Question

Pls help I’m stuck
In Earth's current orbit, it is 1.52 x 1011 meters from the Sun at aphelion and 1.47 x 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

Answer:

0.017 is the right answer from accurate calculation, pls mark me branliest

Answer 2

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

Given the following data:

Distance A = 1.52 × 10¹¹

Distance P = 1.47 × 10¹¹

How to calculate the eccentricity of Earth's orbit.

Mathematically, the eccentricity of Earth's current orbit would be calculated by using this equation:

$e=\frac{a\;-\;p}{a\;+\;p}$

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;

$e=\frac{1.52 \times 10^{11}\;-\;1.47 \times 10^{11}}{1.52 \times 10^{11}\;+\;1.47 \times 10^{11}}\\\\e=\frac{5 \times 10^9}{2.99 \times 10^{11}}$

e = 0.017.

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