# How would you calculate the speed at which the earth revolves around the sun in m/s?

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17
Feb 21, 2016

Calculate from the average distance of the Earth from the sun and the time it takes to complete one orbit as about $3 \times {10}^{4} m {s}^{- 1}$

#### Explanation:

The average distance of the Earth from the Sun is spproximately $150$ million km or $1.5 \times {10}^{11} m$

The orbit of the Earth around the sun is roughly circular (more eliptical really, but let's keep it simple), with a period of roughly $365$ days.

So the path that the Earth traverses in $365$ days is approximately the circumference of a circle of radius $1.5 \times {10}^{11} m$.

Use the formula $C = 2 \pi r$ to find that this is about $9.4 \times {10}^{11} m$.

The number of seconds in $365$ days is:

$365 \cdot 24 \cdot 60 \cdot 60 = 31536000$

So the tangential velocity of the Earth is approximately:

$\frac{9.4 \times {10}^{11}}{3.1536 \times {10}^{7}} \approx 3 \times {10}^{4} m {s}^{- 1}$

That is $30$ km per second.

If you want more accuracy, get more accurate figures for the distance of the Earth from the sun at perihelion/aphelion and use a more accurate figure for the orbital period.

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