Download When astronauts are in the space shuttle

When astronauts are in the space shuttle
they seem to float? Why is that?
Are they so far from Earth that the force of gravity is
zero? Or there is another reason?
Let’s calculate g at 400 km altitude:
W =m g = G ME m/R2
g = G ME/(RE + 400 km)2 = 8.68 m/s2
Compare this to 9.81 m/s2 on the surface of the earth
≈88% - hence, the force of gravity acting on them is not
much different
But they are orbiting the earth – need a centripetal force –
and the force of gravity is that centripetal force!! There
are no other forces – the orbiting spaceship along with its
crew is falling freely – therefore, there is no contact with
a support, no weight, no reaction of the surface WEIGHTLESSNESS
mv 2
= mg − N ,
r
mv 2
= mg ⇒ N = 0
r
“Dependence” of “g” on the latitude
ƒ The Earth is rotating – spins on its axis
with a frequency of 1/(24 hours)=1.16×10-5
s-1.
ƒ The “linear” velocity of a point at the
equator (where it is maximum) is
2πR/T=465 m/s.
ƒ The centripetal acceleration required
for that is 0.03m/s2.
ƒ The centripetal force is a result of action
of two forces – gravity and reaction of
surface, mv2/r=mg-N —> N=mg-mv2/r < mg
ƒ Weight = -N and in inertial frame is equal
mg. The rotating earth is not an completely
inertial system, and an “effective” g is
getting smaller as we get closer to the
equator.
ƒ At the intermediate latitudes, the weight
is not directed to the center of the Earth,
but the force of gravity is!
Force of
Gravity
Force of
Reaction
A net
centripetal
force
Artificial gravity
ƒ In the previous example, you saw that the frame
related to the ground may not always considered to
be inertial. g per se, does not depend on latitude, but
the weight does!
ƒ Sometimes it is needed to “mimic” gravity effect, or
rather weight. One of the ways is to place an object
in a “centrifuge” – a rotating cylinder. Then, to rotate
with a cylinder, a centripetal force is needed, and it is
supplied by the reaction of the wall. The force
pressing on the walls is equivalent to weight, but
certainly not the gravity.
TIDES
Tides are caused by the gravitational force of the
moon (and the sun) on the earth.
The first (nearer) bulge – is intuitively simple: it is due to simple
fact that in the Newton’s gravitational law, the force of gravity is
inversely proportional to the square of distance between the
objects. Point A in the picture is nearer to the Moon than points
C and B. Therefore, the force of gravity acting on the water at a
point A is larger than the force of gravity acting on the similar
sample of water at points C and B. This difference reduces the
relative push of water on the see floor and therefore reduces
the reaction force. Hence, point A can support more water –
nearer bulge.
But where does the second bulge come from?
Tides are caused by the gravitational force of the
moon (and the sun) on the earth.
ƒ To understand the whole picture, however, we need to understand the
motion of the Moon around the Earth.
ƒ Forget the Sun for a moment, then the system Earth-Moon is moving
uniformly – no external forces.
ƒ Both Earth and Moon are massive – act on each other with forces of
gravity. Why the Moon does not fall on Earth? Because the Moon is
revolving and its free fall just supplies the centripetal acceleration. But
we can ask the same about the Earth…
ƒ There must be a symmetry moved only by the difference masses of
Earth and Moon: As a result both of them are revolving around an axis
located about 4600 km form the Earth’s center. (And as a whole falling
on the Sun.)
Mary-go-round
ƒ Inertial frame – ground observer:
ƒ Trajectory is circular – need a
centripetal force
ƒ What force provides it? –
Friction force, f=mv2/r
ƒ The frame of Mary-go-round – accelerating frame:
ƒ The object is at rest, but the force of friction is
acting towards the center of the circle
ƒ What compensates the friction force
ƒ The force of inertia – a toll for using accelerating
frame: the acceleration is a=v2/r towards the
center, the force of inertia is equal to (–ma). In
our case it is equal to mv2/r and directed away
from the center. If we add it with the force of
friction, we will get a desired zero. This force is
called centrifugal.
The second (further) bulge
ƒ Now we can apply our understanding of Mary-goround to tides.
ƒ Consider motion from the earth’s observer’s
reference frame
ƒ The earth is revolving around a center of
ƒ What is the centrifugal force acting on 1 kg of
water? 1×v2/r, where r is the distance from this
water sample and the axis of Earth-Moon revolution.
It turns out that it is proportional to the distance
from the axis adds to the reaction on water, larger on
the far side – can support more water – second bulge
BC > CA
Pressing forces
due to Earth’s g
Resultant
pressing forces
Gravity of
the Moon
=
=
+
+
Inertial
forces –
revolving
Earth
-
The Earth’s revolution
around the Sun and
tides
The Sun effects on
tides as well – if it is
aligned with the Moon,
the tides are stronger
– spring tides, if not,
weaker – neap tides
ƒ Sun’s effect on tides is not as strong than the Moon’s
because the Sun is much farther away – the
difference in relative distances from the Sun to the
near and far side of the Earth is small.
ƒ However, the effect small as it is proves that the
Earth is revolving around the Sun!!
Conservation laws
ƒ Why conservation laws?
ƒ How do we describe motion? – Kinematics – equation
of motion
ƒ How do we get such equations of motion? – Analysis
of forces.
ƒ Is such analysis always possible? No. For example, if a
tennis ball hits a wall, it does not immediately
reflect, it stops, flattens, then flexes, and only then
reflects. It may be and it usually is quite difficult to
follow all forces.
ƒ Because of this, we may ask a question – are there any
quantities that do not change as a result of
interaction or conserve? They may change during the
interaction in ways we could not follow, but at the
end, they would be the same as in the beginning.
CONSERVATION LAWS
1. Conservation of Mass:
In a closed system, mass is conserved.
A. If you weigh components of the system
separately, and then together, you are going to
observe the same result.
B. Chemical reactions – mass balance
Einstein generalized mass conservation and energy
conservation, principle of equivalence, E=mc2
2. Let us consider the (LINEAR)
MOMENTUM that we have introduced
before
Linear momentum or “momentum” is a product of
mass and velocity
p=mv
ƒ Momentum is a vector quantity
ƒ Units: kg m/s
ƒ Notation: p
Consider two parts, A and B, of one system. If they
interact – act on each other with forces, these forces
must be equal and opposite according to the third
Newton’s law:
F
= -F
A on B
Use the
B on A
2nd
law:
∆ (mv)
F=
∆t
∆(mv) B
∆(mv) A
=−
∆t
∆t
Cancel ∆t and find:
∆(mv) B = −∆(mv) A
Or
total (mv)before = total (mv)after
Total momentum of the closed system is conserved!
Applications of momentum conservation law
ƒ Collisions are such interactions of objects in which
the interaction per se can be separated from the
initial and final state. Only initial and final states are
of interest.
ƒ If two (several) particles collide, they are all parts of
one system, so their total initial momentum is equal to
their total final momentum (vector sum).
EXAMPLE:
Car & Truck
A car (1000 kg) moves at 30 m/s hits a truck (4000 kg)
at rest, they “stick” together. What is their velocity
just after the collision ?
Momentum of car:
pcar = mcar vcar = (1000 kg) x (30 m/s) = 30,000 kg m/s
Momentum of truck:
ptruck = 0
Total momentum before:
p = pcar+ptruck = (mv)before = 30,000 kg m/s
Example continued…
total(mv)before = total(mv)after
mcar vcar + mtruck vtruck
= 30,000 kg m/s = (mtruck + mcar) x vafter
30,000 kg m/s = (1,000 kg + 4,000 kg) x vafter
30,000 kg m/s = 5,000 kg x vafter
vafter = 6 m/s in the direction of the car’s motion
Completely inelastic collision: no recoil
QUESTION:
During a collision of two football players, the
total momentum of the system can not be zero.
1. False.
2. Correct.
The total momentum is the sum of the two
individual momenta. If one is directed opposite to
the other, but has the same magnitude, they
would add up to zero.
First exam review
ƒ Measurement: properties, units, vectors, scalars
ƒ Kinematics: position, displacement, time, velocity,
speed, average, instantaneous, d vs.t, v vs.t,
acceleration, free fall, circular motion at a constant
speed, uniform motion.
ƒ Dynamics: inertia, inertial frames, what is relative and
what invariant, forces, Newton’s laws, force of
gravity, weight, friction (rest and kinetic), free fall,
projectile motion, centripetal force, Newton’s law of
gravity, orbits, elastic forces, harmonic motion, air
resistance, weightlessness, tides, and other examples
ƒ Conservation laws: mass conservation, momentum
conservation, collisions