Keplers 3rd Law

  • #1

Homework Statement


An asteroid is located between Mars and Jupiter. It is thought that a planet once orbited here but was somehow destroyed and broken up into small chunks(perhaps by getting hit by a comet or asteroid). If an asteroid in this belt has an average distance from the sun of 500 * 10^6 km what would the orbital period be?


Homework Equations





The Attempt at a Solution


I know you use the keplers formula

t^2 / r^3

and make it proportional. I know the distance between Mars and Jupiter and their orbital periods , how can that help me?
 

Answers and Replies

  • #2
SammyS
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Homework Statement


An asteroid is located between Mars and Jupiter. It is thought that a planet once orbited here but was somehow destroyed and broken up into small chunks(perhaps by getting hit by a comet or asteroid). If an asteroid in this belt has an average distance from the sun of 500 * 10^6 km what would the orbital period be?

Homework Equations



The Attempt at a Solution


I know you use the Kepler's formula

t^2 / r^3

and make it proportional. I know the distance between Mars and Jupiter and their orbital periods , how can that help me?
Hi Havenater23.

Right: T2/R3 is the same for objects orbiting the sun.

What is Earth's orbital period and average distance from the sun?

Note: 1 Astronomical Unit ≈ 150 × 106 kilometers.
 
  • #3
Average distance : 149.6 * 10^6 km

Period : 1.0 earth years
 
  • #4
SammyS
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average distance : 149.6 * 10^6 km

period : 1.0 earth years
T2/R3 = ? for earth. It's the same for the asteroid.
 
  • #6
SammyS
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  • #7
That the period of orbital is squared and that radius is squared, and this proportional between two planets , I believe ?

So it would actually work with distances and period orbitals of other planets correct ?
 
  • #8
SammyS
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That the period of orbital is squared and that radius is squared, and this proportional between two planets , I believe ?

So it would actually work with distances and period orbitals of other planets correct ?
So, it will work for all objects orbiting the sun.
 
  • #9
I need someone to check my work can you please solve this and tell me what you get?

I have it set up like so

t^2 / (500*10^6)^3 = 3.35*10^24
 
  • #10
SammyS
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T2 / (500*10^6)^3 = 1 / (3.35×1024)

Now solve for T.
 
  • #11
Okay I got the answer, but I don't think it's as long as I calculated. I got something like

2.05*10^25

I can see it being 2 years, but what am I doing wrong?
 
  • #12
SammyS
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T2 / (500*10^6)^3 = 1 / (3.35×1024)

Now solve for T.
This is:

[tex]\frac{T^2}{(500\times10^{\,6})^3}=\frac{1}{3.35\times10^{\,24}}[/tex]
 

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