# Could one calculate the relative speed of the Sun viewed from the Earth?

From point of view from the Earth it seems that the Sun is making an orbit around the Earth. Of course this is not right, the Earth orbits the sun and the Earth rotates on his axis. But is it possible to give a speed to the motion of the Sun between sunrise and sunset.

The speed of the rotation is at the equator is 1.600 km/h and the orbit is 200.000 km/u. So could we say that from point of view of the Earth the sun has a speed of 201.600 km/h? Or doesn't this make any sense?

• I don't understand the question exactly – Utkarsh futous Mar 30 '17 at 18:55
• You already have answers so I wont post another, but I did want to comment on "Of course this is not right." Actually, it can be right. It is absolutely possible to treat the earth as stationary and have the sun orbit around the earth. I even have a function in a simulation that does exactly that. You can have the earth orbit the sun or the sun orbit the earth. Either works. However, what we find is that the equations of motion are substantially simpler if we assume the earth orbits the sun in an inertial frame rather than... – Cort Ammon Mar 30 '17 at 19:12
• ... having the sun in a complex multiple component orbit the earth in a rotating frame full of centripetal accelerations and Coreolis effects. Both view points can get you a correct answer, but the earth-around-the-sun view point gets you there with so much less work that we'd rather everyone think "the earth revolves around the sun," because it means less work for us physicists and engineers! – Cort Ammon Mar 30 '17 at 19:14

The correct approach is to say that in our frame of reference, the sun completes one revolution per day around the Earth. The distance between Earth and the sun is roughly 150 million kilometers, so the sun is tracing with circle having this radius per day, therefore, the speed of the sun is:$$v=\dfrac{2\pi\times150\times 10^6}{24}\approx40\times10^6\ km/h$$

• But how is it possible that this speed is in the order of 200 times larger as the Earth orbits the Sun? – Marijn Mar 30 '17 at 19:29
• @Marijn The angular speed, $\omega$ of Earth's rotation will transform into the angular speed of the Sun's motion. The tangential velocity of the Sun's motion is $\omega R_{SE}$. $R_{SE}$ is the Sun-Earth distance. – Bill N Mar 30 '17 at 19:44
• Roughly the same speed as if the earth didn't rotate on it's axis, but made a full revolution around the sun each day. – Ask About Monica Mar 30 '17 at 21:26

Yep, we know the distance from the Earth to the sun, and we know the Earth's rate of rotation. If we imagine the Sun orbited around the Earth, the observed duration of a day wouldn't change, so $\omega$ remains the same.

We know for circular motion $v = r\omega$, so just approximating it as an actual circle gives us $r_{earth-to-sun}\omega_{earth}$ where $r_{earth-to-sun}$ is the separation of their centres of masses.

$r_{earth-to-sun} \approx$ 150 million kilometers or 93 million miles.

$r_{earth-to-sun} = 150000000*10^3m$

$\omega_{earth} = 7.2921159 × 10^{−5}rad/s$

The fake-tangential-velocity is then $150000000*10^3 * 7.2921159 × 10^{-5} = 10938173.85$

So the apparent tangential velocity is then $10938173.85m/s$

No, this makes no sense. The 'speed' that you would obtain (and which other answers provide) is not actually a speed. It is merely an angular speed (that of the Earth's spin) multiplied with a distance (to the Sun), but there is no object moving (relative to whatever) with that speed.

Note that in this contrived way you can easily obtain speeds in excess of the speed of light. Of course, one can compute such quantities with the physical dimension of a speed, but that does not imply that they are actually a speed.