QUESTION:
how much times earth to be compressed to become black hole?

ANSWER:
A black hole is a singularity, of zero size, so compression
is infinite. If the earth were a black hole, its
Schwartzchild radius, R =2GM /c ^{2} ,
would be about 1 cm. R is the radius within which
nothing can escape. So, if you could get the earth down to
that size, it would then collapse the rest of the way to a
black hole, I presume.

QUESTION:
If momentum is applied not on the center of mass on an object, what would its momentum be at the center of mass and is there an angular momentum at the point of impact?

ANSWER:
Linear momentum is not something which is applied, force is
what is applied. And, what happens also depends on the
details of how the force is applied. I will work out one
simple example which will illustrate the principles involved
when the force is applied in a simple collision. A mass
m approaches the end of a thin uniform rod of mass
M and length 2L with velocity v
perpendicular to the rod; when it collides with the end of
the rod it sticks. The center of mass of the rod is at its
geometrical center a distance L from where m
exerts the force. Linear momentum and angular momentum are
both conserved but, as we shall see, energy is not
conserved. After the collision the center of mass has a
velocity u ; conserving linear momentum, (M+m )u=mv
or u=v [m /(M+m )]. Before the
collision m brings in an angular momentum mvL .
After the collision, the rod plus mass has angular momentum
(I +mL ^{2} )ω
where I=M (2L )^{2} /12=ML ^{2} /3
is the moment of inertia of a thin rod about its center of
mass. Conserving angular momentum I find ω [m +(M /3)]L ^{2} =mvL
or
ω=mv /{[m+ (M /3)]L }.

Just for fun I put in some numbers to illustrate what would
happen in a specific situation: m =1 kg, M =3
kg, L =1 m, v =4 m/s. With these I find:
u =1 m/s, ω =2 ra dians/s=19.1 rpm. You can
also calculate the energy before and after the collision:
E _{before} = ½mv ^{2} =8
J, E _{after} =½(M+m )u ^{2} +½([m +(M /3)]L ^{2} ω ^{2} =6
J. So, energy was not conserved in this collision, 2 Joules
were lost.

CORRECTION:
I realized that an error had been made in my analysis. After
the collision the center of mass is no longer at the center
of the rod but displaced a distance x=mL /(M+m )
toward where m is stuck. It is this center
of mass which moves forward with speed u , and u
is still v [m /(M+m )]. The rotation
is about this center of mass but now the moment of
inertia is different; the net (rod plus mass) moment of
inertia is ML ^{2} /3+Mx ^{2} +m (L-x )^{2} ,
so the corrected value of the angular velocity is
ω=mvL /[ML ^{2} /3+Mx ^{2} +m (L-x )^{2} ].
In the example I did, x =1/4 m, so ω= 16/7 radians/s=21.8 rpm and E _{after} =46/7=6.57
J (1.43 J lost).

QUESTION:
I have been trying to figure this out, but having some trouble doing it: In general, let's say I had a vehicle that could accelerate me (according to my perception) at 1G for 1 year (again according to my perception),
then turn around and decelerate (by my perception again) for another year to return home. How far would I have travelled and how old would my twin at home be when I got back?

ANSWER:
The reason you are having trouble is that relativistic
kinematics involving acceleration, even if the acceleration
in one frame is uniform, is not easy; the main reason is
that, unlike Newtonian mechanics, acceleration in one frame
is not the same as acceleration in a different frame. As a
prelude to reading my answer, you should read an
earlier answer
and the earlier answers referred to in that answer. The
graph here shows the results derived or cited in that
earlier answer. The three graphs all show what is observed
of your motion as seen by the twin on earth. The red curve
is your velocity v relative to c as seen
by the earth-bound twin, v /c =(gt /c )/√[1+(gt /c )^{2} ];
the blue curve is your acceleration a
relative to g , a /g =[1+(gt /c )^{2} ]^{-3/2} ;
the black curve is your position expressed (approximately)
in light years, gx /c ^{2} =√[1+(gt /c )^{2} ]-1.
Now, since I have set this up as seen from earth, let me
specify the time you travel as the time of your trip out as
observed from earth: earth-bound twin observes you to arrive
at you destination in two years. Then, as you can read off
the graph, gx /c ^{2} =√[1+(2)^{2} ]-1=1.24
ly; the return trip will be the same length of time, so the
earth-bound twin's clock will have advanced 4 years. Light
would take 2.48 yr to travel that distance, so your clock
will have advanced a time somewhere between 2.48 and 4
years. Your maximum speed would have been about v /c =(2)/√[1+(2)^{2} ]=0.89.
We can get a rough estimate of your clock reading by
approximating your average relative speed as 0.89/2=0.445;
so you would have seen, on average, the distance contracted
by √[1-(0.445)^{2} ]=0.9 and your time would
have been 0.9x4=3.6 years.

QUESTION:
What happend to accelaration due to gravity if the earth stops rotating?

ANSWER:
Because of the centrifugal force, g appears to be
somewhat smaller than it really is, but the amount is very
tiny. This effect depends on your latitude and is zero at
the poles and greatest at the equator. Also because of this
effect, the acceleration is not directly toward the center
of the earth except at the poles and equator.

QUESTION:
Hello, is the mass of a person on a spaceship (to provide a suitable upward force) important in designing a spaceship? My teacher says otherwise but upon researching I found out that additional fuel is required for every kg of payload that will be sent out into space.

ANSWER:
If the ship and all the fuel are much bigger than the mass
of the person, then you could neglect the mass of the person
in any calculations you know; in other words, the rocket
launch would be, as close as you could hope to be measure,
identical whether or not an astronaut was on board. However
you are also right that the amount of fuel you need depends
on the payload you want to deliver, and that would include
everything which is not fuel. You might find the an
earlier answer
illuminating.

QUESTION:
How exactly do electrons in electric discharges generate radio waves?
Whenever someone uses an AM radio while there are electric sparks like lightning strikes nearby the radio waves they emit can be detected.

ANSWER:
Any time an electric charge is accelerated it emits
electromagnetic radiation. There are many electrons in the
spark which accelerate under the influence of the electric
fields they experience.

QUESTION:
I just read about the kilogram being defined by measuring the current of an elector-magnet to balance a scale. How can various electromagnets have exactly the same force to current ratio? Would variations in the purity of the conductor, variations in the exact coil shape, etc. cause variations in this ratio? What am I missing about this new designation?

ANSWER:
You cannot use any old electromagnet. There is an
exquisitely accurate balance called the
Kibble balance which is used. Also see
Wikipedia for more detail

QUESTION:
Is the following statement correct?
Net radiated energy is made up of emitted and absorbed energy.

I stated that it was incorrect based on the fact that
"made up of" usually means putting two or more things together (aka addition). Net radiated energy is the difference between emitted energy and absorbed energy. I have yet to see my exam paper but I believe that the mark I lost was due to this question on basis that “made up of” means that it is a result of a mathematical relation between the two values and not necessarily addition (in this case subtraction). What is the correct answer in this case and why?

ANSWER:
I do not believe that there is a hard and fast definition of
what "net radiated energy" is. It is a matter of semantics.
However, radiated energy usually means energy emitted and
absorbed energy means energy absorbed; therefore, I would
say that this is an incorrect statement because net radiated
energy would mean the sum of all radiated energy, absorbed
energy having nothing to do with it. Your reason is quite
shaky because "made up of" is even more semantically
ambiguous than "net radiated"; it would simply imply that
"net" means total energy flux.

QUESTION:
How is momentum conserved for single or double slit diffraction? in other words:I have a photon gun that I can dial the intensity down to fire one photon at a time. I fire my photon gun, the gun recoils in the -X direction, the photon flies off in the +X direction. The photon then passes thru two closely spaced slits (double slit experiment)instead of appearing on the screen directly behind the slits, the photon ends up to the side (after many, many photons the interference pattern appears) however, for the single event of a single photon, how is momentum conserved? dos the single event photon exchange momentum with the walls of the slit? Since the photon ended up to the side, it was no longer travelling in the purely X direction, therefore shouldn't the photon have exchange d momentum with the slit somehow? If the photon interacts with the slit, that counts as “which way” information, and the interference pattern vanishes. If the interference pattern remains, then how do we account for the photons change in momentum? This is for a single event, one photon, not expectation value of many, many photons.

ANSWER:
You are forgetting that the photon goes through both slits,
is being forced (if you like) to assume its wave-like
identity. It behaves like a wave until something makes it
act like a particle, so it is a wave spreading out in both
directions, so its net momentum in the direction parallel to
the screen is zero. When the screen is encountered it is a
"measurement" which collapses the wave function at some
point.

QUESTION:
I can imagine 2 similar objects rotating around a sphere at the same velocity but at different orbits. One object would rotate around the sphere faster than the other, right? What is puzzling to me is that if I think of both objects going now in a straight line one could not say that one object was going faster than the other. Can someone help me through this? Is there a subtlety here?

ANSWER:
You want to say same speed, not same velocity, but I get the
idea. Say one orbit is twice the radius as the other. Then
the smaller orbit has a circumference half that of the
larger. So each time the object in the larger orbit goes
around once, the one in the smaller orbit goes around twice.
But they still have the same speed. But the angular
speed of the object in the smaller orbit is twice that
of the object in the larger orbit even though their linear
speeds are the same. You cannot do this with planets or
moons, though, because the speed of a satellite in a
circular orbit depends on the radius.

QUESTION:
If space is curved, doesn't it need another dimension to curve into? If a 2d object is curved it curves into the 3rd dimension. So if 3d space is also curved it must curve into 4d.

ANSWER:
This
is really mathematics, not physics; I am not a
mathematician. But the definition of an N-dimensional space
is, I believe, that N quantities are required to specify the
location of a point in that space. Curvature of a space is
defined as a space in which the geometry is not Euclidean.
The physical existence of an N-dimensional space does not
imply the physical existence of an N+1 dimensional space.
You cannot make a generalization that curvature is
necessarily "curving into" a higher dimension. For
example, a planar sheet can be distorted without leaving the
2D. You also might "curve into" a higher dimension than N+1;
for example, a coil spring is a one dimensional space which
exists in a 3D space. You are trying to make generalizations
based on the only spatial dimensions which we know to exist;
but higher dimensions are defined by certain rules which in
no way prove that they exist physically.

While
researching this question I came upon a very interesting
little
essay about general relativity, curvature of
spacetime, gravity. Many nonscientists (and scientists as
well) are familiar with and accept as gospel that general
relativity demonstrates that the distortion of spacetime by
the presence of mass is what gravity is; the "cartoon" which
illustrates this is the "bowling ball on a trampoline"
example in which a mass causes the distortion of the
two-dimensional space defined by the trampoline surface. The
implication is that gravity is just geometry. On the other
hand, many theorists are struggling to find a successful
theory of quantum gravity and to do that one must treat
gravity as a force. The take-away from this essay, though,
is that curved spacetime is not a unique representation of
what the mathematics which are the framework of the very
successful theory of general relativity mean. The author,
Rodney Brooks, notes that "…you can view gravity as a force field that, like the other force fields in QFT
[quantum field theory], exists in three-dimensional space and evolves in time according to the field equations."
He also goes so far as to say "…shame on those who try to foist and force the four-dimensional concept onto the public as essential to the understanding of relativity theory."

QUESTION:
A rear wheel drive car accelerates quickly from rest. The driver observes that the car noses up. Why does it do that? Would a front wheel drive car behave differently?

ANSWER:
The car with rear-wheel drive has an acceleration
a and a mass M . The forces on
the car are its own weight W (W=Mg )
acting at the center of mass a distance d from the
real axle; the normal and frictional forces on the rear
wheels, N _{1} and
f _{1} ; and the normal and
frictional forces on the front wheels, N _{2}
and f _{2} ; the radius of
the wheels is R and the distance between front and
rear axels is D . I will ignore the rotational
motion of the wheels, that is, I will approximate their
masses as zero. If the front wheels are not lifting up from
the ground, the sum of torques about the rear axel is (f _{1} -f _{2} )R +N _{2} D -Mgd= 0;
Newton's first law for the horizontal and vertical
directions are (f _{1} -f _{2} )=Ma
and N _{1} +N _{2} -Mg =0.
From these equations it is easy to show that N _{2} =M (gd-aR )/D .
This is interesting because it shows that N _{2} =0
if a=gd /R ; for larger accelerations, the
front wheels will lift off the ground. (Note that as N _{2}
approaches zero, f _{2} also approaches
zero.) I have ignored the suspension of a real car, so this
lift will not be abrupt; it will happen gradually as the
front springs decompress and the rear springs compress; so
overall the nose goes up but the rear end goes down.

If
it is a front-wheel drive, the directions of the frictional
forces reverse. The corresponding torque equation (this time
about the front axle) is
(f _{2} -f _{1} )R-N _{1} D+Mg (D-d )= 0;
Newton's first law for the horizontal and vertical
directions are (f _{2} -f _{1} )=Ma
and N _{1} +N _{2} -Mg =0.
From these equations it is easy to show that N _{1} =M (aR+g (D-d ))/D .
As a increases in magnitude, N _{1}
gets bigger and bigger until N _{1} =Mg
which necessarily means that N _{2} becomes
zero. It is easy to show that this happens when a=gd /R ,
exactly the same result as for the rear wheel drive.

Finally,
it might appear that there is no difference between front-
and rear-wheel drive. That is not the case in the real world
at all. Keep in mind that the frictional forces are static
friction and the larger the corresponding normal force is,
the more friction you can get. Since in the rear-wheel
situation the normal force gets bigger, the amount of
friction you can get without the wheels slipping can keep
increasing so that eventually (with a powerful enough
engine) you could lift the whole front end far off the
ground. However, when you have front-wheel drive, the normal
force on the front wheels gets smaller as you increase a ,
so most likely the wheels will slip well before N _{2} =0
when f _{2} must be zero. For very large
accelerations, like required for drag racers, rear-wheel
drive is imperative.

QUESTION:
Gravity is what makes objects orbit around other objects,and gravity is a reflection of an object's mass.So why doesn' the mass of the objects appear in kepler's third law??

ANSWER:
For the same reason why all objects have the same
acceleration in the gravitational field, i.e. all
objects near the surface of the earth have an acceleration
of 9.8 m/s^{2} . The acceleration (which is what
describes the orbit) is F /m but the force
is MmG /r ^{2} , so the acceleration
is MG /r ^{2} .

QUESTION:
If kinematics can give you data on a dropped object, would it be valid to use kinematics to calculate the acceleration experienced by the object upon impact with the ground? (the time from when it hits the ground to when it actually stops)

ANSWER:
When the object hits the ground until its velocity reaches
zero is some time, call it Δt . Its
momentum drops from mv (where m is the
mass and v is the speed when it hits) to zero
because an impulse I was delivered to it by the
ground. The impulse may be written as I=F _{avg} Δt .
where F _{avg} is the average
force. Therefore you can find the average acceleration a _{avg}
during the collision, but not the acceleration as a function
of time: a _{avg} =F _{avg} /m =I /(m Δt )=v /Δt.
So you cannot get detailed kinematics about the
collision without measuring how the speed decreases with
time.

QUESTION:
We know that air density decreases as you ascend into space, but what about looking out to the horizon? Is there a atmospheric lensing that occurs? Is there a formula for its calculation?

ANSWER:
Certainly there is refraction due to this density profile of
the air. There is no simple formula because the actual
density at any point is dependent on the pressure and
temperature of the air as well as altitude, and these vary
with time. One could calculate based on average density
profiles, but I would suppose it is a numerical calculation,
not an analytic formula. There is a nice description of this
phenomenon
here .

QUESTION:
I've been interested in physics for years but have never managed to get my head around the concept of space-time. I understand time and i understand space but somehow i could never get this. However today i think it finally clicked when i wasn't even trying. Can you tell me if my understanding of it is correct or not?
I thought of space-time as like if i were on a roller-coaster or train tracks. When i am on the long flats, i travel slower, and when i am on the declines, i travel faster. Is space-time like this analogy? In that, when you are on Earth for example, you experience time moving faster than if you were floating in space millions of miles away. This is because the presence of Earth "warps" space time, much like a bowling ball on a trampoline. This "dent" in space-time causes a gravitational pull, pulling time into it, much like if i were rolling down a hill.
I know that might sound very convoluted but it's the idea i have in my head and it has bothered me for YEARS, but i think i am on to an understanding here.

ANSWER:
Sometimes we make things more difficult than they need to
be. The main takeaway from the theory of relativity is that
time and space are not separate and universal things. In
Newtonian times, it was assumed that any clock in the
universe would run at exactly the same rate; now we know
that if a clock moves with respect to us that it runs more
slowly than ours —doesn't appear to run slower,
it runs slower. Similarly, space depends on the observer; a
meter stick moving by you is actually shorter than 1 m. What
these facts about space and time tell us is that time and
they are not different things, they are really entangled
with each other. You cannot talk about one without talking
about the other. They are two pieces of the same thing. So
the term space-time is simply a way of acknowledging that. I
do not find your roller coaster analogy very helpful. The
bowling ball on a trampoline is a classic but keep in mind
that it is really a cartoon and not rigorous. Generally
trying to picture something in four dimensions futile!

QUESTION:
What does a wind speed (a gust is okay)in miles per hour, need to be to launch a baseball at a speed of 80 miles per hour into an object like a window.
Given:
Ball is 60 feet 6 inches from the window that it hits traveling at 80 mph Baseball is 9.25 inches in circumference and 2.94 inches in diameter and weighs 5.25 ounces
My name is Karin and I'm testing to see if I use a baseball pitching machine from 60 feet 6 inches from a window protected with a protective panel, and set the machine to pitch at 80 mph, what would the hurricane wind speed need to be to get that baseball to fly into the window at 80 mph. I hope this is clear enough.

ANSWER:
If I understand your question, you are essentially asking
what wind speed would be required for a baseball to reach a
speed, starting from rest, of 80 mph in a distance of 60.5
ft due to the force of the wind on the ball. This is a
problem which is very tedious to do in full analytic detail
and not very illuminating. Also, all problems involving air
drag are approximate, so it would be appropriate for me to
find an approximate way to solve this problem. I am
going to work in SI units so I will rephrase your question:
what wind speed would be required for a baseball to reach a
speed, starting from rest, of 35.8 m/s in a distance of 18.4
m due to the force of the wind on the ball? In SI units you
may approximate the force at sea level which the wind exerts
on the ball as F= ¼A (u-v )^{2 }
where u is the speed of the wind, v is the
speed of the ball, and A is the cross sectional
area of the ball. Applying Newton's second law, ¼A (u-v )^{2} =ma =m dv /dt
. This differential equation may be shown to have the
solution t=cv /[u (u-v )] where
c =4m /A and t is the time it takes the
ball to reach the speed v if the wind speed is
u . I find that c =137 s^{2} /m. Note
that after a very long time the speed will be constant at
the wind speed; we are only interested in the time it takes
to get to 35.8 m/s (80 mph). I have plotted t for
three values of u , 70, 80, and 90 m/s.

Now comes the hard part. We are interested in finding the
value of u for which v will be 35.8 m/s when the ball has
gone a distance x =18.4 m. This means that we need
to write the equation for t as
t=c {dx /dt }/[u (u- {dx /dt })]
and solve that differential equation for t as a
function of x . That is actually doable, but the
solution is so messy that I looked for a simpler way to do it.
This took some trial and error, but I got a pretty good
estimate, I believe. Look at the black line in the graph; it
is not a straight line, but it is pretty close, so I am going
to approximate the acceleration as being constant over this
time interval of 1 s, a ≈(35.8 m/s)/(1 s)=35.8
m/s^{2} . Now, an object with constant acceleration
starting from rest goes a distance d = ½at ^{2}
in a time t , so if u =90 m/s=201 mph the ball will have a
speed of 35.8 m/s when d =35.8/2=17.9 m=59 ft. As
far as I am concerned, I have adequately solved this problem—it
takes a wind of approximately 200 mph to accelerate a
baseball from rest to the speed of 80 mph in a distance of
about 60 ft.

QUESTION:
Why is light speed the fastest speed and why would people see a train going backwards if it is traveling at a velocity greater than light speed?

ANSWER:
I will not answer the second question because the site
groundrules state
that I will not answer questions of this sort. Your first
question is equivalent to the question "why can nothing with
mass travel as fast or faster than light?" which is included
in the FAQ page.