Question

What is the velocity of Earth at perihelion and aphelion? How is this information calculated?

Answer 1

Hint: Aphelion is the point of the Earth's circle that is farthest away from the Sun. Perihelion is the point of the Earth's circle that is closest to the Sun. Aphelion consistently occurs toward the beginning of July. Around fourteen days after the June solstice, Earth is farthest from the Sun. Perihelion consistently occurs toward the beginning of January. Around fourteen days after the December Solstice, Earth is nearest to the Sun.

Complete step by step answer:
With the assistance of Newton's condition, the power because of gravity which the Sun applies to the Earth is given by:
$$F={\displaystyle \frac{GMm}{r^2}}$$
Where $$G$$ represents the gravitational constant, $$M$$ represents the mass of the Sun, $$m$$ represents the mass of the Earth, and $$r$$ represents the distance between the center of the Sun and the center of the Earth.
The centripetal force needed to keep Earth in orbit is given by:
$$F={\displaystyle \frac{m{v}^2}{r}}$$
Where $v$ is the orbital velocity.
When we combine the two equations, divide by m and multiply by r obtains:
$$v^2={\displaystyle \frac{GM}{r}}$$
The value of $GM$ = $1.327\times {10}^{11}k{m}^3{s}^{-2}$.
At perihelion the expanse from the Sun to the Earth is $$147,100,000km$$. Replacing the values into the equation gives $v=30km{s}^{-1}$
At aphelion the expanse from the Sun to the Earth is $$152,100,000km$$. Replacing the values into the equation gives v=29.5km/s.
The original values as obtained using the NASA DE430 ephemeris data are $30.28km{s}^{-1}$ and $29.3km{s}^{-1}$.

Note: Another approach is that consider that the average velocity $$29.7848km/s$$is attained when $$r=a=1.496\times {10}^{8}km$$. Then the formula $$v=29.7848\times sqrt\left(2a/r-1\right)$$ gives mini/max 29.22 km/s and 30.29 km/s. At perihelion, $r=a\left(1-e\right)-1.471\times {10}^{8}km$ and at aphelion $r=a\left(1+e\right)=1.521E+08km$. Here, the value of e is $0.01671$.