Density of a planet using orbit of a satellite

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Homework Help Overview

The problem involves determining the density of a planet based on the orbital characteristics of a satellite, specifically its altitude and period. The context is rooted in gravitational physics and orbital mechanics.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the derivation of the density formula and identify potential algebraic errors in the calculations. Questions arise regarding the correct manipulation of terms and the interpretation of the formula components.

Discussion Status

Participants are actively engaging with the problem, identifying errors and clarifying the relationships between variables. Some guidance has been offered regarding algebraic simplifications, and there is a sense of progress towards understanding the correct formulation.

Contextual Notes

There are indications of confusion regarding the algebraic steps taken in deriving the density formula, particularly in relation to the terms involving the radius and altitude of the satellite. Participants are also checking assumptions about the relationships between the variables involved.

XxBollWeevilx
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[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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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].
 
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 ?
 
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.
 
[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!
 
Yes, that looks perfect.
 
Thanks again!
 

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