Monday, June 21, 2010

Lost Ring

I got a question from my sister this past week:

“I recently lost my engagement ring. (It's not the first time I lost it, but I think this time it's really gone. Sigh.). But, I’m not alone—as I just learned that Jupiter lost a band recently… How many times bigger is Jupiter's band than my engagement band?

Jupiter’s radius is about 71,000 km.1  My sister’s ring’s radius was likely about 1.0 cm.  This would make Jupiter’s ring (or band) about 7 billion times bigger if we’re talking about its linear dimension.  However, the word “bigger” has several meanings.  If we’re talking about how many times larger the volume of the ring is, then we need to cube the linear dimension.  In this case, Jupiter’s ring is about 3.6×1029 times bigger.  As antoniseb has pointed out, there’s also the issue of density.  If we want to talk about how much more massive the ring is we need to know the ratio of Jupiter’s ring’s density to the density of gold.  The density of gold is 19.3 g/cm3.  Jupiter is a gas giant with a low average density of about 1.33 g/cm3.  The upper atmosphere where the ring is located is probably much less dense.  Assuming it has the same density as air at the surface Earth (~0.0012 g/cm3), Jupiter’s band will be about 2.2×1025 times bigger. 

Sorry about the ring.  If you want to make sure you don’t lose your next ring, you can always do what I did.2

[1]  Since the ring is not at the equator, it will actually be a little smaller than this.
[2]  See picture above.

Sunday, June 20, 2010

Happy Father’s Day

Happy Father’s Day to all the dads out there.  My Dad suggested a problem the other day: How long would it take to wear down a tire by 0.75 inches if you’re driving a 55 mph?

I’ll assume a tire tread starts about an inch thick.  By the time 0.75 inches have worn off, it’s time to get them replaced.  Tires presumably wear down because of friction between the rubber and the road.  When tires rotate, this frictional force from the ground pushes the car forward while an equal and opposite force pushes back on the road.  The faster you drive, the greater the frictional force is and this presumably causes more wear and tear.  However, if you’re driving at a constant speed, you probably don’t burn as much rubber as you would by slamming on the breaks every five seconds.  It’s not immediately clear which of these effects causes greater erosion, but you’re generally advised to change tires roughly every 50,000 miles1.  If that’s the case for driving at 55 mph, then it’s easy to compute how long it will take before your tires have worn down by 0.75 inches:

time = (distance) / (speed)
= (50,000 miles) / (55 mph)
= 38 days.

It should take a little more than a month to wear off 0.75 inches of your tires by driving at 55 mph.

Thanks, Dad.  Happy Father’s Day!

[1] You can also test it by sticking a penny head first in the tread.  If you can see Lincoln’s head, it’s time to change tires.

Monday, June 14, 2010

A New Magnetic Train

As I was driving the other day, I noticed the little dashboard compass needle spinning whenever the car turned. Perhaps there’s a more energy efficient way to travel with a compass. How fast could you go from Brooklyn to Manhattan by sitting on the point of a giant compass needle?

***********WARNING: Math Ahead***********

Imagine a giant compass needle whose pivot is centered halfway between Manhattan and Brooklyn. Initially the needle points south so that its tip lies in Brooklyn. The needle would have to be about 8.0 km (~5.0 mi.) long. It would need to carry people, so it might look somewhat like a subway car, making it at least 3 m (~9.8 ft) in diameter.1

The torque, or rotational force, on a magnet is given by the equation,

torque = (magnetic moment) · (magnetic field) · sin(angle).

The angle in question is the angle between the magnetic moment and the magnetic field. This angle would change as the needle turns. On average, the sine of the angle would be about 0.63 as the needle passes from Brooklyn to Manhattan. The magnetic field would be the Earth’s magnetic field, which has a magnitude of roughly 5.0×10-5 T.2 The needle could be a large permanent magnet (like the ones on refrigerators), but these are not as strong as electromagnets and aren’t able to flip the north and south poles like an electromagnet can3. An electromagnet can also be turned on and off, so you could control when our magnetic transporter was about to leave the station. We can make an electromagnet by wrapping tons of wire around the needle in a helical fashion and then applying a current through it. The magnetic moment for a cylindrical coil is given by,

magnetic moment = (number of coil turns) · (current) · (area across the cylinder).

From the diameter above, the area would be about 10 m2. If the wire is 1.0 mm thick and there are 10 rows wrapped around each other, then the 8.0 km needle would have about 8 million turns in it. I’ll assume the current in the coils is 1.0 A.4

The time it takes the needle to rotate to Manhattan is given by the equation,

time = [2 π · (moment of inertia) / torque]1/2.

The moment of inertia of the needle can be computed using a formula found here. Plugging in the values for our needle, we can calculate the moment of inertia to be about 3.4×1012 kg·m2.5

By combining equations and plugging all the numbers in from about, we find that it would take about 1.1 days to get from Brooklyn to Manhattan traveling by giant compass needle.


If you made a giant compass needle to harness the Earth’s magnetic field and tried to use this to travel, it would take you a day to get from Brooklyn to Manhattan. Since the subway costs about $2.00 and takes less than an hour, building a giant compass is probably not an efficient use of time or money.

There’s another interesting dilemma in this problem. Nothing in life is free, so you’d suspect that even our poorly functioning magnetic transporter must also sap energy from somewhere. Much like oil, it’s possible the source of Earth’s magnetic field would also eventually run out if we made lots of giant magnetic transporters. How long do you think it would take to sap away the Earth’s magnetic field?

[1] It would also be highly magnetic, so you couldn’t bring your credit cards.
[2] The “T” stands for “Tesla,” a unit of magnetic field.
[3] If you couldn’t flip the poles you’d never be able to get back to Brooklyn in the same way you left.
[4] I don’t have a good reason for assuming this value of current other than the fact that one of my intro physics books uses this value in a problem involving a copper wire. It would be interesting to see how this problem would work with superconducting wires, but you’d have to cool them down to very low temperatures, so the passengers might become a wee bit cranky and hypothermic.
[5] You can estimate this by assuming the needle is a 1.0 ft thick cylindrical shell made of iron.

Tuesday, June 1, 2010

Special Guest: Maryellen Hooper

Today’s question comes from the very funny comedian Maryellen Hooper.  Ms. Hooper has appeared on The Tonight Show with Jay Leno and in her own Comedy Central special, Lounge Lizards.  In 1998, she was awarded “Female Comedian of the Year” at the American Comedy Awards.  She writes,

“I've decided to ask a question on behalf of my son, Nate and his best friend, Noah. They're 5 & 6 years old…How many Legos would it take to build a ladder to the Moon?”

***********WARNING: Math Ahead***********

I’ll assume we’re using the standard 8-peg Lego brick.   A single Lego is 9.6 mm tall, 32 mm long, and 16 mm wide1.  By multiplying these together, we can calculate the Lego brick’s volume to be about 4.9 cm3.  To find the total number of Legos needed, we have to estimate the dimensions of the ladder.

The Moon is about 380,000 km away from the Earth.  I’ll assume the sides and rungs of the ladder are both 6.0 cm wide and 6.0 cm thick.  I’ll also assume the rungs are 60 cm (~2.0 ft) long and that adjacent rungs are separated by 30 cm (~ 1.0 ft).  With this rung separation, there will be about 1.3×109 rungs.  From this we can compute the total volume of the ladder.

volume = (# of sides) · (volume of the sides) + (# of rungs) · (volume of the rungs)
= 2 · (6.0 cm · 6.0 cm · 380,000 km) + 1.3×109 · (6.0 cm · 6.0 cm · 60 cm)
= 5.5×109 m3.

To find the number of Legos we’d need to build the ladder, we just need to divide the volume of a single Lego into volume of the ladder,

# of Legos = (ladder volume) / (Lego volume)
= (5.5×109 m3) / (4.9 cm3 per Lego)
= 1.1×1015 Legos


A Lego ladder to the moon would require 1.1 quadrillion Legos!2  Using prices from Lego’s website, this would cost about $51 trillion dollars.  At his richest, Bill Gates was still 500 times too poor to afford this.  Thanks for the great question, Maryellen!

You can find Maryellen’s tour dates and clips of her doing standup on her website.  You can also buy her comedy CDs Fixer Upper and Dignity Under Duress.

[1] Apparently, the good people at Lego are too good for the metric system, so they actually made up a “Lego Unit”.  One Lego Unit is equal to 1.6 mm.
[2] Yes, “quadrillion” is a real number.