Density question?

Question

This problem looks easy, but for some reason I can't get the right answer. If someone could help me (and explain the answer, if possible, please). Thank you!

The radius of Pluto (assumed spherical) is 1.15x10^3 km, and its mass is 1.5x10^22 kg. What is the density of Pluto in grams per cubic centimeter?

Answer 1

density=mass/volume


1kg=1000grams=10^3grams
mass=1.5x10^22kg= 1.5x10^22grams*10^3 = 1.5x10^25grams


1km=1000 meters
1meter=100centimeters
1km= 100*1000 centimeters =100000 centimeters =10^5 centimeters = 10^5cm
Conversion factor:
1km/1km=10^5cm/1km=1=1^3 =(10^5cm/1km)^3

volume= (4/3)*pi*radius^3
= (4/3) * 3.14 * (1.15x10^3km)^3 * (Conversion Factor)
= (4/3) * 3.14 * (1.15x10^3km)^3 * (10^5cm/1km)^3
= (4/3) * 3.14 * (1.15x10^3)^3 * (10^5)^3 cm^3 * (km/km)^3
= (4/3) * 3.14 * (1.15x10^3)^3 * (10^5)^3 cm^3

You use the conversion factor to cancel out the km^3 on the top and bottom and leave you with just cm^3. (km/km)^3 = 1^3 =1

Density= mass/volume
= 1.5x10^25grams / [(4/3) * 3.14 * (1.15x10^3)^3 * (10^5)^3]cm^3
=2.35 grams/cm^3


Whew!! I hope I punched in the numbers right into my calculator, but this is the correct way you work the problem.

Answer 2

As New Horizons approached Pluto, it increased the precision in the measurement of the two quantities needed to calculate the mass by Kepler's laws: the sidereal period of the Pluto-Charon orbit and the separations of Pluto and Charon from their mutual barycenter. With that done, an improved estimate of the masses of Pluto and Charon became possible.

The subsequent precise measurement of their radii made possible improved estimates of their average densities, which allowed better estimates of the ice- and rock-fractions for each body.

The calculation goes like this. Let subscript 1 denote Pluto. Let subscript 2 denote Charon. d is the distance to the Pluto-Charon barycenter. P is the sidereal period. M is mass. R is radius.

d₁ = 2035 km
d₂ = 17536 km
P = 6.387230 days = 551856.7 sec
G = 6.67384e-11 m³ kg⁻¹ sec⁻²
M₁+M₂ = 4π²(d₁+d₂)³/(GP²) = 1.45603e22 kg
M₁ = (M₁+M₂) d₂ / (d₁+d₂) = 1.3046e22 kg
M₂ = (M₁+M₂) d₁ / (d₁+d₂) = 1.5140e21 kg
R₁ = 1186000 m
ρ₁ = 3M₁/(4πR₁³) = 1.867 g cm⁻³
R₂ = 603500 m
ρ₂ = 3M₁/(4πR₂³) = 1.644 g cm⁻³
b = density of rock = 3 g cm⁻³
a = density of ice = 1 g cm⁻³
fraction ice by volume = (b−ρ)/(b−a) = (3−ρ)/2
→ Pluto: 0.5665
→ Charon: 0.6778
fraction ice by mass = a(b/ρ − 1)/(b−a) = (3/ρ − 1)/2
→ Pluto: 0.3034
→ Charon: 0.4122

If that was done correctly, then Pluto probably has a rocky core with a 897.6 km radius, overlaid by an icy shell of 288.4 km thickness. And Charon has a rocky core with a 413.7 km radius, overlaid by an icy shell of 189.8 km thickness.

I kept the general form of the density calculation visible in case someone wants to adjust the values of a and b. Methane ice has a density of about 1.35 g cm⁻³ and nitrogen ice has a density of about 1.25 g cm⁻³ at Pluto temperatures. Water ice's density is about 0.94 g cm⁻³. Pick the mix of water, nitrogen, and methane ices that you think is the right one, figure its density, and assign that value to a.

Similarly, maybe you think there's a small metal core inside the rocky core, which would increase the value of b. (Earth's moon's average density is 3.34 g cm⁻³ because it has a metal inner core.)

Then refigure the fractions of ice for Pluto and for Charon by mass and by volume.

Answer 3

first comvert the radius into cm and the mass into gm. Density is Mass / Volume. So, find the volume of the SPHERE by using the appropriate equation. Then divide that into the mass in grams and there's your answer.

Answer 4

Density is mass /volume

The volume of a sphere is 4/3(pi)r^3

So the volume of Pluto is (4/3)(pi)(1.15x10^3)^3

V = 6.37x10^9 km^3

Density is 1.5x10^22 kg / 6.37x10^9 km^3 = 2.4x10^12 kg/km^3

Now we need to convert to gm/cm^3

(2.4x10^12 kg/km^3)*(1000gm/kg)*(1km^3/1x10^15cm^3)
= 2.4 gm/cm^3

Answer 5

Give me 10 points please!!