Question

Part b in earth’s current orbit, it is 1.52 × 1011 meters from the sun at aphelion and 1.47 × 1011 meters at perihelion. using these values and the equation given in part a, what is the eccentricity of earth’s current orbit? round your answer to three decimal places. consult the math review for help with scientific notation.

Answer 1

Based on the calculations, the eccentricity of Earth's current orbit is equal to 0.017.

Distance A = 1.52 × 10¹¹

Distance P = 1.47 × 10¹¹

How to calculate the  eccentricity of Earth

The present eccentricity of Earth is e ≈ 0.01671. In the past, it has varied between 0 and ∼0.06. Mathematically, the eccentricity of Earth's current orbit would be calculated by using this equation:

Where:

a is the radius at aphelion (apoapsis).

p is the radius at perihelion (periapsis).

e is the eccentricity of Earth's orbit.

Substituting the given parameters into the formula, we have;

e = 0.017.

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Answer 2

Answer: The constant value (k) is the same for all the planets in the solar system, so use the values for any planet to solve for k. Using the values for Mercury:

K=7,739 over 0.059

~1.3 x 10^5

The constant (k) is useful because it helps scientists compare the planetary motion of all the planets in the solar system.

Explanation:

write in equation for the math problem

Its the answer I recieved