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I was thinking about a planet more or less three times the size of Earth. If it revolved around a star similar to our Sun, with the length of one day consisting of 36 hours, and had a similar distance Mars does to the Sun, would it affect the strength of gravity on the surface of said planet? How? Would that distance make the planet too cold to be habitable or is the size of the planet compensating for that?

If the planet had a ring system like Saturn's, how would that affect the planet?

What if the planet had two moons that revolved around the planet?

When thinking of the size of the fictional planet, take this picture of Gliese 6677Cc into consideration:

enter image description here

Please ask for any details in order to give the best answer. I'm very interested in how this would work. Thank you!

Already assume the planet is habitable (water, atmosphere including an ozone layer, oxygen, vegetation, etc.

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closed as off-topic by JDługosz, Burki, clem steredenn, bowlturner, trichoplax Sep 22 '15 at 10:29

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question does not appear to be about worldbuilding, within the scope defined in the help center." – JDługosz, Burki, bowlturner
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ The relationship between a planet's surface gravity and it's mass is a very intimate one, there's shouldn't have enough room for a third party(Sun). $\endgroup$ – user6760 Sep 20 '15 at 7:16
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    $\begingroup$ This looks like three separate questions $\endgroup$ – user243 Sep 21 '15 at 2:43
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    $\begingroup$ This is a straight astrophysics question. Ask on Physics or perhaps on Space Exploration $\endgroup$ – JDługosz Sep 22 '15 at 8:08
  • $\begingroup$ When you say "three times the size," do you mean 3x mass or 3x radius? Each time you ask "affect the planet" are you still talking about surface acceleration due to gravity? Why does it matter whether there's an atmosphere, water, ozone, &c.? $\endgroup$ – nitsua60 Sep 24 '15 at 3:15
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I'll assume that you want the surface gravity to be the same as earth's.

This means that the average density must be 1/3 that of earth, since for a body, if gs is the surface gravity, $\rho$ the density, and r the radius $$gs=\frac{GM}{r^2} = \frac {GMr}{r^3} = G \rho r$$ and increasing r requires a corresponding decrease in $\rho$ if gs is to remain constant.

Since the density of earth is 5.5 times that of water, the density of Earth3 must be on the order of $$\rho = \frac{5.5}{3} = 1.8 $$ This low density means there is no iron core and no magnetic field. As a result, the surface will have a slightly increased solar radiation dose. Actually, it's not clear that there could be any significant rocky material at all, since virtually no rock has a density that low, even at the low pressures associated with surface conditions. Quartz, for instance (aka "sand") has a density of 2.6. The surface of Earth3 has to be water, and to a very considerable depth, with no reasonable expectation of land anywhere.

Of course, if you allow the planetary makup to remain the same as earth's, the increased pressure at the core will increase the density, so the surface gravity will be somewhere north of 3 g's. I'm not competent to figure the exact increase, so someone else will have to fill this one in.

Now, about those rings. Let's assume that the Earth3 system has the same proportions as Saturn. Saturn has a radius 9.88 times that of earth, while the rings extend from 6600 to 121,000 km above the surface. Scaling this to Earth3's radius (6400 km) gives rings extending from 2,000 km to 37,000 km above the equator.

I've not been able to find the reflectivity of Saturn's rings, but let's assume 100% for incidence angles less than 30 degrees. If Earth3 has an orbital tilt the same as earth's (23 degrees) it's clear that winter's will be ferocious, as the rings will shade much of the hemisphere during deep winter. The "vertical" extent of the ring's shadow will be $$x = 37000 km \times \sin{23} = 14,500 km$$ which, compared to a radius of 19,200 km gives full shadow to $$\theta = cos^{-1}(\frac{14,500}{19,200}) = 41 degrees$$ Since this is virtually all of the tropics (except for about a 2000 km belt north or south of the equator), and much of the temperate zone (especially the warmer parts), the intensification of winter temperatures should be fairly severe.

Finally, the moons. Meh. If Earth3 is like Mars, with dinky little moons, there will obviously be no tides to speak of, although the existence of tidal pools is often speculated to be a possible source of the earliest life forms. Two large moons with much the effect of Luna suggests that the two orbit each other, but I suspect that this is difficult to justify in terms of the mechanics of formation.

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In general, the larger the planet, the greater the gravity. If density is held constant then this is a linear relationship, with twice the mass resulting in twice the gravitational force.

That said, depending on how hard your science will be, you can fudge around with this by decreasing its density. A less massive planet will have less gravitation force at its surface than will a more massive planet of an identical size.

Rate of rotation can also be used to fudge with the apparent gravitational force towards the equator of the planet. Again, the degree to which you can use this will depend on how hard your science will be.

Rings and moons will have no effect on surface gravity.

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    $\begingroup$ "twice the mass resulting in twice the gravitational force." - not remotely true without taking density/radius into account. For constant density, doubling the radius (and increasing the mass by a factor of 8) doubles surface gravity. $\endgroup$ – WhatRoughBeast Sep 20 '15 at 19:07
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If a planet is located in the goldilloc zone, it does not automatically grant it habitable status. For example, we cannot inhabit a gas giant in the goldilloc zone for sure.

Secondly, so far there have been no moon systems with moons revolving around each other as well as the planet. I don't think such an unstable assortment of a lunar system would ever be discovered.

To make a planet habitable, you need some very basic prerequisites:

  • presence in the goldilloc zone

  • presence of a large water body to regulate the temperatures

  • something resembling ozone layer to absorb deadly x-rays and ultriaviolet radiation

  • oxygen in the atmosphere to breath (for life as we know it)

Only with these very basic prerequisites available, you can think of the further things required for life (before venturing into the possibility of large, intelligent life like mammals).

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  • $\begingroup$ It's Goldilocks (means roughly yellow hair), the name of a fictional character famous for constantly finding what fits "just right" at the third attempt. $\endgroup$ – AlexP Mar 22 '17 at 15:15

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