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

Jan 29, 2016

Earth's perihelion velocity is $30.28$km/s and its aphelion velocity is $29.3$km/s.

#### Explanation:

Using Newton's equation, the force due to gravity which the Sun exerts of the Earth is given by:
$F = \frac{G M m}{r} ^ 2$
Where $G$ is the gravitational constant, $M$ is the mass of the Sun, $m$ is the mass of the Earth and $r$ is the distance between the centre of the Sun and the centre of the Earth.

The centripetal force required to keep Earth in orbit is given by:
$F = \frac{m {v}^{2}}{r}$
Where $v$ is the orbital velocity.

Combining the two equations, dividing by $m$ and multiplying by $r$ gives:
${v}^{2} = \frac{G M}{r}$

The value of $G M = 1.327 \cdot {10}^{11} k {m}^{3} {s}^{- 2}$.

At perihelion the distance from the Sun to the Earth is $147 , 100 , 000 k m$. Substituting the values into the equation gives $v = 30 k m {s}^{- 1}$.

At aphelion the distance from the Sun to the Earth is $152 , 100 , 000 k m$. Substituting the values into the equation gives $v = 29.5 k m {s}^{- 1}$.

The actual values as calculated using the NASA DE430 ephemeris data are $30.28 m {s}^{- 1}$ and $29.3 k m {s}^{- 1}$.

Jan 29, 2016

An alternative approach: Assume that the average velocity 29.7848 km/s is attained when r = a = 1.496 E+08 km. Then the formula v = 29.7848Xsqrt (2a/r -1) gives mini/max 29.22 km/s and 30.29 km/s.

#### Explanation:

At perihelion, r = a(1 - e) = 1.471 E+08 km and at aphelion r = a(1 + e) = 1.521 E+08 km. e = 0.01671.