Transcript Universal Law of Gravity

Universal Law of
Gravity
Newton’s Universal Law of
Gravitation
Between every two objects there is an attractive
force, the magnitude of which is directly
proportional to the mass of each object and
inversely proportional to the square of the distance
between the centers of the objects.
Universal Law of Gravity
Force of gravity has magnitude given by
(Gravity Force) = (G) x
( Mass of Object A ) x ( Mass of Object B)
( Distance ) x ( Distance )
DISTANCE
Object
A
Force
Force
Equal and opposite forces
(Newton’s Third law)
Object
B
Universal Gravity Constant, G
In the formula for gravity force, we have
G = 0.0000000000667 N m2 / kg2
= 6.67 x 10–11 N m2 / kg2
The formula and the constant are called
“universal” because, up to now, this theory
predicts gravity anywhere in the universe.
Sample Problem
Here is an example of using the formula
(Gravity Force) = (G) x
( Mass of Object A ) x ( Mass of Object B)
( Distance ) x ( Distance )
Object A (1 kg mass)
Force
Object B (Earth)
DISTANCE = Earth’s Radius
Sample Problem
Find gravity force for a 1 kg mass on surface of
Earth.
Earth’s Mass
(Force) = (6.67 x 10–11) x
Universal Gravity Constant, G
( 1 ) x ( 6 x 1024 )
( 6.38 x 106 )2
Earth’s Radius
Value comes out to 9.8 Newtons
Sample Problem (cont.)
Find gravity acceleration on a 1 kg mass.
Using Newton’s Second Law,
(Acceleration) =
( 9.8 N )
(Force)
=
(1 kg )
(Mass)
Answer is 9.8 m/s2. We just confirmed our
value of a!!!
Sample Problem

Are you attracted to the person sitting next to
you? Calculate the gravitational attraction
between you (assume 70 kg) and the person next
to you (assume 65 kg) if you are 1.2 m apart.
givens
formula
substitution
unknown
Fg

Answer = 2.11 x 10-7 N
0.000000211 N

Very Small!!! But anything with mass has an
attractive force.
Sample Problem

Determine the force of gravitational attraction
between the earth (m = 5.98 x 1024 kg) and a 70kg physics student if the student is standing at
sea level, a distance of 6.38 x 106 m from earth's
center.
Newton and the Moon
Newton realized that
Earth’s gravity was the
centripetal force that
kept the moon in orbit.
Also discovered that
gravity was weaker at
that great distance.
Gravity
force
Gravity & Distance
We don’t notice that
gravity gets weaker as
we move away from
Earth because we
rarely go very far.
Value of g
(acceleration due to gravity; m/s2)
Using Fgrav = m·g we can derive:
Value of g


Can now find g anywhere
g = G•m/r2
Where
G = 6.67 x 10-11 N m2/kg2
m = mass of planet in kg
r = radius of planet in meters
Known as “Inverse Square Law”
Gravity force
weakens with
distance as the
inverse of the
square of the
distance.
Geometric
property of area
and distance.
1/4 Earth Gravity
Earth Gravity
outer circle is twice Earth’s radius
Sample Problem

Find your weight if you have a mass of 60 kg
and are 2.1 x 105 m above the earth’s surface.
Solution
Givens:
r = 2.1 x 105 m + 6.37 x 106 m = 6.58 x 106 m
myou = 60 kg
mearth = 5.98 x 1024 kg
Unknown:
Equations:
Fg
F = ma
for weight Fg = mg
Fg = Fg
= (60kg) (9.2 m/s2)
mg = G m1 m2
= 553 N
r2
g=Gm
r2
g = 9.2 m/s2
Weightlessness
In deep space, far away from
all stars, planets, etc. there is
almost no gravity force.
In orbit near Earth, gravity is
still strong (only 10% less
than on surface).
Why are Shuttle and Space
Station astronauts
“weightless”?
Earth is nearby
NASA’s “Vomit Comet”
NASA has a special
airplane for training
astronauts in free-fall
weightless conditions.
The “Vomit Comet”
nickname tells you it’s
quite a wild rollercoaster ride.
The plane flies between 20,000 and
30,000 feet, same as commercial flights.
Flight of the “Vomit Comet”
At the top of
the arc, the
plane’s
trajectory is
projectile
motion.
Boeing 707 (modified)
Weightless
Freefall
Orbits

Geosynchronous Orbit – orbits above the same
point on the equator of the earth at all times

GPS, cell phones, etc.
Orbits and Centripetal Force
Gravity provides the centripetal force required for
a satellite to move in a circle.
Fg = Fc
mg = mv2/r
g = v2/r
Sample Problem

Calculate the speed needed for one of the Direct
TV satellites to orbit at an altitude of 320,000 m.
Getting into Orbit
Rocket needs to lift above
the atmosphere and then
fire thrusters to acquire the
required orbital speed of
about 8 kilometers per
second.
Returning to Earth,
air resistance slows the
spacecraft during reentry.
Elliptical Orbits
For speeds higher than 8 km/s, the orbit is
elliptical instead of circular.
Escape Speed
If speed exceeds
11.2 km/s then
object escapes
Earth because
gravity weakens (as
object gets further
away) and never
slows the object
enough to return it
back towards Earth.
Hyperbolic
Circular
Elliptical
Using the Law of Universal
Gravitation


Newton’s Law is used and applied to the motion
of the planets about the sun.
Newton derived an equation from the Law of
Universal Gravitation to describe the Period of
Planetary Motion:
T² = (4π²/Gms)r³