Download 1 LECTURE 14 1] NEWTON`S LAW OF GRAVITY

LECTURE 14
1] NEWTON’S LAW OF GRAVITY
-The gravity law is formulated for two point particles with masses m1 and m2 at a distance r between
them. Here are the four steps that bring to universal law of gravitation discovered by NEWTON.
a) Based on experimental information one postulates
that gravitation produces an attractive force.
r
m1
→
→
F 12
F 21
m2
Figure 1
b) Based on the second law of Newton
F12 = m1 a1 _ so _ F12 ≈ m1 similarly F21 ≈ m2
c) Based on the third law of Newton F12 = F12 ≡ F It comes out that
F ≈ m1 m2
(1)
d) Based on experimental results Newton guessed that gravitation force decreases with distance as 1/r2.
mm
mm
So, it comes out that F ≈ 1 2 2 more precisely
(2)
F = G 12 2
r
r
Measurements show that G = 6.67*10-11Nm2/kg2 is the universal constant of gravitation
- The gravitational force is a vector directed versus the source that exerts this force. So, its vector form is
→
F 12
→
F 21
→
m m r 21
= −G 1 2 2
r21
r
→
m m r 12
= −G 1 2 2
r12
r
→
(3)
→
r 21
is the unit vector with tail at mass m2 – the source of F 12 .
r21
→
→
r 12
is the unit vector with tail at mass m1– the source of F 21
r12
- To apply the gravitational law for two bodies close to each other one must use integration techniques and
the difficulty of calculi depends on the form of the two bodies. But, if the bodied are far enough to each
other1, one may model them by point particles and apply the law in its original form. In particular, with
some approximations, we are able to model the interaction of earth with an object on its surface as if the
whole mass of the earth is concentrated at its center and the objects are at distance r =R earth.
-Many experiments have shown that; when several particles interact gravitationally between them the
principle of linear superposition applies. So, inside a system of particles m1,m2,m3,…mn , the force
exerted on mass m1 is
→
n
→
F 1 = ∑ F1i
(4)
i =2
2] THE GRAVITATIONAL AND THE INERTIAL MASS
→
→
-When expressing the second law of Newton we use the inertial mass
F NET = min a
(5)
When formulating the gravitation law, Newton was not sure that the mass of particles in this law is the
same as their inertial mass. Let’s verify this issue. We start by supposing that the mass in the gravitational
law may be different from min. So, we note it mgr. Let’s consider now a body in free fall close to earth
m gr M Earth
Fgr = G
(3)
surface. The earth will exert on it the gravitational force with magnitude
2
REarth
Here we assume that the body is close to the surface so that its distance form earth center is ≅ REarth .
1
Their dimensions are <<< than the distance between them.
1
As this is the net force exerted on the body we apply the second law of Newton
m gr M Earth
M Earth
M
) = m gr g because G 2Earth = g
2
R Earth
R Earth
R
m gr
Then,
a=
g
(6)
min
A big number of measurements show that the acceleration of free bodies is equal to g = 9.8m/s2. This
means that mgr/min =1, and min = mgr.
So, the experiments confirm that the gravitational mass is the same as the inertial mass.
FNET = min a = Fgr = G
2
Earth
So, we get
min a = m gr (G
- Let’s apply the gravitation law for the force exerted by earth over a mass 1kg close to earth.
→
→
M Earth
R Earth →
F = −G
≡g
(7) So, the g-vector is equal to the gravitation force exerted
( R Earth + h) 2 R Earth
on a mass 1kg. By measuring the force exerted on the mass 1kg in different locations on the earth one gets
a whole system of g-vectors (fig.2). The totality of these vectors forms the gravitational field of the earth.
(7)
Note that the g-vector magnitude decreases with the increase of
distance “h” from the earth but it is always directed versus the center of
the earth. The gravitational field of the earth has a spherical symmetry.
In fact, it is not exactly spherical, because the model of earth as a uniform density
sphere is not very precise. Now, the object weight is equal to gravitational
→
→
W = m g (7)
force exerted by this field
So, the weight of the same object is a vector that is different in
different points of gravitational field of the earth.
Figure 2
3] KEPLER’S LAWS ON PLANETARY MOTION
- First law: The planets move on elliptic orbits around the sun that is located at one of its focuses.
The minor axis is long 2b and the
major axis is long 2a. The closest
distance to sun is called perihelion
and the biggest distance to sun is
called aphelion (fig.3).
Perihelion
2b
Sun
Aphelion
2a
Figure 3
Second law: The line sun- planet sweeps out equal areas for equal interval of times.
S1
Figure 4
S2
Third law: The square of the period of planet
motion is proportional to the cube of average
distance from the sun. Calculations show that
the average of distance sun-planet is equal to
half of major axis a. Then, satellite motion (see
Lecture_8.5) tells that
T2 = κ a3
(8)
2
4π
where κ sun =
GM sun
2
Note : Kepler’s laws are valid for elliptical paths of any planet around a central body; for example the
4π 2
.
moon moving around the earth but in this case κ Earth =
GM Earth
THE ENERGY OF PLANETS
υA
rA
rP
rA
υP
Figure 5
-As the mass of other planets is much smaller than the mass of sun
we neglect their action on the motion of the studied planet. We
consider that the system sun-planet is a conservative system, i.e. the
forces originated from outside it are zero. In these circumstances:
a) The torque of exterior force is zero and we can apply the principle
for conservation of angular momentum.
b) The work done by Net exterior force is zero and we can apply the
principle of energy conservation for the system sun-planet
→
→
→
- The principle of angular momentum conservation tells that L A = L P (9)
→
→
→
or r A x p A = r B x p P
(10)
The equality of magnitudes brings to condition rA p A sin 90 0 = rP p P sin 90 0 ⇒ rA m planetυ A = rP m planetυ P
So, we get
-The principle of energy conservation tells that that
K=
As the mechanical energy is E = K + U where
m plυ A2
We get
2
m pl M Sun
rA
2
=
m plυ P2
2
(13)
−G
and
m pl M Sun
rP
(11)
EA=EP
(12)
U = −G
m pl M Sun
rpl − Sun
(14)
(15)
M Sun υ P2
M
1 1
=
− G Sun and finally 2GM Sun ( − ) = υ P2 − υ A2
(16)
2
rA
2
rP
rP rA
By using the equation (11) and the fact that
rA + rP = 2a (17), after some calculations we find that
After cancelling mpl
υ A2
−G
m plυ pl2
rAυ A = rBυ P
−G
GM rP
GM rA
(18)
and
υ P2 =
(19)
a rA
a rP
Finally, by substituting one of this expressions at the expression of total energy (at perihelion or aphelion)
m pl M Sun
E = −G
(20)
we get
2a
υ A2 =
4] THE BOUND AND UNBOUND TRAJECTORIES
-Let see what happens with an object thrown up with high initial velocity υ. The only force exerted on it is
mM Earth
F = −G
(21)
the gravitational force of the earth.
r2
As external net force is zero, the system earth – object conserves its energy. For such a system
E = U + K is a constant all time.
Note: As the earth is the reference frame, generally one talks for object energy.
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mM Earth
(22)
r
The graph of this function is shown in figure 6. Note that the zero value
of potential energy of gravitational interaction is met for big r-values .
The zero-value of potential energy means that there is no interaction
between the system parts, or more precisely the parts are not bound to
each other. When the object is on the earth surface, the distance from
CM of the earth (at origin of graph) is equal to the earth radius RE and
its potential energy is U(RE). When we throw it vertically with initial
velocity υ, the mechanical energy of the object (i.e. of object earth
E (RE) = U(RE)+K(υ)
(23)
system) is
- By using the work done by the gravitation force (expr.21) one may find2 that U (r ) = −G
U( r )
RE
R > RE
U(R)
r
U( RE )
K
Figure 6
As we know, the velocity of the object will decrease with height and will become zero at its maximum
height h, i.e. at distance R = (RE + h). When getting at this distance, all its energy is potential energy. So,
E (R) = U(R)
(24)
We say that the kinetic energy makes it climbs up on a potential well (see fig.6).
The principle of energy conservation tells that
E (R) = E(RE)
By using this equation we can find the maximum height as follows;
As E (RE+h) = U(RE+h) and using (23) we get
So, − G
(25)
E (RE+h) = E(RE)
U(RE+h)= E(RE)= U(RE)+K(υ) (26)
mM Earth
mM Earth mυ 2
= −G
+
2
RE + h
RE
GM Earth (
and
υ2
1
1
)=
−
2
RE RE + h
GM Earth
GM Earth
GM Earth
υ2
υ2
υ2
1
(1 −
)=
_ as _ h << RE _ ⇒
(1 − 1 + h / RE ) =
Then,
⇒
h=
1 + h / RE
2
2
2
RE
RE
RE2
GM Earth
=g
As
Re2
we find that
h=
υ2
2g
(27)
which is a known result from kinematics.
- When the initial velocity υ of the body is such that U(RE) + K(υ) < 0 the object will get to a given
distance from earth but will remain all time bound within the system earth-object;
If U(RE) + K(υ) ≥ 0 the object will get so far that the interaction with earth becomes zero; it becomes an
unbounded object to gravitational field of the earth. The limiting initial velocity necessary to unbind from
gravitational field of earth is known as escape velocity υesc. This velocity can be found by condition
2
mM E mυ esc
2GM E
(28)
U ( R E ) + K (υ esc ) = 0 ⇒ −G
+
= 0 ⇒⇒⇒⇒⇒ υ esc =
RE
2
RE
- Suppose that one must send a rocket out of earth gravitational field; i.e. make it unbound object to
earth. The first requirement is to give to rocket an initial velocity υ ≥ υesc. A simple calculation based on
expression (28) shows that υesc =11190 m/s. This value of initial velocity can be furnished only by
reactive engines.
2
Through a integral calculus.
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What form has the trajectory of an unbound rocket? The mathematical calculations show that if
a) υ > υesc the rocket trajectory will not be closed and it is a hyperbola.
b) υ = υesc the rocket trajectory will not be closed and it is a parabola.
If the initial velocity of rocket υ < υesc the object remains a bound object to the earth gravitation field and
its trajectory will be an ellipse (closed orbit). Note that if the initial velocity is too small υ << υesc , this
orbit may cross the earth, i.e. the rocket will strike on earth surface (figure 7).
-Consider an artificial satellite moving uniformly on a circular
orbit at distance RS from CM of the earth (fig.8). Its motion
has a centripetal acceleration a c =
υ << υesc
υ s2
RS
and the net force
exerted on it is the earth gravitation. So, F = G
υ < υesc
ellips
Earth
We get G
(30)
E(RE) =E(RS)
m sυ L
m M
m υ2
mM
m GM E
mM
−G S E = S s −G s E = S
−G s E
2
RE
2
RS
2 RS
RS
2
υ = υesc
parabolla
υ > υesc
hiperbola
Figure 7
So,
υL2
2
υ L 2 = 2G
υL
Earth
and
υ s2
ms M E
m
=
and
s
RS
RS
Rs2
GM E
υ s2 =
(29)
RS
What is the required launch υL speed for this satellite?
We apply the principle of energy conservation.
F = ms ac = ms
υ s2
ms M E
Rs2
=G
M R
ME
M
1
−G E = G E ( S − )
2 RS
RE
RS RE 2
M E RS 1
M R +h 1
− ) = 2G E ( E
− )=
(
RS RE 2
RS
RE
2
M
M
h 1
2h
= 2G E (1 +
− ) = G E (1 +
)
RS
RE 2
RS
RE
(31)
(32)
For low orbit satellites (say heights100 - 200Km from earth)
RS≈RE(6378km) 2h/RE ≈1 and from (32) we can find that
RS
RE
υL2
2
Figure 8
υS
Circular
orbite of
Satellite.
≅G
ME
2 RS
So, the launch velocity for a stationary
circular low orbit with radius RS is
υL ≈
GM E
RS
(33)
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