Transcript GravMath

“Concern for man himself
and his fate must always
form the chief interest of all
technical endeavors…
Never forget this in the
midst of your diagrams and
equations.”
Albert Einstein
Gravity Math
The History
• We have learned the history
of gravity, the missteps and
false starts, the names of
the movers and shakers in
the field, and the
progression of the ideas
The Math
• The math for gravity is a mix
of the equations for circular
motion and the physics for
acceleration
What could we want to
know?
• Let us make a list of the
possible bits of information
we could want to know
about an object orbiting
around another, or about the
attraction between two
objects
The List
• Constants
• Acceleration
• Force
• Period
• Velocity
• Mass of either object
The Constant
• There is one constant
involved, the gravitational
constant


N

m
11
G  6.67 10 
2 
kg


2
Acceleration
• We know that a change in
direction requires an
acceleration
• We saw this for circular
motion
Acceleration
GM
a 2
r
• r is the distance from the
center of the mass
• M is the mass of the
attracting object
Note…
• The mass of the falling
object is not a factor
• Only the mass of the planet,
Sun, or other large,
attracting body and distance
of the item from the center
counts
Example 1
• At sea-level the radius of the
Earth is 6.38 × 106 m. If the
mass of the Earth is 5.98 ×
1024 kg, what is gravitational
acceleration at sea-level?
What is it atop Mt. Everest,
8848 m above sea-level?
What Do We Know
• G = 6.67 ×
• r1 = 6.38 × 106 m
• m = 5.98 × 1024 kg
• r2 = ((6.38 × 106)+ 8848) m
-11
10
2
2
Nm /kg
Answer 1
•a=
• a1 = 9.799 m/s2 ≈ 9.80 m/s2
• a2 = 9.772 m/s2 ≈ 9.77 m/s2
2
Gm/r
Example 2
• Jupiter orbits the sun at a
radius of 7.78 ×1011 m. The
Sun has a mass of 1.99
×1030 kg. What is the
acceleration holding Jupiter
in its orbit?
What Do We Know
• G = 6.67 ×
• r = 7.78 ×1011 m
• m = 1.99 ×1030 kg
-11
10
2
2
Nm /kg
Answer 2
•a=
• a = 2.1929 × 10-4 m/s2 ≈
-4
2
2.19 × 10 m/s
2
Gm/r
Force
• We already know the basic
equation
F  ma
Force
• If we know the mass of the
object, we need only find the
acceleration to calculate the
force
Force
GMm
F 2
r
Force
• G is the gravitational
constant
• M is the larger of the two
masses
• m is the smaller of the two
• r is the radius between them
Force
• This was Newton’s big
achievement
• He was able to find out what
Kepler’s constant was
Example 3
• Pluto has a mass of 1.25 ×
1022 kg and an orbital radius
of 5.87 × 1012 m. What is
the force, in Newtons,
between Pluto and the Sun?
What Do We Know
• G = 6.67 ×
• r = 5.87 × 1012 m
• m = 1.25 ×1022 kg
• M = 1.99 ×1030 kg
-11
10
2
2
Nm /kg
Answer 3
•F=
• F = 4.82 × 1016 N
2
GMm/r
Period
3
r
T  2
GM
Period
• Remember, this is the time it
takes to make one full
revolution
• The time is measured in
seconds
Example 4
• What is the orbital period of
Pluto, given the information
from the pervious problems?
What Do We Know
• G = 6.67 ×
• r = 5.87 × 1012 m
• M = 1.99 ×1030 kg
-11
10
2
2
Nm /kg
Answer 4
•T=
• T = 7.76 × 109 s
Converted to years
• T = 246 years
3
1/2
2π(r /(GM))
Example 5
• The Earth has a period of
365 days. If it orbits at a
radius of 1.50×1011 m, then
what is the mass of the Sun.
What Do We Know
• G = 6.67 ×
• r = 1.50 × 1011 m
• T = 365 days = 3.15 × 107 s
-11
10
2
2
Nm /kg
Answer 5
•T=
• M = (4π2r3)/(GT2)
• M = 2.01 × 1030 kg
3
1/2
2π(r /(GM))
Velocity
GM
v
r
Velocity
• The velocity of the orbiting
object depends only on the
mass of the larger body, not
the mass of the orbiting
object
Example 6
• A satellite orbits the Earth at
a distance of 340000 m.
The mass of the Earth is
5.97 × 1024 kg. What
velocity must it have to
maintain that orbit?
What Do We Know
• G = 6.67 ×
• M = 5.97 × 1024 kg
• r = 340000 m
-11
10
2
2
Nm /kg
Answer 6
•v=
• v = 34222 m/s ≈ 34200 m/s
1/2
((GM)/r)