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Answer :
Final answer:
Using Kepler's Third Law of Planetary Motion, we can calculate that it takes Pluto approximately 92,446 Earth days to complete one orbit around the Sun.
Explanation:
To calculate the time taken by Pluto to orbit the Sun, we use Kepler's Third Law of Planetary Motion, which states that the square of the orbital period of a planet is directly proportional to the cube of the semi-major axis of its orbit. Given that the distance between Pluto and the Sun is approximately 39.1 times more than the distance between the Earth and the Sun, we could simplify and use the approximation that Pluto's distance in astronomical units (AU) is roughly 40, as given in the reference information.
Pluto's orbital period (P) can be calculated using the average distance from the Sun (a) with the formula P² = a³. Here, we assume Earth's orbital period to be 1 Earth year and its distance as 1 AU by definition. Therefore, for Pluto:
- a = 40 AU (Pluto's distance from the Sun)
- P² = 40³
- P² = 64,000
- P = √64,000
- P ≈ 253 (Pluto's orbital period in Earth years)
To convert this period into Earth days, we multiply by the number of days in one Earth year (365.25 days, accounting for the leap year cycle):
- P (in days) = 253 years × 365.25 days/year
- P (in days) ≈ 92,446.25 days
Hence, it takes Pluto approximately 92,446 Earth days to complete one orbit around the Sun.
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Answer:
Explanation:
Given
Distance between Pluto and sun is 39.1 times more than the distance between earth and sun
According to Kepler's Law
[tex]T^2=kR^3[/tex]
where k=constant
T=time period
R=Radius of orbit
Suppose [tex]R_1[/tex] is the radius of orbit of earth and sun
so Distance between Pluto and sun is [tex]R_2=39.1\cdot R_1[/tex]
[tex]T_1[/tex] and [tex]T_2[/tex] is the time period corresponding to [tex]R_1[/tex] and R_2[/tex]
[tex](T_1)^2=k(R_1)^3---1[/tex]
[tex](T_2)^2=k(R_2)^3---2[/tex]
divide 1 and 2
[tex](\frac{365}{T_2})^2=(\frac{R_1}{39.1})^3[/tex]
[tex]T_2^2=365^2\times 39.1^3[/tex]
[tex]T_2=89239.67\ Earth\ days[/tex]
