Calculating Orbital Period of Asteroid Between Mars & Jupiter

AI Thread Summary
To calculate the orbital period of an asteroid located between Mars and Jupiter at an average distance of 500 million kilometers from the Sun, Kepler's Third Law is applied, which states that the square of the orbital period (T) is proportional to the cube of the average radius (r) of the orbit. The formula used is T²/r³ = constant, where the constant can be derived from known values of other celestial bodies, such as Earth. The discussion highlights the need to correctly set up the equation using the asteroid's distance and the proportionality constant derived from Earth’s orbital data. A participant expresses uncertainty about their calculations, suggesting they obtained an orbital period that seems too long, indicating a potential error in their computations. The conversation emphasizes the importance of correctly applying Kepler's Law to find accurate results for orbital periods.
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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