How many kilograms are in one cubic kilometre?

Question

I'm trying to solve a math riddle.

Oh, right. Well, here's the entire question.
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.

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.

Yes, don't mind the Skeiths and Neopia thing. It's a strange riddle with strange details, but I'm trying to figure it out. Please and thank you.

Answer 1

I haven't seen this problem since I was in high school back in the 1970s. I assume that you are taking physics. Let's think a bit. The gravitational force between the planet and an object is given by Newton's law of gravity.

F=G*mObject*mPlanet/R^2

Where G is the gravitational constant (=6.67*10^-11m^3/kg*sec^2)
R is the planet radius (R=2100/2 km)
mObject is the mass of an object on the planet surface
mPlanet is the mass of the planet

Using one of Newton's laws of motion, we can say that the acceleration due to gravity is

a=F/mObject=G*mPlanet/R^2

We can also say that density d = M/V, where V = 4/3 *pi*R^3

We can combine these relationships to show that

d=(3/4)*a/(G*pi*R)

Notice that we now know all the important factors. Substituting in the appropriate values, for your world, d = 34.1 gm/cm^3.

We can now solve for the mass of 1 km^3. Since M=d*V, we can compute the mass as 3.41*10^13 kg.

Now that we have the mass of the cubic km, we can proceed to determine how long it will take those hungry Skeiths to eat it all.

T=M/(N*R)

Where T is the consumption time (rounded up to the nearest year)
M is the mass of the cubic km (above)
N is the number of Skeiths (=10^6)
R is the consumption rate (=0.4 kg/min)

T=3.41*10^13 kg/(10^6*0.4*kg/min)
= 8.53*10^7 min
=163 years (actually closer to 162 years, but I rounded up as directed)

Answer 2

It depends on the density of the substance you are woking with

1 ml =1 cm^3

1000 liters = 1 m^3

1km^3 = 10^3*10^3 *10^3 = 10^9 m^3 = 10^12 liters

If d= 1, then 1 g = 1 ml and 1 kg= 1 l

So, you will have 10^12 kg

Ana

Answer 3

All of the above answers are correct. There is no direct relationship between volume and weight - if you don't know the density.

If you knew the density (g/cc) of your one cubic kilometer, the formula to solve for kg would be:

W=D*10^12

where W=weight in kg
and D=Density in grams per cc

If it was water (D=1 g/cc or SG=1), each cube of 1x1x1 cm would weigh exactly 1 g, each cube of 10x10x10 cm would weigh exactly 1 kg, each cubic meter would weigh eaxactly 1 metric tonne, and each cubic kilometer would weigh 1 billion tonnes, or 1 trillion kg.

Answer 4

Well like everyone said I think you would need to find the volume and the density for this problem.And we would probally also need to find the radius of a sphere.And plus we need to have a formula. Mass=Volume*Density
But hey I'm probally wrong lol cause I'm only 11 years old.

Oh and by the way,good luck winning the jackpot for all the neopoints.I was tryin to fgure out too.

Answer 5

um one of those is a measure of mass, the other one of volume. They don't work together...

Maybe if you knew the density of the cubic km you could find out how many kgs there are, but there isn't enough info to solve the problem.

Answer 6

It depends on what you have. The answer will be different if you have hydrogen gas or plutonium.

But don't try this with plutonium! That would melt down before you got a dozen kg in one place.

Answer 7

It depends on the density of mass in the space.

Answer 8

Weight does NOT equal distance