Question

the dwarf planet pluto has an elliptical orbit with a semi-major axis of 5.91×1012m and eccentricity 0.249. (a) Calculate Pluto's orbital period. Express your answer in seconds and in earth years. (b) During Pluto's orbit around the sun, what are its closest and farthest distances from the sun?

Answer 1

To calculate Pluto's orbital period, we can use Kepler's third law. Pluto's orbital period is approximately 245.6 Earth years. The closest distance of Pluto from the Sun is approximately 4.44×10^12m and the farthest distance is approximately 7.38×10^12m.

Step-by-step explanation:

To calculate Pluto's orbital period, we can use Kepler's third law, which states that a planet's orbital period squared is proportional to the semimajor axis of its orbit cubed. Pluto's semimajor axis is 5.91×10^12m, so we can set up the equation as follows: (T^2)/(5.91×10^12)^3 = 1, where T is the orbital period we are trying to find. Solving for T, we get T = √(5.91×10^12)^3 = 7.746×10^9 seconds. To convert this to Earth years, we divide by the number of seconds in a year, which is approximately 31,536,000. Therefore, Pluto's orbital period is approximately 245.6 Earth years.

To find Pluto's closest and farthest distances from the Sun during its orbit, we can use the equation for the distance of a point on an ellipse from its center, which is given by r = a(1 - e) and r = a(1 + e), where r is the distance, a is the semi-major axis, and e is the eccentricity. Substituting the values for Pluto's semi-major axis (5.91×10^12m) and eccentricity (0.249), we get the closest distance as r = 5.91×10^12m(1 - 0.249) = 4.44×10^12m and the farthest distance as r = 5.91×10^12m(1 + 0.249) = 7.38×10^12m.