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Related to If Mars Were a Superearth

In terms of diameter and mass, Venus is one step close to identical to Earth.

But in this scenario, Venus is actually the size of the one exoplanet that astronomers had been calling a "Super-Venus"--Gliese 832c.

enter image description here

The mass of the planet varies between 5.5 and 10.4 times higher than Earth's. Unlike the previous planet enlargement question, which put Kepler-10b 225 million miles away from Earth, this super-rock is considerably closer to Earth--almost 27 million miles away. So if the planet Venus were enlarged to the size of Gliese 832c, would it affect Earth's orbit in any way? What would the nightscape look like with a larger Venus?

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  • $\begingroup$ What's the significant difference between this and the linked question, besides the subquestion about nighttime? $\endgroup$ – HDE 226868 Feb 2 '16 at 15:57
  • $\begingroup$ The last link has enlarged Mars. This one has enlarged Venus. That's the difference. $\endgroup$ – JohnWDailey Feb 2 '16 at 16:00
  • $\begingroup$ I understand; I'm wondering what makes you think the answers will be much different. $\endgroup$ – HDE 226868 Feb 2 '16 at 16:00
  • $\begingroup$ Venus is much closer to Earth than Mars. $\endgroup$ – JohnWDailey Feb 2 '16 at 16:19
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    $\begingroup$ By less than a factor of two. $\endgroup$ – HDE 226868 Feb 2 '16 at 16:38
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It would perturb the orbit a bit but not enough to change much. i.e. the change would be calculatable but would not change the weather or seasons enough to matter.

Phosphorous and Heosphoros$^1$ would be brighter and would be visible later and earlier respectively. Making them even more important in mythology and popular culture. If it were big enough to be seen as a disk heliocentrism would have had better traction earlier. Unfortunately, it would just be marginally brighter.

$^1$ the morning and evening stars i.e. Venus in the morning and evening. the stars you wish on.

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  • $\begingroup$ "It would perturb the orbit a bit but not enough to change much." How do you mean? And who are Phosphorous and Heosphoros? $\endgroup$ – JohnWDailey Feb 2 '16 at 17:02
  • $\begingroup$ clarified with an edit $\endgroup$ – King-Ink Feb 2 '16 at 17:08
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Calculating the exact outcome is the n body problem for n = 9 or thereabouts. (I'm counting the sun but not Pluto.) Solving it is too broad to answer.


Follow-up: John, I see several problems with many of your questions. I'll try to explain, and I hope the others will bear with me for an answer which goes way beyond the original question ...

Analytic vs. Numeric Solutions

Some scientific problems can be solved by "solving the equations." One example is the movement of two objects in orbit around each other, e.g. a planet orbiting a sun. (Actually both orbit around their common center of gravity. A slight wobble for the sun, while the planet circles the sun.) This is known as the two-body problem.

For many other problems there are no equations, or they cannot be solved. What is possible is to take initial situation at some point in time and calculate a very good approximation for the situation one "time step" later. The problem with this is that a "very good approximation" isn't the exact answer, and the errors will add up over time.

The Butterfly Effect

Imagine you take two lumps of bread dough and try your very best to make them exactly the same size and shape. Then you put a raisin into each, again trying to put the raisins into exactly the same spot. Then you knead each lump couple dozen times, again trying to knead both lumps with exactly the same movements. Bake the breads and slice them. Count which slice contains the raisin. Repeat the experiment a few times.

  • In theory, if you had exactly the same lumps of dough and did exactly the same to them, then the raisins should always end up in the same slice, right? You could even have a mathematical model of the dough and the raisin and calculate where the raisin ends up before you start to knead.

  • In practice, the raisin could show up in any slice of bread. When you shape the dough and place the raisin, there will always be little differences. When you knead the dough there will be little differences how you push. The effect of those differences will add up with each repetition of the kneading movement. After a dozen repetitions, tiny differences in the initial position of the raisins will be magnified. The mathematical model from your theory is sensitive to the initial conditions.

One way to solve a problem like that in climate models is to make many runs with slightly different initial conditions. The researchers look at the average outcome, but also how close the others come to the average. If most results are close to the average, perhaps the average is a good prediction. If the results are all over the place, the average is not all that useful.

Calibration vs. Calculation

Another issue with some types of climate models is calibration. Researchers create a climate model and use it to predict the weather of last year from the weather the year before that. If the predicted weather doesn't match what really happened, then the model must be adjusted. Repeat until the prediction matches. Careful researchers will then take another historical set of initial conditions and see if their model works for those, too. If not, something is wrong with the model.

Very simplified, if your model doesn't have the Gulf Stream, simply note that Europe is a little warmer than the same latitude in America and you might get reasonably good predictions.

Other climate models are less dependent on calibration, but they need more initial data. If they don't want to calibrate for the effect of the Gulf Stream, they have to calculate those effects. That requires data on the shape of the sea floor, the shape of the coast, and so on. The number of islands in the Caribbean and the channels between them matters. For the atmosphere, they need the shape of the mountains, etc.

Either way, simply stating "I make this ocean a thousand miles wider" won't work. The model can't be calibrated on historical data, because it is a fictional world, and the model can't be calculated because so many details are missing.

False Precison

Imagine I eyeball a box and say that it is 9 meters long, 6 meters wide, and 4 meters high. When I say "9 meters" then my estimate has one significant digit. So I would give my estimate of the volume with one significant digit, too -- I estimate that the box is 200 cubic meters, not 216 cubic meters.

That means when you talk about an ocean with an average depth of 3,767 meters, you imply that you're working with four digits of precision. You won't get answers with that kind of precision on Worldbuilding.

For reasons like that, I consider many of your questions unsuitable for Worldbuilding. To get a good answer to your climate questions, first you need a much more detailed question, and then somebody has to do a lot of hard work. I'm not currently in academia, but I guess a good and substantiated answer might well be worthy of a BSc or MSc thesis in mathematics or Earth sciences. Anything less than such a thesis is "primarily opinion-based" ...

The orbital question is slightly easier to solve using numeric methods, but the inaccuracy of the methods would make the answer worthless for any practical purpose.

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    $\begingroup$ To elaborate on @o.m.'s answer: for a n body problem like the Solar System, there's no guarantees about the outcome. Even simulations of the same problem give different answer. As mentioned in the Mars question and here, it probably has no effect. However, if it does have an effect, we can't tell you what that would look like. The range of outcomes probably goes from "it clears all bodies out of the inner solar system" to "no change". $\endgroup$ – Jim2B Feb 2 '16 at 17:29
  • $\begingroup$ @o.m. Google could not understand what I was saying--just throwing in random articles because they highlighted individual words instead of the whole thing. This is why I post them here on stackexchange--because they are relevant to the subject at hand. Dismissing anything as "broad" instead of actually providing an answer is just a lazy wave of dismissal. $\endgroup$ – JohnWDailey Feb 2 '16 at 19:25

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