Determining the density of a stellar object

In summary, the satellite Callisto has a mean radius of 2403 km and a mass of 1.076 × 1023 kg. Its density is 1.851238398 x 109 kg m-3. If you divide this answer by 1 million, you get the correct answer. But you don't know why!
  • #1
Johnboy
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0

Homework Statement


Hello all. I am new to physics but am quite competent in maths and other sciences. I am studying Open University's Exploring Science module and this one question has me stumped. I follow all of the instructions but always come to a completely improbable answer. The interesting thing is, that if I divide my answer by 1 million I get the correct answer, which makes me think that I have missed out a line of working somewhere along the line. I looked at the densities of the planets in our Solar System and tried to get the correct answer by the same calculations. Once again, my answer is one million times to large! I was hoping somone with a bit more experience in the field could shed some light for me. The question is:

Callisto, one of the satellites of the planet Jupiter, has a mean radius of
2403 km and a mass of 1.076 × 10 to the power 23 kg. Calculate its mean density in kg m
(i.e. SI units), giving your answer in scientific notation to the appropriate
number of significant figures and showing all of your working


Homework Equations



V= 4/3πr3 and p = m/v


The Attempt at a Solution




Cube the radius: (2403km)3 = 1.387590483 x 1010
Multiply that answer by 4: 1.387590483 x 4 = 5.550361931 x 1010
Multiply that answer by π: 5.550361931 x π = 1.743697627 x 1011
Divide that answer by 3: 1.743697627 x 1011 / 3 = 5.812325422 x 1010
Volume of Callisto = 5.812325422 x 1010 km3 = 5.812325422 x 1013 m3

Now that the volume is known - and the mass is provided in the question (1.076 x 1023 kg) - the density can be calculated by the following formula:

Density = mass/volume = (1.076 ×〖10〗^23 kg)/(5.812325422 × 〖10〗^(13 ) m^3 ) = 1.851238398 x 109 kg m-3

If I divide this answer by 1 million, I get the correct answer. But I don't know why!
 
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  • #2
One km^3 is (10^3 m)^3 = 10^9 m^3, not 10^3 m^3 This is your problem.
 
  • #3
Thanks phyzguy, feel a bit stupid now!
 

FAQ: Determining the density of a stellar object

What is density and why is it important to determine the density of a stellar object?

Density is the measure of how much mass is contained within a certain volume. In terms of a stellar object, it refers to how much matter is packed into a given space. Determining the density of a stellar object is important because it can tell us about the object's composition, age, and evolutionary stage. It can also help us understand the object's internal structure and how it behaves under extreme conditions.

How is the density of a stellar object calculated?

The density of a stellar object is calculated by dividing its mass by its volume. The mass can be determined through the object's gravitational pull on surrounding objects, while the volume can be estimated based on the object's size and shape.

What units are used to measure the density of a stellar object?

The density of a stellar object is typically measured in grams per cubic centimeter (g/cm³) or kilograms per cubic meter (kg/m³), depending on the size of the object. Other units such as solar masses per cubic meter (M☉/m³) may also be used for larger objects.

How do scientists use the density of a stellar object to classify it?

The density of a stellar object can be used to classify it into different types, such as brown dwarfs, main sequence stars, or white dwarfs. These classifications are based on the object's mass and density, which can give insight into its physical properties and evolutionary stage.

Can the density of a stellar object change over time?

Yes, the density of a stellar object can change over time due to various factors such as nuclear fusion, mass loss, or gravitational interactions with other objects. For example, as a star evolves and burns through its fuel, it may expand and become less dense. Or, if a star loses mass through stellar winds, its density may increase as the remaining matter is more tightly packed.

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