In Round 213, the 'Lenny' authors wrote:
Neopia is a strange little planet. Its gravitational acceleration at its surface is exactly 10.0 metres per second per second, and its diameter is exactly 2100 kilometres.
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:
. . . . .G = 6.673 × 10^(-11) N m^2 / kg^2 = ( 6.673 / 10^11 ) m^3 / kg s^2
. . . . .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.)
In Round 213, the 'Lenny' authors wrote:
Also, a completely unrelated fact, Skeiths are able to consume about 0.4 kg of pretty much anything they want to eat, every minute, nonstop.
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? Please round up to the nearest year
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)
The mass consumed in one Neopian year is the product of one million, "0.4", and the number of minutes in one Neopian year. Dividing the mass by the mass-per-year consumption rate should give you the number of years.
I also get a three-digit number.
Eliz.
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