Thursday, July 17, 2008

Round 271

Date Released: July 17th, 2008.
Date Rewarded: July 24th, 2008.

Assume the world of Neopia is a spherical planet with a diameter of 1600 km with a uniform density of 5200 kilograms per cubic metre. If the moon of Kreludor is in a geostationary orbit around Neopia, and assuming a Neopian sidereal day is exactly 24 hours, how far is that geostationary orbit above the surface of Neopia?

Please submit your answer in kilometres, and round to the nearest kilometre. Submit only a number; any additional information may disqualify your entry.


Kreludan Defender Plushie

Click to show/hide solution

This puzzle is about Geostationary orbit. You can find a full explanation about this topic in [1]. The main equation we need to focus on is:
r3 = G*M / ω2

where r is the distance between the Neopia's center of mass to the Kreludor moon's geostationary orbit, G = 6.674 28 * 10-11 m3/(kg*s2) is the gravitational constant, M is the mass of the Neopia and ω is the angular speed of the moon.

The mass of the Neopia planet can be calculated by the following equation:
M = ρ * V = ρ * 4/3 * pi * R3

where ρ = 5200 kg/m3 is the density of Neopia, V is the volume of the planet, R = d/2 = 800 km = 800,000 m is the radius of the planet.

The angular speed of the Kredulor moon can be calculated by the following equation:
ω = (2 * pi) / 1 sidereal day = (2 * pi) / (24 * 60 * 60) = pi/43200 rad/s

Therefore, we have:
r3 = G * ρ * 4/3 * pi * R3 / (pi/43200)2 = (86400)2 * G * ρ * R3 / (3*pi)

Replace numbers into the above equation, then take cubic root, we can get r = 5201692.7355. The answer is the distance from the Neopia planet surface to the orbit. Therefore, it should be:
distance = (r - R)/1000 = 4401.69

We use 1000 factor in the above equation because we want to get in kilometers. Round the distance and we get 4402 as the answer.

Results: 526 people guessed the correct answer earning themselves 3803 NP each.

[1] Geostationary orbit


DarkAngel said...

this is what i did but i saw so many number i was like no

philip said...

so..... whats the answer? 1000km?

Freaky Gerbil said...

I do love pages like this that spoil the chances of anyone getting the avatar through genuine hard work.

Ron said...

Some kids would like to have the trophy and are not so good in solving lenny. So for them I'm glad there is a help site like this.

shayybevsx3 said...

i'm really confused.
i read and re-read that 7 times and i still don't get it.

LC Solver said...

I do love pages like this that spoil the chances of anyone getting the avatar through genuine hard work.
Thanks :)

thejunkfaeries said...

Freaky Gerbil.... and you're here because...

thank you for the help lc solver :D

beniix3gj said...

What is the answer exactly? lol
i only got the math not the answer

Rodrunner said...

is your answer a 4-digit number?

lisah746 said...

I got a 4 digit number as well.

LC, thanks so much for helping us with the steps to solve these. I like that you give us the steps make make it so that we still have to figure some things out for ourselves.

LC Solver said...

yes it is a 4 digit #

k2thee98 said...

What number is r? Sorry I'm not the best at math and I don't have 3 hours to waist on confusing myself even more.

xxxanglekittyxxx said...

huh? cant u just tell the #.. i get lost i really did try but i dont no...

skyler said...

oh shut the hell up freakygerbil, try to stop being such an insufferable asshat.


Anyway, LC, I appreciate you helping out with us solving this. I'm not very good at math related Lenny Conundrums.

savannahjones21 said...

im so confused i havent done an equation with so many unknown variables in like 10 years. i figured out the volume and speed but im totally lost on everything else.

Michael said...

Humm... I think I can do it, but where does the variable "n" come from?

LC Solver said...

It isn't n, it's pi. I guess blogger doesn't display well the pi character. I'll edit in a min.

Michael said...

Lol, maybe it's Firefox. Anyway, thanks LC. Smarty pants engineer.

Kevin Dong said... what is the pattan in the link above because i cannot solve it

Destiny said...

what is 's' in the G= equation? i dont quite understand the gravitational constant.

LC Solver said...

s is the second unit, not a variable.

Kevin, it's an interesting puzzle. I wish I had time to solve it (which I don't have much now)

Destiny said...

ohhhh....! thank you! :]

Jennifer said...

so is little m the unit meters or is it supposed to be a number? i cant find what number to plug in for little m. I know big M's formula.

Jennifer said...

i dont think i did it right. my answer is a negative 3 digit number.


r^3 = G*M / w^2

0.000000000667428 m^3/(kg*s^2) * 5200 kg/m^3 * 4/3 * 3.14 * 800,000 m^3 / 3.14/43200 rad/s

0.0000034706256 * 3349333.3333333333333333333333333 / 3.14/43200

1.1624282009599999999999999999999 / 7.2685185185185185185185185185185 = 0.15992642764799999999999999999987

1.16242820096 / 3.14/43200 = 8.5694459259259259259259259259259~

1.16242820096 / 7.2685185185185185185185185185185~ = 0.15992642764800000000000000000001

r= 6.2528752424273184371618344241282

distance = (r - R)/1000

6.2528752424273184371618344241282 - 800,000 m / 1000 = -799.99374712475757268156283816558

Matt said...

Since this is after the fact: 4402; and I didn't find this site until AFTER I answered. The wiki page shows how they calculated the answer for earth, just plug in your numbers and solve. (Make sure your answer is in km, not m.)