Is the distance between the Sun and Venus increasing?
What is the measurement?
May I request for a formula?
2 Answers
As you are new here, welcome. It would be good if you could give a justification to why you are asking this question.
A relevant thread to this is Why does the moon drift away from earth? .
You will find there the simple mechanics formulae of why the moon, because of the earth tides it induces, drifts away slowly from the earth.
Now Venus does induce tides on the sun, but the sun also induces tides on Venus so one has a more complicated problem. I expect though that the effect would be the same over centuries, that the average distance between Venus and Sun would grow.
Most Probably Yes.
The Earth is moving away from the Sun (see PSE-Question) at a rate $0.57H_{0}$, the most recent reference AFAIK, that can not be explained by any of the current official models, one can expect that the same happens with the Earth-Venus distance.
A 'simple' triangulation when the 'transit of Venus' (in the line of sight Earth-Sun) happens will reveal if the answer is yes or no. The natural answer is yes.
References:
by G. A. Krasinsky and V. A. Brumberg, 2004
Secular Increase of astronomical unit from analysis of the major planet motions, and Its Interpretation
$\frac{dAU}{dt}=15\pm4\: m/cy$
at present there is no satisfactory explanation of the detected secular increase of AU
WeiJia Zhang, ZhengBin Li and Yang Lei, 2010
Experimental measurement of growth patterns on fossil corals: Secular variation in short distances ancient Earth-Sun
both the modern and ancient leaving rates could be measured with high precision, and it was found that the Earth has been leaving the Sun over the past 0.53 billion years. The Earth’s semi-major axis was 146 million kilometers at the beginning of the Phanerozoic Eon, equating to 97.6% of its current value. Measured modern leaving rates are 5–14 m/cy, whereas the ancient rates were much higher. Experimental results indicate a special expansion with an average expansion coefficient of $0.57H_{0}$
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