The Circle of Life: the Earth’s Orbit around the Sun

The central miracle is that there are simple underlying laws, expressed in the precise language of mathematics.”

Nima Arkani-Hamed, professor in the School of Natural Sciences at the Institute of Advanced Study, Princeton, NJ.

Earth is a Goldilocks kind of place: not too hot, not too cold. Things here are just right. We have a solid rock to stand on, liquid water to sustain us, and an atmosphere to shield us from radiation. Our cosy planet happens to lie just the right distance from the sun, in what astronomers call the “habitable zone. But that’s not all. Something else must be in place, and it has to do with the rate at which the Earth rotates around the Sun. Luckily for us, this rate is also just right. If it is any slower, we will be incinerated by the Sun’s inferno. If it is any faster, we will be flung out of its habitable zone and reduced to cosmic dust.

Newton’s fundamental laws of physics can explain this wonder. His fundamental equations are not only elegant, they link important quantities such as the Earth’s rotational velocity around the Sun, the distance of the Earth from the Sun, and the gravitational constant. If we know the values of these parameters (we do), we can calculate things like mass of the Sun, and given that, the Earth’s velocity around the Sun. The existential implication of Newton’s law for life on Earth will then become clear.

Let’s begin.

Consider a satellite with mass Msat orbiting a central body with a mass of mass MCentral. The central body could be a planet, the Sun or some other large mass capable of causing sufficient acceleration on a less massive nearby object. For today’s post, the Earth is the satellite and the Sun is the central body.

If the Earth moves in circular motion, then the net centripetal force acting upon Earth is given by the following equation:

Fnet = (Msat • v) / R

where v is the Earth’s rotational velocity and R is the earth’s distance from the Sun. This net centripetal force is the result of the gravitational force that attracts the satellite towards the central body:

Fgrav = ( G • Msat • MCentral ) / R2

Since Fgrav = Fnet, we can equate the two expressions.

(Msat • v2) / R = (G • Msat • MCentral ) / R2

Simple manipulations give the following equation for the square of the rotational velocity of the earth:

v2 = (G • MCentral ) / R

Taking the square root of each side, we have:

We can also deduce the mass of the Sun (MCentral) which is perhaps more fascinating than v, provided we know the values for the other three parameters.

We know that:

G is 6.673 x 10-11 N•m2/kg2, where N (for Newton) is the force needed to accelerate one kilogram of mass at the rate of one meter per second squared in the direction of the applied force,

R is 1.5 x 1011 m, and

v is 29,889.65 per second (it is equal to the total distance travelled by the earth in its circular orbit which is 2 $\large {\pi}$ R divided by the time required for the Earth to make one complete orbit around the Sun, which is 365 days x 24 hours x 3600 seconds).

Using these numbers give the mass of the Sun as 2 x 1030 kg which is way more than that of the Earth (more precisely, the ratio of the Sun’s mass to that of the earth is more than 333,000!)

Now, the Sun’s mass and the Earth’s orbital velocity implies that Earth must be going around the Sun in an exact rate of acceleration (by the way, acceleration is the rate of velocity). The equation that describes this connection between the Earth’s acceleration and those other quantities is:

R’’ = (G • MCentral ) / R2 + (v2 / R)

where R’’ denotes the Earth’s acceleration. This equation says that given the sun’s specified mass, a v larger than 29,889.65 per second would force a larger R’’, introducing the possibility that the Earth might escape from the solar system, in which case we will turn into cosmic dust. Luckily for us, this does not happen.