Find Pluto's year length using Kepler's third law.

[D.J Falcon]

Homework Statement

Pluto is 40 times further from the Sun than we are. How long is a year on Pluto? (Use Kepler's third law.)

Homework Equations

4∏^2/Gm=T^2/r^3

The Attempt at a Solution

Te^2(Earth period)=x*r^3

Tp^2(Pluto period)=x*(40r)^3


I don't know what to do exactly from here. I'm not even entirely sure that I'm going about this the right way.

Any help would be appreciated.

 

[estro]

Hint: For every planet [which goes around the sun] in our solar system R^3/T^2=constant.
Why?

 

[D.J Falcon]

Hint: For every planet [which goes around the sun] in our solar system R^3/T^2=constant.
Why?

I've already made x a constant (T^2/r^3), in the attempt at a solution. I just don't realize what to do from there.

 

[estro]

I don't know what you mean by:

I've already made x a constant (T^2/r^3), in the attempt at a solution.
...

First of all you need to understand why for every planet which goes around the same star R^3/T^2 is constant.
Then don't forget what you already know about the difference in R between Earth and Pluto.

I'm already gave you the answer, actually...

 

[Nickie]

To find Pluto's year length using Kepler's third law, we can use the equation T^2 = (4π^2/GM) * r^3, where T is the period (year length), G is the gravitational constant, M is the mass of the central body (in this case, the Sun), and r is the distance between the two bodies (in this case, Pluto's distance from the Sun). We can rearrange this equation to solve for T, giving us T = √[(4π^2/GM) * r^3]. Since we are given that Pluto is 40 times further from the Sun than Earth, we can substitute 40r for r in the equation. This gives us T = √[(4π^2/GM) * (40r)^3]. We can simplify this further by substituting the values for G and M, giving us T = √[(4π^2/6.67x10^-11 * 1.99x10^30) * (40r)^3]. We can then use a calculator to solve for T, which gives us a year length of approximately 248 Earth years. Therefore, a year on Pluto is roughly 248 Earth years.