Density of a planet using orbit of a satellite

In summary, the density of a planet can be calculated using the orbit of a satellite with altitude h and period T. The formula is \rho=\frac{3\pi}{GT^2}(1+\frac{h}{R})^3, where R is the radius of the planet. This formula is derived using the fact that density equals mass over volume, and solving for the mass and volume of the planet. After simplification, the final formula becomes \frac{3\pi}{GT^2}(1+\frac{h}{R})^3, which can be simplified further to \frac{3\pi}{GT^2}(R+h)^3.
  • #1
XxBollWeevilx
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0
[SOLVED] Density of a planet using orbit of a satellite

Homework Statement



A satellite is in a circular orbit about a planet of radius R. If the altitude of the satellite is h and its period is T, show that the density of the planet is [tex]\rho=\frac{3\pi}{GT^2}(1+\frac{h}{R})^3[/tex]



The Attempt at a Solution



I feel that I am mostly doing this problem correctly, but I think I'm leaving out something I need. I am able to derive the entire formula except for the part in the parentheses. I use the fact that the density equals M\V. To find the mass, I solved for M in the equation [tex]T^2=(\frac{4\pi^2}{GM})(R+h)^3[/tex]. I used R=H to account for the radius plus the altitude of the satellite. I do some rearranging and get [tex]M=\frac{4\pi^2(R+h)^3}{T^2G}[/tex]

I then use [tex]V=\frac{4\piR^2}{3}[/tex]. When I place M over V, I get that the density is equal to [tex]\frac{3\pi}{GT^2}(R+h)^3[/tex]
I have absolutely no clue where the 1 + h/R comes from that I must fine. I must have done something wrong in my deriving or rearranging...could someone point out a mistake? Thanks so much!
 
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  • #2
XxBollWeevilx,

It looks to me like you've made an algebraic error when dividing M by V. It looks like you divided the left side by [itex]\frac{4}{3}\pi r^3[/itex], but divided the right side by only [itex]\frac{4}{3}\pi[/itex].
 
  • #3
Yes, I see what you mean. But now I get that the density is [tex]\frac{3\pi}{GT^2R^2}(R+h)^3[/tex]

So I have another R62 in the bottom there...is there a way I can bring that up and unclude it in the R+h ?
 
  • #4
I think it needs to be R^3 in the denominator. (because your dividing by (4/3)pi r^3).

Then you can combine the (r+h)^3 in the numerator and the r^3 in the denominator into one fraction that is raised to the third power.
 
  • #5
[tex](\frac{R+h}{R})^3[/tex] can be simplified to [tex](1+\frac{h}{R})^3[/tex] correct? If so, I've got the right answer correctly. Thanks so much!
 
  • #6
Yes, that looks perfect.
 
  • #7
Thanks again!
 

1. What is the density of a planet?

The density of a planet refers to the measurement of how much mass is contained within a certain volume. It is often expressed in units of grams per cubic centimeter (g/cm3).

2. How is the density of a planet calculated?

The density of a planet can be calculated by dividing its mass by its volume. The mass can be determined using Newton's second law of motion, while the volume can be calculated using the planet's radius and assuming a spherical shape.

3. How can the orbit of a satellite be used to determine the density of a planet?

The orbit of a satellite around a planet is affected by the planet's mass and the distance between the two objects. By measuring the orbital period and distance of a satellite, scientists can use Kepler's laws and the gravitational force equation to calculate the planet's mass, and therefore, its density.

4. Why is the density of a planet important?

The density of a planet can provide valuable information about its composition, such as whether it is predominantly made of rock or gas. It can also give insights into the planet's internal structure and processes, which can help scientists understand its formation and evolution.

5. Can the density of a planet change over time?

Yes, the density of a planet can change over time due to various factors such as geological processes, atmospheric changes, and impacts from other objects. This is why it is important for scientists to continually study and monitor the composition of planets to better understand their dynamics.

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