newton's law of gravitation
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Newton’s Question: If the force of gravity is being exerted on objects on Earth, what is the origin of that force?
Newton’s realization was that this force must come from
the Earth.
He further realized that this force must be what keeps the
Moon in its orbit.
Must be true from Newton’s 3rd Law
The gravitational force on you is half of a Newton’s 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 exerts on a 2nd body , is directed toward the first body, and is equal and opposite to the force exerted by the second body on the first
Newton’s Universal Law of Gravitation
Every particle of matter in the universe attracts every other particle with a force that is directly proportional to the product of the masses of the particles and inversely proportional to the square of the distance between them.
F12 = -F21 [(m1m2)/r2]
Direction of this force: Along the line joining the 2 masses
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!
The Universal Law of Gravitation is an example of an inverse square law› The magnitude of the force varies as the
inverse square of the separation of the particles
The law can also be expressed in vector form
The negative sign means it’s an attractive force Aren’t we glad it’s not repulsive?
Comments
F12 Force exerted by particle 1 on particle 2
F21 Force exerted by particle 2 on particle 1
This tells us that the forces form a Newton’s 3rd Law action-reaction pair, as expected.
The negative sign in the above vector equation tells us thatparticle 2 is attracted toward particle 1
F21 = - F12
More Comments
Gravity is a field force that always exists between 2 masses, regardless of the medium between them.
The gravitational force decreases rapidly as the distance between the 2 masses increases› This is an obvious consequence
of the inverse square law
Example : Spacecraft at 2rE
• Earth Radius: rE = 6320 km
Earth Mass: ME = 5.98 1024 kg
FG = G(mME/r2)
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 (¼)mg = 4900 N
• A spacecraft at an altitude of twice the Earth radius
Example : Force on the Moon
Find the net force on the Moon due to the gravitational attraction of both the Earth & the Sun, assuming they are at right angles to each other.
ME = 5.99 1024kg
MM = 7.35 1022kg
MS = 1.99 1030 kg
rME = 3.85 108 m
rMS = 1.5 1011 m
F = FME + FMS
F = FME + FMS
(vector sum)
FME = G [(MMME)/ (rME)2]
= 1.99 1020 N FMS = G [(MMMS)/(rMS)2]
= 4.34 1020 NF = [ (FME)2 + (FMS)2]
= 4.77 1020 Ntan(θ) = 1.99/4.34 θ = 24.6º
Gravity Near Earth’s Surface
Gravitational Acceleration g
and
Gravitational Constant G
Obviously, it’s 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 N∙m2/kg2
Always same on every location g The Acceleration due to Gravity
g = 9.80 m/s2 (approx) on Earth’s surfaceg varies with location
G vs. g
Consider an object on Earth’s surface:
mE = mass of the Earth
rE = radius of the Earth
m = mass of object Let us the Earth is a uniform, perfect sphere.
The weight of m: 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/s2 All quantities on the right are measured!
Using the same process, we can Weigh Earth (Determine it’s 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 Earth rE, the mass of the Earth can be calculated:
mE
m
All quantities on the right are measured!
Effective Acceleration Due to Gravity
Acceleration due to gravity at adistance r from Earth’s center.
Write gravitational force as:FG = G[(mME)/r2] mg
(effective weight)
g the effective acceleration due to gravity.
SO : g = G (ME)/r2
ME
Altitude Dependence of g If an object is some distance h
above the Earth’s surface, r becomes RE + h. Again, set the gravitational force equal to mg: G[(m ME)/r2] mg This gives:
This shows that g decreases with increasing altitude
As r ® , the weight of the object approaches zero
2E
E
GMg
R h
ME
The End
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