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Fun With Fundamentals: Problem 196

July 1, 2000
Present values and interest do indeed determine future values, as this month’s problem by J. Mehta of Wilmington, Del., demonstrates

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Problem 196 — Present values and interest do indeed determine future values, as this month’s problem by J. Mehta of Wilmington, Del., demonstrates.

It is the year 2196, and in the kitchen of Phineas Gotrocks IX, an argument is brewing.

“I tell you, it’s a lovely little planet,” chirped Marcia Gotrocks. “Just perfect for our holiday!”

“It’s parsecs from civilization,” retorted Phineas. “I’d just as soon find a resort and not have to camp out in the portable house.”

“We’ll just send our inquiry satellite and find out, replied Marcia. She waved her hand against a panel set in the wall, and the kitchen table instantly became a computer console. “Let’s see, I never can remember how to program this thing.” Then she entered the name and location of the planet as well as data for the gravitational constant, (3.44 x 108 ft4lb-l - sec-4), and specific gravity, which was 10. After this was done she pushed the launch button.

“I’m not going to wait all day for this,” snapped Phineas, as he opened his news monitor. “I’m playing tennis on Jupiter at 1600 hours, and my gravity suit still has to go to the cleaners.”

“It’ll just take a few minutes for the satellite to complete an orbit and then relay the data,” placated Marcia. “I’ve set it for a circular orbit whose radius is equal to the planet’s diameter. We’ll know in an hour.”

“The last time you said that, it took two days to get any information back!”

How long it will take the Gotrocks’ satellite to complete an orbit?

Assume the planet is a homogeneous sphere of uniform density, and neglect all other gravitational attractions caused by adjacent planets or stars. (You may need an engineering reference book for some of the calculations!)

Send your answer to:

Fun With Fundamentals
1100 Superior Ave.
Cleveland, OH 44114-2543
Deadline is August 10. Good luck!

Technical consultant, Jack Couillard, Menasha, Wis.

Solution to last month’s problem 194— You know the limits if you answered 32.9 mph. Here’s DePouff’s technicality:


Ff = Frictional force between driver wheels and pavement, lb
= Weight of car, lb
= Mass of car, lb
g = Gravitational constant, 32.2 ft/sec2
= Coefficient of friction, given as 0.375
= Force of acceleration, lb a = Rate of acceleration, ft/sec2 T

he problem states that 40% of the car’s weight is supported by the driver wheels. Therefore,
= μN
= m(0.4W) = .375(.4W) = .15W

But w = mg
Therefore, Ff = .15mg
From Newton’s Laws, Fa = ma, and v = at

In order to accelerate, Ff > Fa .

Assume DePouff accelerated at his maximum possible rate. Then, Ff = Fa, and ma = 0.15mg

Dividing through by m, a = .15g = .15(32.2 ft/sec2)

Now v = at = (4.83 ft/sec2)(10 sec)

The speed DePouff attained after 10 sec was at most 48.3 fps

The judge voided the ticket and had a word with his son-in-law.

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