Questions & Answers

With the recent publication of PHYSICS IS... there are now three Ask the Physicist books! Click on the book images below for information on the content of the books and for information on ordering.

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/c², 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+mL²)ω where I=M(2L)²/12=ML²/3 is the moment of inertia of a thin rod about its center of mass. Conserving angular momentum I find ω[m+(M/3)]L²=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 radians/s=19.1 rpm. You can also calculate the energy before and after the collision: E_before=½mv²=8 J, E_after=½(M+m)u²+½([m+(M/3)]L²ω²=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 ML²/3+Mx²+m(L-x)², so the corrected value of the angular velocity is ω=mvL/[ML²/3+Mx²+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/c²=√[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/c²=√[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 exchanged 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, N₁ and f₁; and the normal and frictional forces on the front wheels, N₂ and f₂; 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 (f₁-f₂)R+N₂D-Mgd=0; Newton's first law for the horizontal and vertical directions are (f₁-f₂)=Ma and N₁+N_₂-Mg=0. From these equations it is easy to show that N₂=M(gd-aR)/D. This is interesting because it shows that N₂=0 if a=gd/R; for larger accelerations, the front wheels will lift off the ground. (Note that as N₂ approaches zero, f₂ 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 (f₂-f₁)R-N₁D+Mg(D-d)=0; Newton's first law for the horizontal and vertical directions are (f₂-f₁)=Ma and N₁+N_₂-Mg=0. From these equations it is easy to show that N₁=M(aR+g(D-d))/D. As a increases in magnitude, N₁ gets bigger and bigger until N₁=Mg which necessarily means that N₂ 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 N₂=0 when f₂ 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/s². The acceleration (which is what describes the orbit) is F/m but the force is MmG/r², so the acceleration is MG/r².

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_avgΔt. 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/(mΔt)=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=mdv/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 s²/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/s². Now, an object with constant acceleration starting from rest goes a distance d=½at² 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.