X 106 m)2
g = 9.80 m/s2
Calculate the value of g at the top of Mt.
Everest, 8848 m above the earth’s surface.
g = GmE
r2
g = (6.67 X 10-11 N-m2/kg2)(5.98 X 1024 kg)
(6.38 X 106 m + 8848 m)2
g = 9.77 m/s2
g varies with:
• Altitude
• Location
– Earth is not a perfect sphere
– Different mineral deposits can change density
– “salt domes” are low density salt regions near
petroleum deposits
Gravitational Potential Energy
• mgh is not valid except near the surface of the
earth
• Zero point is an infinite distance from the earth
• Need calculus since the force varies with distance
U = - ∫F dx
U = - ∫ Gm1m2 dr
r2
U = -Gm1m2
r
A 1000 kg rocket is launched from the surface of
the earth. Calculate the escape velocity. Hint:
set the final K and U equal to zero. (11,200
m/s)
A 1000 kg rocket is launched from the surface of
the moon. Calculate the escape velocity. The
mass of the moon is 7.35 X 1022 kg and the
radius is 1.74 X 106 m. (2383 m/s)
A tile of a rocket falls off at an altitude of 1500
km above the surface of the earth. The
rocket is travelling upwards at 2000 m/s.
a. Calculate the speed of the tile as it hits the
ground. (5276 m/s)
b. In reality. Could the tile actually reach this
speed?
Suppose the earth stops orbiting and falls into
the sun (see page 395)
a. Use conservation of energy to calculate the
speed of the earth as it just crashes into the sun.
(6.13 X 105 m/s)
b. Why doesn’t our earth get pulled into the sun?
What speed should a satellite be launched if it
needs to have a speed of 500 m/s at 400 km.
Remember to include the radius of the earth
in your calculations. (2755 m/s)
Why don’t satellites fall back onto the earth?
• Speed
• They are “falling”
due to the pull of
gravity
• Can feel
“weightless” if you
are on board (just
like in the elevator)
Speed of a Satellite
F= GMm = mv2
r2
r
GM = v2
r
v = \/GM/r
The Starship Enterprise wishes to orbit the earth
at a 300 km height.
a. Calculate the total distance of the Enterprise
from the center of the earth. (6.68 x 106 m)
b. Calculate the proper orbital speed for the
Enterprise. (7730 m/s)
c. Would a heavier starship require a greater orbital
speed?
Kepler’s Laws (1571-1630)
1. The orbit of each planet is an ellipse, with
the sun at one focus
2. Each planet sweeps out equal areas in
equal time
3. T1 2 =
r1 3
T2
r2
1. The orbit of each planet is an
ellipse, with the sun at one focus
http://csep10.phys.utk.edu/astr161/lect/history/kepler.html
2. Each planet sweeps out equal areas in
equal time
• Suppose the travel time in both cases is
three days.
• Shaded areas are exactly the same area
Mars has a year that is about 1.88 earth years. What
is the distance from Mars to the Sun, using the
Earth as a reference (rES = 1.496 X 108 m)
T12 = r13
T22
r23
TM2 = rM3
TE2
rE3
rM3 = TM2rE3
TE2
rM3 = (1.88y)2(1.496 X 108 m)3
(1 y)2
rM3 = 1.18 X 1025 m3
rM = 2.28 X 108 m
How long is a year on Jupiter if Jupiter is 5.2
times farther from the Sun than the earth?
TJ2 = rJ3
TE 2 r E 3
TJ2 = rJ3 TE2
rE 3
TJ2 = rJ3 TE2 = (5.2)3(1 y)2
rE 3
(1)3
TJ2 = 141 y2
TJ = 11.8 y
How high should a geosynchronous satellite
be placed above the earth? Assume the
satellite’s period is 1 day, and compare it to
the moon, whose period is 27 days.
Ts2 = rs3
TM2 rM3
rs3 = rM3 Ts2
TM2
rs3 = rM3 Ts2
TM2
rS3 = rM3
729
=
rM3 (1 day)2
(27 day)2
Take the 3rd root of
both sides
rs = rM
9
The satellite must orbit 1/9 the distance to the
moon.
Deriving the Third Law
To derive Kepler’s Law, we will need two
formulas.
F= Gm1mJ
r2
mJ
m1
F=m1v2
r
GMm = mv2
r2
r
GM = v2
r
GM = 4p2r2
r
T2
T2 = 4p2
r3 GM
Substitute v=2pr
T
T2 = 4p2
r3 GM
T12 = 4p2
r13 GM
T12 = T22
r13 r23
We can do this for two
different moons
T22 = 4p2
r23 GM
A Useful Form
This form of the equation:
T2 = 4p2
r3 GM
S could be the Sun,
Earth, or other body
with satellites.
Is useful for determining the mass of the
central planet, using only the period and
distance of one of the satellites.
Calculate the mass of the sun, knowing that
the earth is 1.496 X 108 m from the sun.
(2.0 X 1030 kg)
A student is given the following data and asked to
calculate the mass of Saturn. The data describes
the orbital periods and radii of several of
Saturn’s moons.
Orbital Period, T
Orbital Radius, R
(seconds)
(m)
4p2
8.14 X 104
1.85 X 108
1.18 X 105
2.38 X 108
1.63 X 105
2.95 X 108
2.37 X 105
3.77 X 108
Let’s use this equation:
T2 = 4p2
r3
GmS
And rearrange it:
GmS= 1
4p2 r3 T2
Once more:
1 = GmS
T2
4p2 r3
Calculate the following values and graph them.
1
T2
G
4p2 r3
1.60E-10
1.40E-10
1.20E-10
1.00E-10
8.00E-11
6.00E-11
4.00E-11
2.00E-11
0.00E+00
0.00E+00 5.00E-38 1.00E-37 1.50E-37 2.00E-37 2.50E-37 3.00E-37
Calculate the slope of the graph
y
= m
x
1 = mS G
T2
4p2 r3
ms = 5.9 X 1026 kg
+b
Orbital Energies
GMm = mv2
r2
r
GMm = mv2
r
1 mv2 = GMm = K
2
2r
K = -½ Ug
E = K + Ug
E = - ½ Ug + U g
E = ½ Ug
A 1500 kg satellite orbits at an altitude of 4,000
km above the earth’s surface. Calculate the
work required to move that satellite to an orbit
of 12,000 km above the earth’s surface. (1.25 X
1010 J)