The E Tutor - Newton's Law of Gravitation

8/2/2019 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

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The E Tutor - Newton's Law of Gravitation