The E Tutor - Newton's Law of Gravitation
Transcript of The E Tutor - Newton's Law of Gravitation
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Newtons Question:If the force of gravity is being exerted
on objects on Earth,what is the origin of that force?
Newtons realization was thatthis force must come from
the Earth.
He further realized that this
force must be what keeps the
Moon in its orbit.
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Must be true from
Newtons 3rd Law
The gravitational force on you is half of a Newtons 3rd Law pair: Earth
exerts a downward force on you, & you exert an upward force on Earth.
When there is such a large difference in the 2 masses, the reaction force(force you exert on the Earth) is undetectable, but for 2 objects with masses
closer in size to each other, it can be significant.
The gravitational force one body exertson a 2nd body , is directed toward the first body, and is equal and opposite to theforce exerted by the second body on thefirst
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Every particle of matter in the universeattracts every other particle with a force that isdirectly proportional to the product of themasses of the particles and inversely proportional to the square of the distancebetween them.
F12 = -F21 [(m1m2)/r2]Direction of this force:Along the line joiningthe 2 masses
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G = the Universal Gravitational constant
Measurements in SI Units:
The force given above is strictly valid only for:
Very small masses m1 & m2(point asses)
Uniform spheres
For other objects: Need integral calculus!
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TheUniversal Law of Gravitationis an example of an inverse square law
The magnitude of the force varies as the inversesquare of the separation of the particles
The law can also be expressed in vector form
The negative sign means its an attractive force
Arent we glad its not repulsive?
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Comments
F12 Force exerted by particle 1
on particle 2
F21 Force exerted by particle 2on particle 1
This tells us that the forces form a Newtons 3rd Law action-reaction pair, as expected.
The negative sign in the above vector equation tells us that
particle 2 is attracted toward particle 1
F21 = - F12
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More Comments
Gravity is a field force that alwaysexists between 2 masses, regardlessof the medium between them.
The gravitational forcedecreases rapidly as the distancebetween the 2 masses increases This is an obvious consequence of
the inverse square law
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Earth Radius: rE = 6320 km
Earth Mass: ME = 5.98 1024 kgFG = G(mME/r
2)
Mass of the Space craft m At surface r = rE
FG = weight
or mg = G[mME/(rE)2]
At r = 2rE
FG = G[mME/(2rE)2]
or ()mg = 4900 N
A spacecraft at an altitude of twice the Earth radius
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Find the net force on theMoon due to the gravitationalattraction of both the Earth &the Sun, assuming they are atright angles to each other.
ME = 5.99 1024kgMM = 7.35 1022kgMS = 1.99 1030 kgrME = 3.85 108 mrMS = 1.5 1011 mF = FME + FMS
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F = FME + FMS
(vector sum)
FME = G [(MMME)/ (rME)2]
= 1.99 1020 NFMS = G [(MMMS)/(rMS)2]
= 4.34 1020 NF = [ (FME)
2 + (FMS)2]
= 4.77 1020 Ntan() = 1.99/4.34
= 24.6
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Gravity Near Earths Surface
GravitationalAcceleration g
and
Gravitational
Constant G
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Obviously, its very important to distinguish
between G and g They are obviously very different physical
quantities
G The Universal Gravitational Constant It is the same everywhere in the Universe
G = 6.673 10-11 Nm2/kg2 Always same on every location
g The Acceleration due to Gravityg = 9.80 m/s2(approx) on Earths surface
g varies with location
Gvs.g
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Consider an object on Earths surface:
mE =mass of the Earth
rE =radius of the Earth
m =mass of object
Let us the Earth is a uniform,perfectsphere.
The weight ofm:FG = mg
The Gravitational force on m:
FG = G[(mmE)/(rE)2]
Setting these equal gives:
g in terms of G m
mE
g = 9.8 m/s2All quantities on the right are measured!
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Using the same process, we can Weigh
Earth (Determine its mass).
On the surface of the Earth, equate the
usual weight of mass m to the Newton
Gravitation Law form for the
gravitational force:
Knowing g = 9.8 m/s2 & the radius of
the EarthrE, the mass of the Earth can
be calculated:
mE
m
All quantities on the right are measured!
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Acceleration due to gravity at a
distancerfrom Earths center.
Write gravitational force as:
FG = G[(mME)/r2] mg
(effective weight)
g the effective accelerationdue to gravity.
SO : g = G (ME)/r2
ME
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If an object is some distance h
above the Earths surface, rbecomes RE + h. Again, set the
gravitational force equal tomg :G[(m ME)/r
2] mg This gives:
This shows that gdecreases with increasing altitude As r , the weight of the object approaches zero
( )2
E
E
GMg
R h=
+
ME
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Altitude Dependence of g