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Universal Gravitation
M
E
Ah, the same force that pulls
the apple to the ground pulls
moon out of its inertial path
into circular motion around
the earth!
Therefore, the forces must be
proportional to each other!
M
60re
E
Now, if the earth pulls the apple
at a rate of 9.8 m/s2, then, the
same earth must pull to moon at
a proportional rate to that.
If the moon is 60 x further from the
apple, and all forms of energy obey
the Inverse Square Law, then, the
acceleration of the moon should be
1/602 of that of the apple, or
9.8 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
M
FM
Force of
Earth on
moon
FE
Force of
Moon on
Earth
E
FE = FM
3rd Law
Universal Gravitation
M
Force of
Earth on
moon E
FM
FE
Force of
Moon on
Earth
FE = FM
3rd Law
Because of the
3rd Law and the
Inverse Square
Law :
F = Gm1m2/r2
Universal Gravitation
F = Gm1m2/r2
If “F” is the weight of an object, “Fw”, then,
Fw = m2g
and, m2g = Gm1m2/r2
or,
g = Gm1/r2 (m2’s cancel)
or,
m1 = gr2/G
Universal Gravitation
If gravity is the force that causes an
object to travel in circular motion, then,
F = Fc
or, Gm1m2/r2 = m2v2/r
or, (m2’s cancel) m1 = v2r/G
or,
or,
v2 = Gm1/r
v = Gm1/r
m1 = v2r/G
or,
r = Gm1/v2
Universal Gravitation
If gravity is the force that causes an
object to travel in circular motion, then,
F = Fc
or, Gm1m2/r2 = m2v2/r
Gm1m2/r2 = m24p2r/T2
or,
T2/r2 = m24p2r/Gm1m2
T2/r3 = 4p2/Gm1
transpose extremes
divide by “r”
and cancel m2
or, r3/T2 = Gm1/4p2
This equation is called
“Newton Variation
of Kepler’s 3rd Law”
Universal Gravitation
r3/T2
=
Gm1/4p2
Therefore,
Note that for any object
circling a superior object
that Gm1/4p2 remains
constant!!!! (k).
r3/T2 is also constant
for all objects circling
that superior object
Universal Gravitation
r3/ T2 = “k” for all circling objects
Therefore,
for two objects circling the
same superior object...
r13/ T12= r23/T22
Kepler’s Laws
1st Law…all planets circle the Sun in
ellipital paths with the Sun at one
focus
planet
F2
Sun
F2
Kepler’s Laws
1st Law…all planets circle the Sun in
ellipital paths with the Sun at one
focus
2nd Law…Each planet moves around
the sun in equal area sweep in
equal periods of time
Kepler’s Laws
2nd Law…Each planet moves around
the sun in equal area sweep in
equal periods of time
1
4
a
b
2
3
Area 12a = Area 43b
Kepler’s Laws
1st Law…all planets circle the Sun in
ellipital paths with the Sun at one
focus
2nd Law…Each planet moves around
the sun in equal area sweep in
equal periods of time
3rd Law…the ratio of the squares of the
periods to the cube of their orbital
radii is a constant
Kepler’s Laws
3rd Law…the ratio of the squares of the
periods to the cube of their orbital
radii is a constant
r3/T2 = “k” for all circling objects
Therefore,
for two objects circling the
same superior object...
r13/ T12= r23/T22
Sample Problems
What is the gravitational attraction between the Sun
and Mars?
F=?
ms = 1.99 x 1030 kg
mm = 6.42 x 1023 kg
2
11
F
=
Gm
m
/r
rm = 2.28 x 10 m
s m m
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 kg
mm = 6.42 x 1023 kg
rm = 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 kg
mm = 6.42 x 1023 kg
rm = 2.28 x 1011 m
F = Fc
Gmsmm/rm2 = mm4p2r/T2
T2 = 4p2r3/Gms
T2 = 4p2(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 da
re = 1.5 x 1011 m
rm = 2.28 x 1011 m
F = Fc
Gmsmm/rm2 = mm4p2r/T2
r3/T2 = Gms/4p2 = k
rm3/Tm2 = /re3/Te2
(2.28 x 1011 m)3/ Tm2 = (1.5 x 1011m)3/ (365.25 da)2
Tm= (2.28 x 1011 m)3 x (365.25 da)2
(1.5 x 1011m)3
Tm = 684 days
Sample Problems
What is the force of attraction (gravitational attraction)
between a 85 kg person and a 55 kg person 0.90 m
apart?
F = Gm1m2/rm2
55 kg
F = 6.67 x 10-11 Nm2/kg2(85 kg)(55 kg)
(0.90 m)2
F = 3.85 x 10-7 N
85 kg
0.90 m