Physics 320 

Prof. Dale E. Gary
NJIT 
Newtonian Mechanics
A: Dynamics
We are now going to study orbits in some detail, but first we need to review some basic mechanics such as you learned in your Freshman Physics class. In particular, we need to review:
1. Coordinates and Vectors
Recall that in three dimensions, a vector equation really represents three equations, one for each spatial dimension. A vector equation likeB. Newton's Lawsr = r_{o} + v_{o}treally represents the three equationsx = x_{o} + v_{ox }twhere the coordinates of the vectors are
y = y_{o} + v_{oy }t
z = z_{o} + v_{oz }tr = (x, y, z)Legal vector operations are addition and subtraction, e.g.,
r_{o} = (x_{o}, y_{o}, z_{o})
v_{o} = (v_{ox}, v_{oy}, v_{oz}).v + u = u + vof multiplication by a scalar, e.g.,
(v  u) + w = v  (uw)av = vaOrdinary multiplication of two vectors generally has no meaning, but there are two special ways to "multiply" vectors that are defined: the dot product and the cross product.
a(v + u) = av + au , etc.a. Dot Product (or scalar product)
v . u = v_{x}u_{x} + v_{y}u_{y} + v_{z}u_{z} = vu cos q (a scalar)The meaning is " the component of v in the direction of u times the magnitude of u," or equivalently, " the component of u in the direction of v times the magnitude of v."Example: A box sliding down an incline
b. Cross Product (or vector product)
The magnitude is
 x y z  v x u =  v_{x }v_{y }v_{z}  = (v_{y}u_{z}u_{y}v_{z}) x (v_{x}u_{z}u_{x}v_{z}) y + (v_{x}u_{y}u_{x}v_{y}) z (a vector)  u_{x }u_{y }u_{z}  where x, y, and z are unit vectors.
and the direction is perpendicular to both v and u.
 v x u  = vu sin q (area of parallelogram) c. Unit Vectors
Note that polar coordinates, (r, q), are related to rectangular (2D) coordinates (x, y) by
r = ( x^{2} + y^{2 }) ^{1/2} ; q = tan^{1} ( y / x )or converselyx = r cos q ; y = r sin q.Unit vectors are vectors of length 1, e.g. the x, y, and z above. Unit vectors in polar coordinates are:r = cos q x + sin q y ;and have directions in the r and q directions, respectively. Graphically:
q = sin q x + cos q y
Note that dr /dq = qand dq/dq = rd. Time Derivatives, Velocity and Acceleration
In rectangular coordinates, the 2dimensional position, velocity, and acceleration are as follows:
r = xx + yy (position)In terms of polar coordinates, things are a little more complicated:
v = dr / dt = r' (prime notation) =dx/dt x + dy/dt y = v_{x }x+ v_{y }y (velocity)
a = dv / dt = d^{2}r / dt^{2} = r" (prime notation) =d^{2}x/dt^{2}x + d^{2}y/dt^{2}y = a_{x }x+ a_{y }y (acceleration) r = rr
v = (d/dt) rr = r'r + r dr/dt = rr + r dr/dq dq/dt = r'r + rq'q ==>v_{r} = r' ; v_{q} = rq'
a = dv / dt = (d/dt)(r'r + rq'q) = r"r + r' dr/dq dq/dt + r'q'q+ rq"q+ rq'dq/dq dq/dt =r"r + r'q'q+ r'q'q+ rq"q  rq'^{2}r = (r"  rq'^{2})r + (rq" + 2r'q')q ==>a_{r} = r"  rq'^{2} ; a_{q} = rq" + 2r'q'
I. Law of Inertia
Newton's first law is basically a statement of conservation of linear momentum, p = mv. The law states:II. Force Law
"The velocity of a body remains constant unless the body is acted on by an outside force." or "A body at rest tends to remain at rest, a body in motion tends to remain in motion, unless acted on by an outside force.."
or
dp/dt = 0 ==> m dv/dt = 0 ==> ma = 0
Newton's second law defines the force on a body in terms of its effect in accelerating the body. The law states:III. Action and Reaction
"The acceleration imparted to a body is proportional to and in the direction of the force applied, and inversely proportional to the mass of the body." or
F = ma
Note that if m = constant, this can be written
F = m dv/dt = d/dt mv = dp/dtThis can be thought of as the definition of force. If you pull or push a 1 kg body, and it is observed to accelerate by 1 m/s, you have applied a force of 1 N (newton).
Newton's third law is basically a statement of conservation of total linear momentum for a system of particles. The law states:C. Law of Universal Gravitation
"For every force acting on a body, there is an equal and opposite force exerted by the body." or F_{1} =  F_{2}
or P = S p_{i }= constant
Newton was interested in explaining the motion of the Moon. He knew its distance fairly accurately, and that it orbited the Earth with a nearly uniform circular motion.
In uniform circular motion, the magnitude of v is constant, but the direction changes. This turning of the velocity is due to a central force, called the centripetal force, which gives the body a centripetal acceleration (towards the center of its circular path). The centripetal acceleration can be found by considering the velocity at two points on the circle, and taking the limit,
Dv dv lim = = a Dt=>0 Dt dt
Diagram showing centripetal acceleration for an object in uniform circular motion. The centri petal acceleration needed to "turn" the velocity from v to v' is Dv/Dt, where the relationships between s, Dv and Dq are as shown. 
From the drawings above, we have
s = r Dq = v Dt and Dv = v Dq ==> Dv/Dt = v^{2}/rso
a = v^{2}/r (centripetal acceleration)Note that the velocity here is only in the theta direction, v_{q} = r dq/dt, so we can recognize the second term in our earlier, general expression for radial acceleration as the centripetal acceleration, complete with minus sign to indicate that it points inward toward the center:
The force is F = ma, so the Moon must experience a force F = mv^{2}/r, where m = M_{Moon} = mass of the Moon, v = orbital velocity of the Moon, and r = D_{Moon} = distance to the Moon from the center of the Earth (actually from the center of the EarthMoon system, as we will see later).
a_{r} = r"  rq'^{2} = r"  v_{q}^{2}/r
It may seem hard to measure the velocity of the Moonhow would you do it?
Just use its period and the length of its orbit (2pD_{Moon}) to get v = 2pD_{Moon}/P, which gives
But recall that Kepler gave the relationship between the period and radius (semimajor axis) of an orbit, which in this case is: P^{2} = kD_{Moon}^{3}, which gives
mv^{2} M_{Moon}(2pD_{Moon}/P)^{2} M_{Moon}4p^{2}D_{Moon} F = = = r D_{Moon} P^{2}
From Newton's third law, he knew that the force of the Earth on the Moon must be balanced by an equal and opposite force of the Moon on the Earth, so the force had to be proportional to both masses:
M_{Moon}4p^{2} F = ==> inverse square law Fa 1/r^{2}. kD_{Moon}^{2}
Of course, the force is attractive, directed along the radius vector, so the vector force is
M_{Moon}M_{Earth} GMm F a = r^{2} r^{2}
Newton now took at leap of insight, and considered this law to be valid everywhere, i.e. it is a Universal Law. In particular, he could measure the force on an object at the surface of the Earth:
GMm F =  r r^{2}
which allowed him to relate his constant, G, to the acceleration of gravity at Earth's surface:
GM_{Earth}m F =  r = mgr R_{Earth}^{2}
If his law was Universal, then the magnitude of the force on the Moon would be
GM_{Earth} g = R_{Earth}^{2}
Could it be so simple? Is the Moon merely "falling" with this value of acceleration? To find out, Newton needed to show that the acceleration needed to keep the Moon in its orbit was related to the acceleration of a stone falling at the Earth's surface, according to this relationship (working with accelerations rather than forces avoids the need to know the Moon's mass). To show this, Newton considered the distance the Moon must "fall" towards Earth in one second.
GM_{Earth}M_{Moon} R_{Earth}^{2} F = = gM_{Moon} = g' M_{Moon} D_{Moon}^{2} D_{Moon}^{2}
The value of g' , above, is
g' = 9.8 m/s^{2} (R_{Earth}/D_{Moon})^{2} = 2.71 x 10^{3} m/s^{2}.Thus, in 1 s, the Moon will fall a distance
d = 1/2 g't^{2} = g'/2 = 1.36 x 10^{3} m (1.36 mm).Let's see if this matches the orbit of the Moon. The sidereal period of the Moon is P = 27.32 days = 2.36 x 10^{6} s, so the angular velocity is 2p/P = 2.66 x 10^{6} s^{1}. Thus, in 1 s, the Moon moves through an angle of 2.66 x 10^{6} radians. The distance that the Moon falls is shown in the figure below, and is related to the distance to the Moon by
d = D_{Moon}/cos q  D_{Moon} = D_{Moon} (1  1/cos q) .
We cannot use most calculators with such a small angle (most will just give cos 2.66 x 10^{6}= 1), but we can expand as a power series, which for such a small angle will allow us to keep only the leading terms. We will need two expansions:
cos q = 1 q^{2}/2 + q^{4}/24  ...where the second expansion is used to expand the result of the first expansion, in the form (1 q^{2}/2)^{1}. Performing these expansions and keeping only the leading term, we find that
(1  x)^{n} = 1 + nx + 1/2n(n1)x^{2} + ...
d = D_{Moon}q^{2}/2 = (384,000 km) (2.66 x 10^{6})^{2 }/ 2 = 1.36 x 10^{3} m.
The same answer! So Newton finally understood the truth! The Moon is falling towards Earth just like any stone or apple would do. The Law of Gravitation truly is Universal!