A Neopian wrote:I have no idea what Im supposed to put down, can someone help please. Ive answered 10 lenny conundrums correctly and am yet to get the avatar o_0 lol Thanks!
Plenty of "help" has already
been provided, making quite clear (with links, formulas, etc) "what [one is] supposed to do" -- if one cares to invest some effort of one's own.
However, for those for whom "help" means "give me the answer, and do it quick enough that i[sic] get a prize this time!!!!!!!", the following may be "helpful":
The 'Lenny' authors, in Round 271, wrote: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.
Let a planetary body have mass "m" (in kilograms) and radius "r" (in meters), and let the body have an object in geostationary orbit some distance "d" above the surface. Suppose the body is in our universe, thus sharing the universal gravitational constant "G":
. . . . .G = (6.67428 * 10^(-11)) m^3 / kg-s^2
(To clarify, the units on the gravitational contant are meters-cubed, divided by kilograms and by seconds-squared.)
Let "R" be the orbital radius of the object, so R = r + d. Note that the time "t" required for one orbit is (24 hours)(60 min/hr)(60 s/min) = 86 400 seconds.
Using radians for the angle measure, one revolution (that is, one orbit) is 2pi radians. The object's angular velocity "w" is then:
. . . . .w = (2pi radians) / (86 400 s) = (pi/43 200) rad/s
(Leave this value in exact form, for the time being.)
The planetary body has a radius of r = 800 km = 800 000 m, giving a volume V = (4/3)(pi)(r^3) of:
. . . . .V = (4/3)(pi)(800 000)^3 m^3
. . . . .. .= ((2.048 * 10^18)(pi/3)) m^3
(Leave this in "exact" form for now; do not approximate in your calculator.)
The mass of any body is found by multiplying its density (its mass-per-volume constant) by its volume. In our case, the planetary body then has a mass m of:
. . . . .m = (5 200 kg/m^3)(((2.048 * 10^18)(pi/3)) m^3)
. . . . . . .= (1.06496 * 10^22)(pi/3) kg
(Leave this in exact form for now.)
The orbital radius R of the object is related to the mass and radius of the body by:
. . . . .R = cbrt[(Gm)/w^2]
...where "cbrt[value]" indicates "the cube root of [the argument value]". In this case, the mass is found
. . . . .R = cbrt[(((6.67428 * 10^(-11)) m^3 / kg-s^2)((1.06496 * 10^22)(pi/3) kg)/((pi/43 200) rad/s)^2)]
. . . .. . .= cbrt[((6.67428 * 1.06496 * 10^19)/(pi)) m^3]
. . .. . . .= cbrt[2.26249613255 * 10^19 m^3] (approx.)
. .. . . . .= 2 828 324.801 36... m (approx)
Keeping in mind that this is R, where R = r + d, that r = 800 000 m, we get a value for d of:
. . . . .d = R - r = 2 828 324.801 36 - 800 000 = 2 028 324.801 36... m (approx)
...or about 2 028 325 meters. Converting to kilometers, we get a value of:
. . . . .2 028.324 801 km (approx)
Rounding as directed, we get a value of 2028 kilometers.
Please check my work (or, if you just blindly sumbit "2028", don't complain later if a typo somewhere is found to mean that the final value above is incorrect).
Eliz.
thanks so much! 
