Venus Transit - What should be a simple problem

[Mazin Nasralla]

I was given a question which I have worked out, but not to my satisfaction...

"if the inclination of Earth and Venus orbit was the same, how often would the transit occur?"

I am given 4 options and I have got the right answer, but I did it by trial and error, and I am sure that there must be a better solution.

I looked up Venus's Orbital Period and it's 224.7 days. Call it 224 days.

So if we start off in a transit with the planets aligned on an axis, the two planets trace out different angular displacements over time:

Earth 2Pi/365 days and Venus 2Pi/365 days. I thought that I should be able to create a simple equation and solve, but I can't. Can anyone help?

I worked out by putting days in that the answer is 581 days (maybe small error due to rounding), but I am more interested in how I should go about solving this.
Please note that I am a pre-University student so go easy on me! Thanks

 

[willem2]

both planets start out at an angle of 0, there will be an alignment again when the difference in angles is a multiple of 2pi.

\frac {2 \pi } {224} t - \frac {2 \pi } {365} t = 2 k \pi

divide out 2pi and solve for t

 

[Mazin Nasralla]

Thanks Willem.

I realized after posting how to do it.

 

[Janus]

The simple equation is

\frac{1}{\frac{1}{period_{Venus}}- \frac{1}{period_{Earth}}} = period_{synodic}

Where the synodic period is the time between conjunctions of Venus and the Sun.

This will be the average value, as due to the eccentricities of both Venus' and Earth's orbits, there will be some variation in the timing between actual transits. What is interesting in this is that this average value is an even multiple of the mean Venereal solar day (the time it takes Venus to rotate once with respect to the Sun.). This means is that Venus always presents the same side towards Earth when it is in conjunction ( or during a transit).

 

[Mazin Nasralla]

The simple equation is

\frac{1}{\frac{1}{period_{Venus}}- \frac{1}{period_{Earth}}} = period_{synodic}

Where the synodic period is the time between conjunctions of Venus and the Sun.

This will be the average value, as due to the eccentricities of both Venus' and Earth's orbits, there will be some variation in the timing between actual transits. What is interesting in this is that this average value is an even multiple of the mean Venereal solar day (the time it takes Venus to rotate once with respect to the Sun.). This means is that Venus always presents the same side towards Earth when it is in conjunction ( or during a transit).

Thanks that's even simpler! The first equation has pi as a factor on both sides and has a constant which could be awkward - especially for me, but it hints at where it came from.

How do you remember that, or derive it. Willem's equation makes sense. I can see where it comes from with reference to the 2pi. Your equation is simpler but it's not quite the equivalent. I don't think I can get to it from Willem's. The 2Pi is a factor on both sides but that's about it. Is it a stupid question to ask why it works?

 

[Janus]

Thanks that's even simpler! The first equation has pi as a factor on both sides and has a constant which could be awkward - especially for me, but it hints at where it came from.

How do you remember that, or derive it. Willem's equation makes sense. I can see where it comes from with reference to the 2pi. Your equation is simpler but it's not quite the equivalent. I don't think I can get to it from Willem's. The 2Pi is a factor on both sides but that's about it. Is it a stupid question to ask why it works?

You can get to it along lines very much like Willem's

You are staring out with two planets in line with the Sun. The inner faster planet is is going to travel around the Sun and catch up with the outer planet. If the period of the inner planet is P1 then its angular velocity will be 2 pi/P1. If if the outer planet's period is P2 then its angular velocity is P 2 pi/P2. Thus their relative angular speed to respect to each other is \frac{2 \pi}{P1}- \frac{2 \pi}{P2}

Now from the inner planet's perspective, this is like having the outer planet stay still while it travels 2 pi radians and returns to the outer planet's position. The time it takes to do this at the above relative angular speed between the planets is
T=\frac{2 \pi}{ \frac{2 \pi}{P1}- \frac{2 \pi}{P2}}

2 pi cancels out on both top and bottom of the fraction, giving you the simple equation.

 

[Mazin Nasralla]

You can get to it along lines very much like Willem's

You are staring out with two planets in line with the Sun. The inner faster planet is is going to travel around the Sun and catch up with the outer planet. If the period of the inner planet is P1 then its angular velocity will be 2 pi/P1. If if the outer planet's period is P2 then its angular velocity is P 2 pi/P2. Thus their relative angular speed to respect to each other is \frac{2 \pi}{P1}- \frac{2 \pi}{P2}

Now from the inner planet's perspective, this is like having the outer planet stay still while it travels 2 pi radians and returns to the outer planet's position. The time it takes to do this at the above relative angular speed between the planets is
T=\frac{2 \pi}{ \frac{2 \pi}{P1}- \frac{2 \pi}{P2}}

2 pi cancels out on both top and bottom of the fraction, giving you the simple equation.

I like that explanation. I found a snippet on wiki under synodic period which confirms the formula.

https://en.wikipedia.org/wiki/Orbital_period#Synodic_period

Thanks very much to all who contributed. I am happy now, and I know a little bit more!