Kepler's 3rd Law Question finding how long a year is for Pluto

In summary, to calculate the length of a year for Pluto, we can use Kepler's Constant (3.36x1018m3/s2) and the mean radius of orbit (5.9x1012m) to solve for T, the length of the year. By using the equation T2 = (4π2R3)/K, we can plug in the values and find that the length of a year for Pluto is approximately 7.82x109 seconds.
  • #1
Jon Bori
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Homework Statement



Kepler's Constant for any object or planet orbiting the sun is 3.36x1018m3/s2. Calculate how long a year is for Pluto given the "radius of object" is 3.0x106m and the "mean radius of orbit" is 5.9x1012

Answer: 7.82x109s

Homework Equations


K = T2/R3

The Attempt at a Solution


I tried plugging the values into the equation

T2 = sqrt (5.9x1012m)+(3x106)/3.36x1018m)

= 6.1x1019[/B]

 
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  • #2
Jon Bori said:

Homework Statement



Kepler's Constant for any object or planet orbiting the sun is 3.36x1018m3/s2. Calculate how long a year is for Pluto given the "radius of object" is 3.0x106m and the "mean radius of orbit" is 5.9x1012

Answer: 7.82x109s

Homework Equations


K = T2/R3

The Attempt at a Solution


I tried plugging the values into the equation

T2 = sqrt (5.9x1012m)+(3x106)/3.36x1018m)

= 6.1x1019[/B]
What equation is that? The one you listed in the Relevant Equations only involved one distance value, but you seem to have used two ("radius of object" and "mean radius of orbit"). The listed equation also cubed the distance, but you've used a square root to resolve it?

How does the given value of Kepler's constant relate to your Relevant equation? Do the units match?
 
  • #3
I got it. You only use the mean radius of orbit and I forgot to cube the distance.. lol
 
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FAQ: Kepler's 3rd Law Question finding how long a year is for Pluto

How does Kepler's 3rd Law apply to finding the length of a year for Pluto?

Kepler's 3rd Law states that the square of a planet's orbital period is directly proportional to the cube of its semi-major axis. In other words, the farther a planet is from the sun, the longer it takes to complete one orbit. This law can be applied to determine the length of a year for Pluto by measuring its semi-major axis and using the formula:
T2 = (4π2/GM) a3, where T is the orbital period, G is the gravitational constant, M is the mass of the sun, and a is the semi-major axis.

How do scientists calculate the semi-major axis of Pluto's orbit?

The semi-major axis of an orbit is the longest radius of an ellipse. In order to calculate this for Pluto, scientists use data from spacecraft missions and observations from telescopes to determine the average distance between Pluto and the sun. This distance is then used in the Kepler's 3rd Law equation to calculate the semi-major axis.

Why is it important to determine the length of a year for Pluto?

Knowing the length of a year for Pluto helps us better understand its orbital dynamics and the relationship between its distance from the sun and its orbital period. It also allows scientists to accurately predict future positions of Pluto in its orbit and plan for future spacecraft missions.

How does the length of a year for Pluto compare to that of Earth?

The length of a year for Pluto is significantly longer than that of Earth. While it takes Earth approximately 365.24 days to complete one orbit around the sun, it takes Pluto 248 Earth years to do the same. This is due to Pluto's much larger distance from the sun and its slower orbital speed.

Has the length of a year for Pluto always been the same?

No, the length of a year for Pluto has not always been the same. As with all planets, Pluto's orbit is affected by other celestial bodies, causing slight variations in its orbital period. Additionally, Pluto's orbit is slightly elliptical, so its distance from the sun can vary slightly, leading to variations in its orbital period.

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