NeoPets Riddles (Lenny Conundrums) and Answers Here

I just don't know where to go without knowing how long a year is for Neopia. Could it be the same as Earth, because on Neopets, the calendar has the same number of days as us??? Maybe the answer lies in there

The formula which relates mass, radius, and gravitational acceleration is given as:

. . . . .a = (GM) / R^2

...where "a" is the acceleration, "G" is the universal gravitational constant, "M" is the mass, and "R" is the radius. We have the following known values:

. . . . .. . . . .R = D/2 = (2100 km) / 2 = 1050 km = 1 050 000 m

. . . . .a = 10 m / s^2

...where "N" is "Newtons", which is "force" expressed as kilogram-meters per second squared (similar to foot-pounds).

We know that the volume of the sphere is:

. . . . .V = (4/3)(pi)(R^3)

...and that "density" is "mass per unit volume". With the mass and the volume, we can find the density. Since we are given the consumption rate in terms of mass per unit time, we need the density. Then we can compute how much mass (that is, how many kilograms) would be in one cubic kilometer. (One cubic kilometer of packing peanuts would not have nearly the mass of one cubic kilometer of lead, for instance.)


If one works symbolically, simplifying as much as possible first, one can check that the units are correct, and perhaps avoid some typoes and round-off errors.

. . . . .formula: a = (GM) / R^2

Solving this formula for the mass in terms of the other variables, we get:

. . . . .mass (kg): (aR^2) / G = M

Since the units on G are N-m/kg^2 (that is, (kg-m/s^2)(m/kg^2) = m^3 / kg s^2), the units on "a" are m / s^2, and the units on R^2 are m^2, then the units on M are kilograms, as required. Continuing:

. . . . .density (kg / m^3): M / V = (a R^2) / (G (4/3) pi R^3) = (3 a) / (4 pi G R)

Then the mass of one cubic meter is given by the numerical value of the above, and, since 1000 m = 1 km so 1 km^3 = 1000^3 m^3 = 10^9 m^3, we have:

. . . . .mass (kg) of 1 km^3: (M / V) (10^9)

. . . . . . . . .= (3 a 10^9) / (4 pi G R)

Assuming a "standard" year of 365.25 days, we get:

. . . . .. . . . .[ (3)(10)(10^9) ] / [ (8.41536) (10^11) (pi) (6.673 / 10^11) (1.05) (10^6)]

. . . . . . . .= [ (3) (10^10) ] / [ (8.41536) (pi) (6.673) (1.05) (10^6) ]

. . . . . . . .= [ (3) (10^4) ] / [ (8.41536) (pi) (6.673) (1.05) ]

....or about 161.95727 years.

Of course, this depends upon my having used the correct formula, having used it correctly, and having manipulated the constants and the variables correctly.

Since my answer is about twice what the previous poster's was, one of us may be off by a factor of 2. That poster appears to have used the diameter -- twice the radius -- for the value of "R", which could be the source of the difference. Or we both may be entirely at sea...

As always, check my work!!

Eliz.

When i used anneska's calculations for consumption rate (23.46m^3/min) and the number she got for the amount of minutes it takes to eat 1km...
there are 1440 minutes in a day, thus 525600minutes in one year...
when you divide the number of minutes it takes to eat one km by the number of minutes in a year, i got 81.099 years.
If i round that, i get 81 years.
check my calculations please, and if there's something wrong with my reasoning, let me know.

Just to put the record strate I didn't do the calculation, someone on yahoo answer helped me... I don't even get it really!

Yip thanks I got it 81.322 thats
81 years!

81 or 161 and I can't tell which math is off.
It is basically a 50/50 guess.

It could be 162... I am not sure about the calculation at all!

Too late for me in any case, I submitted 81...if that's wrong, ah well, there's always next week!

I also submitted 81 o well if its wrong it wrong...

Do you think that gravity, density or rotation are important to do the calculation? Because lenny conundrum says:

Assuming that the density of the planet is uniform, and that orbiting bodies don't significantly affect the planet's gravity, how many years will it take one million Skeiths to consume one cubic kilometre of Neopia?

The facts are:

Skeith eats, but gravity is not important, they just eat 0.4 dm3 every minute, maybe they put that clue to distract us...

my point is, 1 millon skeiths eat 400000 dm3/0.0000004 km3 every minute

If one year has 525600 minutes

in one year skeith eats 0.21024 km3
2 years 0.42048 km3
3 years 0.63072 km3
4 years 0.84096 km3
5 years 1.0512 km3 - Correct answer 5 years

I'm sure that this is not the answer but could someone explain to me why this can't be correct?

the reason that can't be right is because they tell us that skeith's eat 0.4 KG, not KM

Your sound right too... I am so cunfused now.. I sent 81 so its too late anywhy!

Sahar thanks, now I get it aswell!

I guess that could be the right answer as well...

oh, I get it now. nevermind my previous post.

0.4kg equal to 0.4dm3

The volume of 0.4kg is 0.4dm3

On what basis do you conclude that 0.4 kilograms of this planet's mass has a volume of 0.4 cubic decimeters?


The density information is crucial, because there is otherwise no way to know how much mass (that is, how many kilograms) are contained within the volume of one cubic kilometer.

The fact that the density is uniform is a simplifying consideration. For instance, the density of Earth is not uniform. The molten-iron core is much more dense (7.87 grams per cubic centimeter) than is the silica-alumina-rich crust (silicon: 2.33 g / cm^3; aluminum: 2.7 g / cm^3). If the density of this planet were not uniform, we would have to account for where the consumption occurred.

The rotational rate is neither given nor needed, though one can do the computations by going that route.

The surface acceleration due to gravity is crucial, and none of the other computations could occur without it.

Note 1: I have no idea whether the days, hours, or years on Neopia are as on Earth. But if they differ, the only values that would change (in my computations above) would be the 24, the 60, and the 365.25. Using "h" as the number of hours in a Neopian day, "m" as the number of minutes per hour, and "d" and the number of days per year, and simplifying and plugging in, the formulation would be:

. . . . .(10^10) / (37.3688 pi m h d)

If m = 60, h = 24, or d = 365.25 are not valid for Neopia, then plug in the corrected value and compute the answer.

Note 2: The diameter, 2100 kilometers, is not the same as the radius. By definition, the radius is half of the diameter. Since some are doing computations which, somewhere in there, divide by R but use "2100", this would explain why those computations end up with only about half my value: They divided by an extra "2". (The rest of the difference can probably be laid to the accumulation of round-off errors.)

Eliz.

I actually found out the mass....then divided it...but i dont know if thats right.

-
I'm feeling dizzy, lot of numbers.

Thanks for the explanation!

My bases were:

1 kg of feather and 1 kg of iron have the same weight Maybe not density, but the weight is the same so, if they eat 0.4kg they eat 0.4 kg, no matter what material or what thing they are eating.

but the question is not how many kilograms would they eat...its in terms of kilometers. If you don't know how many kilograms are contained in said kilometer....well, it doesn't really matter how much one kilogram is here vs. neopia