Calculating Orbital Period of Asteroid Between Mars & Jupiter

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Homework Help Overview

The discussion revolves around calculating the orbital period of an asteroid located in the asteroid belt between Mars and Jupiter, with a specified average distance from the sun of 500 × 10^6 km. The context involves applying Kepler's laws of planetary motion to determine the relationship between the distance and the orbital period.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the application of Kepler's third law, specifically the formula t²/r³, to relate the orbital period and distance of the asteroid to known values for Earth. There are inquiries about how the distances and periods of other planets might influence the calculations.

Discussion Status

Participants are actively exploring the relationships defined by Kepler's laws and attempting to set up equations based on the given parameters. Some have provided values for Earth's distance and period, while others are questioning the calculations and seeking verification of their work. There is no explicit consensus on the results yet.

Contextual Notes

Participants note the average distance and period of Earth as reference points but express uncertainty about the calculations and the implications of their setups. There are indications of confusion regarding the application of the formula and the expected results.

Havenater23
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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 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?
 
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Havenater23 said:

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.
 
Average distance : 149.6 * 10^6 km

Period : 1.0 Earth years
 
havenater23 said:
average distance : 149.6 * 10^6 km

period : 1.0 Earth years
T2/R3 = ? for earth. It's the same for the asteroid.
 
How come ?
 
Havenater23 said:
How come ?
What is Kepler's 3rd Law?
 
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 ?
 
Havenater23 said:
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.
 
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
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 said:
T2 / (500*10^6)^3 = 1 / (3.35×1024)

Now solve for T.
This is:

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

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