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:
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.