# Challenging Physics Problems

This is a sample of the problems from the book “200 Puzzling Physics Problems” published by Cambridge University Press:

http://catdir.loc.gov/catdir/sam ples/cam034/00053005.pdf

This above resource contains 89 of the problems, but no solutions.

The problems here are simple to state, but seem to get at profound mathematical and physical ideas. For example, if the Earth and Sun were scaled down as to be 1 meter apart on average, how long would a year be?

I can see a lot of productive student struggle coming from this book!

## 12 Comments

## Chelsea M. · January 24, 2013 at 11:09 am

This really is a great problem: If the Earth and Sun were scaled down as to be 1 meter apart on average, how long would a year be?

My initial thought would be that since the gravitational force between the earth and sun must remain constant despite the changes in size and distance, I would use the equation F=G(m1*m2)/R^2 with the original masses and 1 AU as the radius to calculate the force between them. Then this could be set this equal to G(m1*m2)/R^2 with our new radius of 1m such that the force is just equal to G(m1*m2). Since G is a constant we then solve for (m1*m2). Here’s where I get stuck. I would assume the next logical step would be to apply the extension of Kepler’s 3rd law which states P^2= 4(pi^2)(a^3)/G(m1+m2) where P is the period and a is the semi-major axis. However since I previously only solved for the product of the masses, rather than the sum which is needed here, I’m not sure how to proceed. I believe the earth’s mass would be negligible in this case since the mass of the sun is so much greater than that of the earth, yet with only their product, I’ve hit a roadblock. Am I missing something obvious here? I haven’t had my coffee yet…

## Riley · January 27, 2013 at 7:51 am

I don’t see why the gravitational force must remain constant. I’m convinced it will decrease dramatically, since mass changes by the cube of the scale factor (1m/1AU), but the effect on gravity from distance increases only by the square of the scale factor. So the force would change by a factor of 1m/1AU.

I’m not sure we need to calculate the force though. Looking at the simplified version of Kepler’s third law, since Earth is so small compared to the Sun, the fraction inside the radical seems to imply that the changes balance out.

(a1)^3 = (a0 * 1m/1AU)^3 = (a0)^3 * (1m/1AU)^3

M1 = M0 * (1m/1AU)^3

[a1 is the orbital period after scaling, and a0 is the orbital period before scaling. Likewise M1 and M0 for mass of the Sun.]

When we substitute these into the Kepler’s equation, the factor (1m/1AU)^3 gets cancelled out in the numerator and denominator. This happens with the full version of Kepler’s Third Law as well after you factor the denominator.

My conclusion: Using Kepler’s Third Law, orbital period is independent of scale.

## Riley · January 27, 2013 at 11:06 am

Derf. I meant that the force changes by a factor of (1m/1AU)^4, since the numerator is a product of masses, not a sum or difference.

## Jacob · January 31, 2013 at 10:22 pm

Yeah, but to get the acceleration we divide by the mass of the earth, so that just cancels out. It takes the same amount of time.

## Riley · February 1, 2013 at 5:47 am

We are in agreement. I was correcting only my claim about change of force, not my conclusion, since I did not use force to find my answer.

## Chelsea M. · January 24, 2013 at 2:05 pm

On another note, if you only used the basic form of Kepler’s third law, P^2= a^3, in which the period P is in units of sidereal years and the semi-major axis a is in astronomical units, then you would get an orbital period of 1.72829 × 10^-17 years. This is roughly 5.4503×10^-10 seconds, or 0.54503 nanoseconds. Not sure if this equation is applicable here, but either way that’s a really short year!

## MrHonner · January 25, 2013 at 6:37 pm

Chelsea-

I agree that this is a great problem.

I lack the physical intuition and knowledge of astronomy that you are bringing to the analysis. My initial thoughts were to assume a circular orbit and look for a “simple” solution.

I’m not there yet, but you’ve definitely made me rethink my make-a-wish approach!

## Bob Mrotek · January 25, 2013 at 8:58 pm

A year would still be a year but it would only be 3 hours and 59 minutes long…(I think).

## Bo · January 28, 2013 at 12:00 pm

Thank you for keep linking such resources. I myself am thinking and solving these little puzzles all the time, and just posted three little ones on my blog.

http://methmath.wordpress.com/2013/01/25/3-puzzles/

## Jacob · January 31, 2013 at 6:56 pm

I believe that a year would remain constant. The problem with Kepler’s law, is that it’s for comparing bodies orbiting the same star. We can’t however use the same constant of proportionality for earth and some distant planet. Here’s my reasoning:

The average acceleration of the earth is based on three things: a constant of gravity, the sun’s mass, and the mean distance to the sun (I’m approximating that the earth moves in a perfect circle.) Since all distances are being scaled down, including the size of the sun, the sun’s mass decreases by a cube of the factor, but since the force is inversely proportional to the square of the mean distance, the total acceleration decreases by a factor of 1 AU/1 m, the same amount that distance is being scaled down.

Manipulating the centripetal force equation, we have the period T, as

T=4Π2r/a,

where r is the mean distance, and a is the acceleration of earth. Since r and a are being factored down by the same amount the period of revolution, which we know as the time one year takes, will remain constant.

A very nice problem.

## Bob Mrotek · January 31, 2013 at 9:02 pm

Jacob,

When a spinning ice skater pulls in her arms what happens? Yes the year I suppose would remain constant in relation to the dimensions but a revolution of the earth around the sun would only take a few of our “present” Earth hours. What say you Mr. Honner?

## Jacob · January 31, 2013 at 9:22 pm

Bob, time wouldn’t change, only space.

The reason that an ice skater speeds up is because only the radius is changed. His or her angular momentum will remain constant, speeding him or her up. However, there are several other factors to consider in this case. By shrinking the system you actually change the force that’s present on the earth, changing its angular momentum. Remember, the force of gravity exerted by the sun on the earth will change, as both the sun and the earth now have less mass.