Question

**Calculate the Density of an steroid**

An extremely massive (spherical) asteroid with a mass of 8.7 x
10^{15} kg passes through the atmosphere and hits the
Caspian Sea. The diameter of the asteroid is determined to be 20.5
km and there are initially 78200 km^{3} of water in the
Caspian Sea. Assuming that this asteroid is denser than water,
calculate the water displacement in cubic meter. What is the
density of the asteroid in g/cm^{3}? (Hint: use the
equation for the volume of a sphere)

Answer #1

**Solution :-**

Mass of asteroid =8.7*10^15 kg * 1000 g / 1 kg = 8.7*10^18 g

radius =diameter / 2

= 20.5 km / 2 = 10.25 km

10.25 km * 1000 m / 1 km = 10250 m

Now lets calculate the volume of the asteroid

V= 4/3 pi r3

V = 4/3 * 3.14 * (10250 m)^3

= 4.509*10^12 m3

**The volume of water displaced is same as the volume of
the asteroid so it displaces 4.509*10^12 m3 water**

Now lets convert the volume of asteroid from m3 to cm3

4.509*10^12 m3 * 1*10^6 cm3 / 1 m3 = 4.509*10^18 cm3

Now lets calculate the density

Density = mass / volume

= 8.7*10^18 g / 4.509*10^18 cm3

= 1.929 g/cm3

**So the density of the asteroid isd 1.929
g/cm3**

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