### Ch 5: Universal Gravitation

```Barry Latham
Honors Physics
Bloom High School
CH 7: UNIVERSAL GRAVITATION
7.1: Kepler’s Laws of Motion
 Based on years of observations, Kepler devised a series of laws
known to us as his “Laws of Planetary Motion.”
 1st Law: The path of each planet about the sun is an ellipse with
the sun at the focus.
 2nd Law: Each planet moves so that an imaginary line drawn from
the sun to the planet sweeps out equal areas in equal periods of
time.
 3rd Law: The ratio of the squares of the periods T of any two
planets revolving about the sun is equal to the ratio of the cubes
of their mean distances s to the sun: (T1/T2)2=(s1/s2)3.
Kepler’s Laws
Problem Solving
 As always, picture/data/formula/algebra!
 Keep the unknown as a numerator
 Rearrange the formula BEFORE you plug the values
in.
 Watch parenthesis, squares, cubes and their roots.
 Recall that:
x½ = 2√(x)
x1/3 = 3√(x)
x2/3 = 3√(x2)
x3/2 = 2√(x3)
Distance to Mars?
 According to Kepler, Mars’ “year” is 687 Earth-days.
Using what we know about the distance and period of
the earth, we can find Mars’ distance. The mean
distance of the Earth to the Sun is 1.50x1011 m.
 Picture/Data/Formula/Algebra
 (T1/T2)2=(s1/s2)3
 T1=Mars=687 days




T2=Earth=365.24 days s2=Earth orbital radius= 1.50x1011 m
(687 days/365.24 days)2=(s1/1.50x1011 m)3
(1.5x1011 m)(687 days/365.24 days)2/3=s1
s1=2.28x1011 m
Newton’s Law of Universal Gravitation
 Every particle in the universe attracts every
other particle with a force that is
proportional to the product of their masses
and inversely proportional to the square of
the distance between them. This force acts
along the line joining the two particles.
 F=Gm1m2/r2
Universal Gravitation Equation
 F=Gm1m2/r2
 F= the force of gravity (in Newton’s)
 m1 & m2= the two individual masses that we are
measuring
 r= the distance between the masses, measured in a
straight line and using their center of mass
 G= the universal gravitation constant
 6.67x10-11 N m2/kg2
 F=(N m2/kg2)(kg)(kg)/(m)2
 F=N (big surprise)
The Romance of Gravity 
 Find the force of attraction between two people. A 74
kg person is sitting 1.2 m away from a 84 kg person at a
lab bench. Estimate the force attracting them together.
 Picture/Data/Formula/Algebra







F=Gm1m2/r2
G= 6.67x10-11 N m2/kg2
m1=74 kg
m2=84 kg
r=1.2 m
F=G(74 kg)(84 kg)/(1.2 m)2
F= 2.879x10-7 N
(equal to a 0.03 mg object)
Less Romantic Gravity, part 1
 As you get far away from Earth, does your weight
decrease?
 What is the weight of a 2000kg satellite when it is in
the lab at NASA?
 Fg=mg
 Fg=(2000 kg)(9.80 m/s2)
 Fg=19,600 N
Less Romantic Gravity, part 2
 What is the force of gravity acting on a 2000
kg satellite when it orbits two Earth radii
from the Earth’s center (rEarth = 6.38x106 m).
 F=Gm1m2/r2
 F=G(2000 kg)(5.97x1024 kg)/(2x6.38x106 m)2
 Fg=4,891 N
 This is 1/4th of what it was in the lab!
Universal Gravitation &
Kepler’s 3rd Law
 If we take Newton’s law of universal gravitation,
 F=Gm1m2/r2
 And Kepler’s 3rd Law (with centripetal acceleration
included)
 F=m4p2r/T2
 m=mass of central body
 T=period of orbit (time of one revolution)
 Set them equal (they are both F) and solve for T
 T=√(r3(4p2)/(Gm))
 This represents the time it takes for an object to go
around a central body one time.
Cavendish’s Device
 Using a device shown to the
right, Cavendish was able to
accurately measure the
force of gravity.
 By twisting the string
slightly, the scale displayed
the force of attraction
between the sphere on the
rod and sphere A.
Geophysical Applications
 When discussing objects on the earth’s surface, m1
becomes mE and r is the radius of the earth (rE).
 Original formula
 substituted m2 g for F
 cancel m2
F=Gm1m2/r2
m2g=GmEm2/rE2
g=GmE/rE2
 So the acceleration due to gravity of an object is
dependant upon the mass of the object and the radius
of the object (in this case, the earth)
 In lieu of Earth, we can use the data from any object!
What’s the mass of Earth?
 Unless you have a HUGE scale, we need another
way to determine it.
 g=GmE/rE2
but solve for mE
 mE=g rE2/G
 mE=(9.80 m/s2)(6.38x106 m)2/(6.67x10-11 N m2/kg2)
 mE=5.98x1024 kg
Is “g” really 9.8?
 Find g at sea level on Earth.
 g=GmE/rE2
 g=(6.67x10-11 N m2/kg2)(5.98x1024 kg)/(6.38x106 m)2
 g=9.79909 m/s2
 So, yes, it’s basically 9.8.
Is “g” really 9.8?
 Gravity at the top of Mt. Everest. Mt. Everest has a
height of 8,850 m above sea level.
 g=GmE/rE2
 g=(6.67x10-11 N m2/kg2)(5.98x1024 kg)/(6.38x106 m + 8,850 m)2
 g=9.77195 m/s2
 So, yes, it’s basically still 9.8.
Is g 9.8 everywhere on Earth?
 Explain the
losses or
gains in
gravity.
 HW p. 178
q. 6-10
7.2: Satellites
 Artificial satellites are put
into orbit with three
different final velocities.
 Circular orbit (space shuttles)
 27,000 km/h
 Elliptical orbit
 30,000 km/h
 If an object needs to leave
earth’s orbit, such as solar
probes, and extra-solar
objects
 40,000 km/h
Satellite Orbital Velocity
 Geostationary satellite (communications & weather)
 Stays above the same point on earth
Satellite Equation Derivation
 To remain above the same point, the satellite must
have a period (T) of one day (one cycle per day).
 Apply Newton’s Second law and a new formula
(centripetal acceleration)
 Assume the orbit is circular (which it pretty much is)
 The only force on the satellite is the force of universal
gravitation. Plug our new gravity equation and
centripetal acceleration into F=ma.
 F=ma
F=Gm1m2/r2
ac=v2/r
 GmsatmE/r2=(msat)(v2/r)
Speed (v) of a Satellite Orbiting a
Central Body in Circular Orbit
 Since the satellite revolves around the earth at the same rate the
earth turns, v=2pr/T
 T is the period of the movement, 1 day (86,400 s)
 GmsatmE/r2=msatv2/r
 GmE/r=v2
 v=√(GmE/r)
cancel msat and r
solve for v
mE is the mass of the Earth
but can represent the central body
 √((6.67x10-11)(5.98x1024)/(4.23x107))=3070 m/s
 We find the same result if we use, v=2pr/T.
 v=2p (4.23x107)/(86,400)=3076 m/s
Period (T) of a Satellite Orbiting a
Central Body in Circular Orbit
 We saw this on Slide 11!
 T=√(r3(4p2)/(Gm))
 This represents the time it takes for an object to go around a
central body one time.
 If the mass (m) is Earth, then m=mE
Orbital height (r) of a Satellite
Orbiting a Central Body in Circular Orbit
 Using any number of equations containing r, we can find the orbital
height

Remember that height is from the center of the central object to the center of the
satellite
 F=Gm1m2/r2

r= √( Gm1m2/F)
 F=m4p2r/T2

Newton’s version of Kepler’s 3rd Law
r=
 v=√(GmE/r)

Kepler’s Third Law
r=FT2/(m4p2)
 T= 2p√(r3/(Gm))

Newton’s Universal Gravitation
r=GmE/v2
v of satellite in circular orbit
Inertial Mass
 According to F=ma, the mass of an object is the
relationship between the Force applied and the
acceleration of the object.
 Because every object with mass has inertia, we can
rearrange and use subscripts.
 minertia=Fnet/a
 More mass means less reaction to the same force
 Inertial mass is independent of location or a
reference body
 It’s how they measure the astronauts mass in space!
 Lab 7-2
Gravitational Mass
 Unlike Inertial Mass, gravitational mass is
dependant on a reference material
(sometime a whole planet!)
 Using Newton’s Universal Gravitation
formula, we can rearrange to find the mass of
an object
 Fgrav=GmgravmE/r2
 mgrav=r2Fgrav/(GmE)
 To measure this, we use a triple beam or
digital balance in the presence of gravity
Relativity & Gravity
 According to Einstein,
gravity is not a force, but an
effect of space itself
 Mass causes space to curve,
so objects are pulled
toward the “dimple” in
space-time (even light)
 Larger objects make larger
“dimples” hence larger
gravitational pull
Earth’s gravitational field
distorting space-time
Blackholes, defined
 Black hole- a concentration of mass so dense that nothing
can escape the pull of it’s gravity, not even light
 Formed by the severe gravitational pull overcoming the
thermonuclear reaction pushing the material out
 Black hole formation
 Size? Black holes can be measured by mass or area
 Mass- 1031 to 1036 kg (10-1,000,000 times as massive as our Sun)
 Volume- based on Schwartzchild radius or radius of the horizon
 A black hole with a mass the same as the Sun would be 3 km in radius
 The Sun has a radius of 700,000 km now!
 A 10-solar-mass black hole would be 30km and so on
 Event Horizon- the point of no return, the Schwartzchild radius
Black Holes evaporate?
 Black Holes will not eventually suck up
everything in the universe
 They emit “Hawking radiation”
 At the Event Horizon, a particle-antiparticle pair
can appear from within the black hole
 One particle gets sucked back in, the other escapes
 Momentum & Energy are conserved
 The emitted particle is detected
 This happens so frequently, it is observed as a
matter/energy stream
Black Hole Images
A wormhole
End of Unit Review, 1
 What would make you weigh more: an earth twice as
massive with the same diameter or one having half its
diameter with the same mass?
 Solution: Find out using F=Gm1m2/r2
 Fg1=GmEmyou/rE2
 Fg2=G(2mE)myou/rE2
 Fg3=GmEmyou/(½rE)2
normally, what is your Fg?
now, what is your Fg?
now, what is your Fg?
 How do they all compare?
End of Unit Review, 2
 Would one need a longer rocket to take off from the moon
vs. earth?
 What is needed to escape the Earth? The Moon?
 Gravity holds us in place, so which has less gravity?
 Solution: Find out using g=GmE/rE2
 For earth, g=9. 79909 m/s2
 For the Moon, g=GmM/rM2
 g=G(7.349x1022 kg)/(1.738x106 m)2
 g=1.62 m/s2
End of Unit Review, 3
 Based on our prior work, what would your
weight be on the Moon?
 FgE=m(9. 80 m/s2)
 FgM=m(1.62 m/s2)
 FgE=(84.1 kg)(9. 80 m/s2)=824 N
 FgM=(84.1 kg)(1.62 m/s2)=136 N
End of Unit Review, 4
 If an astronaut was to break the connection rope
between him and the space shuttle, will gravity
quickly pull him back to the shuttle?
 F=Gm1m2/r1-22
 g=Gm/r2
 What happens to the acceleration of an asteroid as
it approaches a planet?
 If g based on distance?
 g=Gm/r2
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