From a comfortable position on Earth’s surface, Venus does look beautiful. It is one of the brightest objects in the sky; only the moon and the sun and some future nearby nova or supernova can outshine it. When Earth turns away from the Sun, it also turns away from Venus. Being an “inferior” planet, meaning that it is closer to the sun than we are, Venus can only be seen shortly before and after sunrise or sunset, depending on where it is on its journey around the sun.

There is one of a few special positions during such a revolution around the sun, which has led to a revolution in scientific thought and to another aspect of beauty, a beauty where observations, mathematics and astronomy combine to reveal simple relationships. One of those positions occurs when our line of sight from Earth to Venus form a tangent to the approximate circle of Venus’ orbit. As we know, a circle’s radius is always perpendicular to a tangent. In this case the radius is the distance between the sun and Venus. (In reality the orbit is slightly elliptical, but we’ll assume it’s a circle to simplify the math and to get the general ideas across.)

Shown to the right is what it looks like on paper, but in the sky, how do we recognise such an event? When Venus is at that position, it’s as far as it can be seen from the sun, here on Earth. We dub it the* greatest eastern elongation* because it’s stretching away from the sunset in the west. That we are indeed at that point is verified by measuring the maximum angle between Venus and the setting sun (easier said than done!) On paper that would be the angle at the vertex of the earth, close to 46º . Since the tangent angle at Venus is 90° , we can also easily compute the angle at the sun.

With either angle, we could get the Venus-sun distance relative to Earth’s, which is set at 1 astronomical unit (AU). Using a simple ratio from trigonometry, specifically sine, we use the angle directly measured from Earth, and it yields a relative Venus-sun distance, r_{v}, of 0.723 AU.

We will come back to relative distances. For now, let’s move on to another astronomers’ measurement from the distant past. How long does it take for the greatest eastern elongation to reappear? It takes longer than a year on Earth, about 583.9 days. Dubbed the synodic period, this is not the length of the Venusian year. Venus has a shorter path to complete and it experiences a greater force from the sun, so why does sit take so long for the same alignment to reappear?

It’s because the earth keeps moving along its orbit while Venus does likewise, so Venus has to *gain* 360° on Earth, not merely move 360º. But from that realisation and synodic period-measurement we can deduce its period. How?

First lets’ figure out how many degrees, measured from the sun, Venus gains on earth. That’s 360 ° ÷ ( *x* ° /day) = 583.9 days, where x turns out to be 0.6165 ° /day.

Now of course the reason that Venus gains that many degrees on Earth daily is that it’s moving faster, so that number is the difference between Earth’s orbital angular velocity and that of Venus. The orbital angular velocity is 360 ° /P, where P is the period of a planet. Writing the previous sentence mathematically we obtain,

360 ° /P – 360 π /365.25 = 0.6165 ° /day

Solving the simple equation we obtain the period of Venus to be 224.7 days.

The period , P is also equal to the circumference of 2πr divided by the planet’s velocity, v, expressed in whatever units the radius has per second.

P = 2πr/v … call that equation (1)

This implies that the angular velocity = 360 ° /(2πr/v)= 180° v/ πr. But if we convert the degrees to radians to make the numerator dimensionless, given that π represents 180° in a unit circle, angular velocity = v/r in days ^{-1.}

Venus has a mass, m_{v} , and it has an approximately fixed speed v_{v}. Since it’s moving in an almost circular ellipse, its direction constantly changes. Its speed is constant, but its velocity , a vector quantity is not. Thus Venus is accelerated by a force. How do we express that force?

Its momentum (product of mass and velocity) is equal to its impulse, expressed as the integral sum of the product of force and time, acting over that period of time. To simplify:

m_{v} v_{v}= Ft

Dividing by t,

the force , F = m_{v} v_{v}/t =m_{v} v_{v}(t^{-1}). The (t^{-1}) is in essence a frequency which in this context is none other than the dimensionless angular velocity of Venus, v_{v}/r.

Substituting we obtain, F = m_{v} v_{v} ^{2}/r, which is the expression for the centripetal force experienced by Venus—the force which keeps it moving around the sun.

Historically, Newton based his Law of Gravitation on Kepler’s laws, which in turn were based on observations of the planets. But to keep the narrative going, let’s time-travel backwards and arrive at Kepler’s laws, which will still reveal the consistency between the two.

The gravitational force between the sun and Venus is also given by the product of their masses(m_{s} and m_{v}) and the universal constant, G, divided by the square of their separation distance, r_{v}, which we determined by trigonometry earlier. Equating centripetal force to Newton’s law, we obtain:

If we solve for the orbital velocity for Venus (v_{v}) we get v_{v} = √(Gm_{s} /r_{v}) . With a similar treatment Earth’s orbital velocity = v_{e} = √(Gm_{s} /r_{e}). If we want to know how much faster Venus revolves relative to earth, we could simply divide the two expressions and we see that

v_{v} / v_{e} = √(r_{e}/r_{v}) … call that equation (2)

Rearranging equation (1) , v = 2πr/P. Now we could also express the ratio of velocities as the following:

v_{v} / v_{e} = (2πr_{v}/P_{v} ) / (2πr_{e}/P_{e} ), where P_{v}= period of Venus and Pe is the period of Earth.

Simplifying the above, v_{v} / v_{e} = r_{v }P_{e} /r_{e} P_{v}. …call that equation (3)

We first equate equations 2 and 3 , and after squaring both sides we get:

(r_{e}/r_{v}) = (r_{v }² P_{e}² )/ ( r_{e}² P_{v}²)

Finally, we cross multiply and we see Kepler’s third Law emerge:

P_{v}² r_{e}³ = P_{e}² r_{v} ³

If we set r_{e }to 1 astronomical unit (AU) and use our value of r_{v} = 0.723 AU from trigonometry then

P_{v} = √ [(365.25)² (0.723) ³] = 224.5 days, fairly close to the value that we calculated from the synodic period.

Venus has a runaway greenhouse effect. Being too close to the sun, early in its evolution its water was dissociated by ultraviolet radiation. As a result the CO2 that was out-gassed was not kept in control by a cycle in which the majority of carbon dioxide could be dissolved. That made the planet unbearably hot. Interestingly, in the 19th century Svante Arrhenius, who was able to to foresee that Earth could one day suffer from excess industrial CO2 output, imagined Venus as a lush, tropical planet in his book, *Destinies of the Stars*. Instead, due to oven-like surface temperatures (~460 ºC) but mostly due to the high density of CO2, the pressure at the surface is a crushing 10 000 kPa. That in turn creates strong tidal forces, slowing its rotation to the point that its solar day is longer than its year. But doesn’t all that mathematical and physics- harmony compensate for the fact that Venus is not a green paradise filled with beautiful women?

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