universal gravitation e m ah, the same force that pulls the apple to the ground pulls moon out of...

Universal Gravitation

E

M

Ah, the same force that pullsthe apple to the ground pulls moon out of its inertial pathinto circular motion aroundthe earth!

Therefore, the forces must beproportional to each other!

E

M Now, if the earth pulls the appleat a rate of 9.8 m/s2, then, the same earth must pull to moon at a proportional rate to that.

60re

If the moon is 60 x further from the apple, and all forms of energy obeythe Inverse Square Law, then, the acceleration of the moon should be1/602 of that of the apple, or9.8 m/s2 x 1/602 = 0.0027 m/s2

And, in one second it should fall d = 1/2 (0.0027m/s2)(1sec)2

or, 0.0014 m (d = ½ at2 )

Universal Gravitation

FE

Force ofEarth onmoon

FM Force ofMoon onEarth

FE = FM

3rd Law

Universal Gravitation

FE

Force ofEarth onmoon

FM Force ofMoon onEarth

FE = FM

3rd Law

Because of the3rd Law and theInverse SquareLaw :

F = Gm1m2/r2

Universal Gravitation

F = Gm1m2/r2

If “F” is the weight of an object, “Fw”, then,Fw = m2gand, m2g = Gm1m2/r2

or, g = Gm1/r2 (m2’s cancel)

or, m1 = gr2/G

Universal GravitationIf gravity is the force that causes anobject to travel in circular motion, then,

F = Fc or, Gm1m2/r2 = m2v2/r

or, (m2’s cancel) m1 = v2r/G or, r = Gm1/v2

or, v2 = Gm1/r

or,m1 = v2r/G

v = Gm1/r

Universal GravitationIf gravity is the force that causes anobject to travel in circular motion, then,

F = Fc or, Gm1m2/r2 = m2v2/r

or, Gm1m2/r2 = m242r/T2

transpose extremes T2/r2 = m242r/Gm1m2divide by “r” and cancel m2

T2/r3 = 42/Gm1

or, r3/T2 = Gm1/42

This equation is called “Newton Variationof Kepler’s 3rd Law”

Universal Gravitation

r3/T2 = Gm1/42 Note that for any objectcircling a superior objectthat Gm1/42 remainsconstant!!!! (k).

Therefore,r3/T2 is also constantfor all objects circlingthat superior object

Universal Gravitation

r3/ T2 = “k” for all circling objects

Therefore, for two objects circling the same superior object...

r13/ T1

2= r23/T2

2

Kepler’s Laws1st Law…all planets circle the Sun in

ellipital paths with the Sun at onefocus

Sun

planet

F2F2

Kepler’s Laws1st Law…all planets circle the Sun in

ellipital paths with the Sun at onefocus

2nd Law…Each planet moves around the sun in equal area sweep inequal periods of time

Kepler’s Laws

2nd Law…Each planet moves around the sun in equal area sweep inequal periods of time

1

2 3

4

ba

Area 12a = Area 43b

Kepler’s Laws1st Law…all planets circle the Sun in

ellipital paths with the Sun at onefocus

2nd Law…Each planet moves around the sun in equal area sweep inequal periods of time

3rd Law…the ratio of the squares of theperiods to the cube of their orbitalradii is a constant

Kepler’s Laws

3rd Law…the ratio of the squares of theperiods to the cube of their orbitalradii is a constant

r3/T2 = “k” for all circling objects

Therefore, for two objects circling the same superior object...

r13/ T1

2= r23/T2

2

Sample Problems

What is the gravitational attraction between the Sunand Mars?

F = ?ms = 1.99 x 1030 kgmm = 6.42 x 1023 kgrm = 2.28 x 1011 m F = Gmsmm/rm

2

F = 6.67 x 10-11 N m2/kg2(1.99 x 1030kg)(6.42 x 1023kg)(2.28 x 1011 m)2

F = 1.64 x 1021 N

Sample Problems

What velocity does Mars circle the Sun at?

v = ?ms = 1.99 x 1030 kgmm = 6.42 x 1023 kgrm = 2.28 x 1011 m

F = Fc

Gmsmm/rm2 = mmv2/rm

v2 = Gms/r

v2 = 6.67 x 10-11Nm2/kg2(1.99 x 1030 kg)/2.28 x 1011m

v = 2.4 x 104 m/s

Sample Problems

What is the period of Mars as it circles the Sun?

T = ?ms = 1.99 x 1030 kgmm = 6.42 x 1023 kgrm = 2.28 x 1011 m

F = Fc

Gmsmm/rm2 = mm42r/T2

T2 = 42r3/Gms

T2 = 42(2.28 x 1011 m)3 /6.67 x 10-11)1.99 x 1030 kg

T = 5.9 x 107 s

or, T = 685 days

Sample Problems

What is the period of Mars? This time use Kepler’s 3rd Law to find it!

Tm = ?Te = 365.25 dare = 1.5 x 1011 mrm = 2.28 x 1011 m

F = Fc

Gmsmm/rm2 = mm42r/T2

r3/T2 = Gms/42 = k

rm3/Tm

2 = /re3/Te

2

(2.28 x 1011 m)3/ Tm2 = (1.5 x 1011m)3/ (365.25 da)2

Tm = 684 days

Tm= (2.28 x 1011 m)3 x (365.25 da)2

(1.5 x 1011m)3

Sample Problems

What is the force of attraction (gravitational attraction) between a 85 kg person and a 55 kg person 0.90 mapart?

0.90 m

55 kg

85 kg

F = Gm1m2/rm2

F = 6.67 x 10-11 Nm2/kg2(85 kg)(55 kg) (0.90 m)2

F = 3.85 x 10-7 N