For this question, assume that the planet is Earth-like in almost every way, and is in a system like ours...the distance from the sun, orbital period, rotation, axial tilt, etc is all pretty much the same as Earth now - except this planet's radius and mass are reduced such that the gravity on the surface of the planet is half that of our Earth's. That means that the radius of this planet is $\frac12\ a_\oplus$ (1/2 Earth radius). Keeping the ratio of land mass to bodies of water, let's assume that oceans and continents are about 1/4 the size that they are on Earth.

The air density/atmospheric pressure at any given altitude is about 1/2 that of Earth's.

Could storms on this planet be as violent as storms on Earth? What would it be like in a hurricane?

  • $\begingroup$ At first, i thought it strange to think that gravity played a role in weather. Then I learned that gravity can make tornados. $\endgroup$
    – Frostfyre
    Oct 21, 2015 at 14:51
  • $\begingroup$ You must specify the size of water bodies and the density of its atmosphere too, if you want a reliable answer. $\endgroup$ Oct 21, 2015 at 14:54
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    $\begingroup$ You have some conflicting requirements that you probably don't need to include. If you want half the surface gravity and half the radius, the planet needs to be eight times more massive than our Earth. Obviously the geology and density will be quite different to meet that requirement. You will also have a fourth of the surface area, not a half. $\endgroup$
    – Samuel
    Oct 21, 2015 at 16:37
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    $\begingroup$ @Samuel, I've removed the density and geology requirements. But by my calculations, the mass will be 1/8th that of Earth's, not 8 times. $M_p=\frac{g×r^2}{G}=\frac{4.9×3190^2}{6.67×10^{-11}}=7.47×10^{23} kg$ $\endgroup$
    – Seth
    Oct 21, 2015 at 18:40
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    $\begingroup$ @Seth That's correct. Sorry I meant to say "eight times less massive". $\endgroup$
    – Samuel
    Oct 21, 2015 at 18:55

5 Answers 5


Leaving aside the issues Youstay has raised we can look at answering the core question:

There is one factor here that is going to over-ride everything:

  • Reduced atmospheric pressure. With atmospheric pressure of 50% earths you may well have massive fast moving impressive looking storms. However their ability to carry debris and generally cause damage will be drastically reduced. It is objects being carried by the storm that normally cause the damage, not the wind speed by itself. Lower pressure air will both do less damage in the first place, and also find it harder to carry and accelerate objects.

The other reasons I expect the weather to be less severe is:

  • Lower gravitational gradients mean there is less difference between layers of the atmosphere and the layers are further apart.

  • The smaller world probably means there is less temperature difference between poles and equator, again reducing the power available to drive weather systems.

However the actual effects of the weather might be felt more:

  • Lower gravity means it's easier to lift up debris and carry it with the storm.

  • Even things that cannot be lifted can be moved more easily.

  • Lifeforms would be evolved to suit a lower-gravity environment. This may well make them tall and fragile, more vulnerable to storms.

One thing that would not be different is:

So storms would not form as often and would be weaker. When they do form they may be visually impressive or very large but the reduced atmospheric pressure will vastly reduce how much damage they can do.

  • $\begingroup$ Mars has a much lesser gravity than Earth and it has storms enveloping the whole planet. Now that's some thing to consider! I think the main factor is the size of water bodies. These are the key to temperature regulation on a planet. $\endgroup$ Oct 21, 2015 at 17:48
  • $\begingroup$ Coriolis Force only depends on angular velocity. Temperature at equator and poles will likely be the same (depending on solar irradiance), but the energy transport system between them (Hadley Cells, Ferrel Cells, etc) will be different. The wind can't necessarily lift things easier because the wind itself will have less power as the air has less density. Can't say which effect will be stronger though. $\endgroup$
    – Khris
    Nov 14, 2017 at 12:25
  • $\begingroup$ @Khris Interesting, I did not know that. Thanks. $\endgroup$
    – Tim B
    Nov 14, 2017 at 14:31
  • $\begingroup$ @YoustayIgo Very weak storms though, The Martian for example got it very wrong :) $\endgroup$
    – Tim B
    Nov 14, 2017 at 14:34
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    $\begingroup$ If storms are common, life-forms would have adapted to deal with them, so likely would not grow tall and fragile unless something forced them to.' $\endgroup$
    – bendl
    Nov 15, 2017 at 15:03

First off, I'm afraid that the details you mentioned in your question statement are not logically possible. I should have stated that in a comment, but considering how lengthy it would get and how many laws of physics are involved, I thought it better to post it as a complete answer.

Error #1

this planet's radius and mass are reduced such that the gravity on the surface of the planet is half that of our Earth's. That means that the radius of this planet is $\frac12\ a_\oplus $ (1/2 Earth radius).

If the planet has 1/2 of Earth's radius then its gravity would be 1/8 of Earth's gravity, not 1/2 of it!! This is why:

Volume = $\frac43\pi r^3$

So if the radius of the planet is 1/2, the difference in volume would be $(\frac12)^3 = \frac18$

Considering the aggregate density to be the same (as stated in your question body), the total mass of the said planet would be 1/8 the mass of Earth which would give it 1/8 the gravity of Earth.

Error #2

Keeping the ratio of land mass to bodies of water, let's assume that oceans and continents are about half the size that they are on Earth.

The same thing applies here, only to a factor of 1/4 instead of 1/8.

Surface area of a sphere = $4\pi r^2$

If you keep the same ratio of sizes, then the size of water bodies would be 1/4 the size of water bodies on Earth, not 1/2.

Well, this was just a cosmetic issue and I brought it up only for the sake of correction. This is not as serious as the error in gravitational force.

Error #3

The air density/atmospheric pressure at any given altitude is about 99.5% that of Earth's.

For a planet 1/8 the size of Earth, the atmospheric pressure cannot be the same as Earth's at any two given altitudes! The altitude versus atmospheric pressure graph for the given planet would be vastly different than that of Earth, no matter how thin or thick you make its layer of atmosphere.

So ... correction in planetary details, perhaps?

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    $\begingroup$ Original gravity estimate was correct. You didn't account for $\frac{1}{r^2}$ factor, and in fact calculated the gravity at the Earth radius. At the planet radius the gravity is $\frac{1}{2}$ indeed. $\endgroup$
    – user58697
    Oct 21, 2015 at 17:11
  • $\begingroup$ This doesn't answer the question. As a matter of fact, you even admitted such. $\endgroup$ Oct 21, 2015 at 17:42
  • $\begingroup$ How is $\frac1{r^2}$ relevant here? Considering the density is the same, the mass of the planet would depend on its volume, not surface area. $\endgroup$ Oct 21, 2015 at 17:43
  • $\begingroup$ @DJMethaneMan: I pointed that out in the very start that this answer is about the problems in the question details. How can you answer a question that is contradictory in itself? $\endgroup$ Oct 21, 2015 at 17:45
  • $\begingroup$ @YoustayIgo Unless the question is tagged reality-check answers should assume that premises put forth by OP are true. That doesn't mean you should not correct OP if it is totally illogical, but that you should correct the OP, then answer based off of the correction. IMO merely correcting someone is not an answer unless reality-check/science-based/hard-science is used, and even then it is usually not satisfactory. Corrections should generally be made in the comments, regardless of length. $\endgroup$ Oct 21, 2015 at 18:04

Take a look at the equations for the geostrophic wind.

First the hydrostatic equation:

$0 = -g - \frac{1}{\rho} \frac{dP}{dZ}$

Rewrite it as:

$g \rho = -\frac{dP}{dZ}$

This tells us that the density of the air multiplied with the gravitational acceleration equals the negative vertical pressure gradient.

Therefore having only half of Earth's gravity makes the vertical pressure gradient only half as steep. Added to that less gravity would mean that the planet won't be able to hold as much atmospheric mass so the density of the air will be lower as well decreasing the vertical pressure gradient even further.

Second the geostrophic equations:

$fv=\frac{1}{\rho} \frac{dP}{dX}$

$fu=-\frac{1}{\rho} \frac{dP}{dY}$

The first one says that the meridional wind speed times the Coriolis Parameter equals the zonal pressure gradient times the inverse of the density.

The second one says that the zonal wind speed times the Coriolis Parameter equals the negative meridional pressure gradient times the inverse of the density.

The Coriolis Parameter only depends on the angular speed of the planet's rotation and the geographic latitude. As your planet should be rotating as fast as Earth the Coriolis Parameter will be the same.

If you reduce the density in those formulas while keeping the rest the same you get an increase in wind speeds with the same horizontal pressure gradients.

This makes sense since a less dense volume of air has less mass and therefore less energy is needed to accelerate it to a certain speed.

But in this situation the vertical pressure gradient is flatter so it's questionable if the same horizontal pressure gradienst as on Earth are even possible or if those are decreased as well, and I think they are. If they are the situation is similar to Earth, otherwise you'd get higher wind speeds.

Keep in mind however that lesser density - which is certain on your planet - means that the winds have less impact overall, since there is less mass in a volume of air there is less impulse. So in general I'd say your windstorms might get faster than on Earth but certainly not more devastating, more likely the opposite.

With Hurricanes there's also the question of vertical convection. With the flatter vertical pressure gradient convection will generally be weaker, so thunderstorms and tropical storms will generally be weaker as well.


Because the question is messed up it does make it harder to answer. Half the mass and half the radius gives a g for the new planet a 1.96 m/s^2 value instead of 9.8 m/s^2. That's less than the acceleration due to gravity on Mars with 3.71 m/s^s. Something important to note on a planet with less gravity is that it is less capable of holding the atmosphere. Mars atmosphere is so thin the wind hardly exerts a force at all. It does have massive storms that can last for months, but the particles picked up are very fine. The max winds witnessed on Mars were 175 kph. This would feel like approximately a 10 mph wind. It's one of the flaws in my favorite book, the Martian. What you want is a planet with the same radius and half the mass of Earth, which would give you half the acceleration due to gravity as that of Earth, but I'm not really certain how thin the atmosphere would get. But that would make everything else about the same, right? Coriolis effect is the same assuming the same rotational speed. Amount of the sun's energy incident on the planet would be the same. But with less gravity, water particles would evaporate easier, due to less pressure.

I'm guessing that the storms would be less damaging. They might be as severe, meaning wind velocity, but less damaging. With half the gravity fixed items like concrete, steel, wood, could potentially have the same strength and therefore be less effected by it. Loose items would be lighter, though and more easily picked up. I think the bigger problem relates to how dense the air is on this planet. Drag is directly proportional to density of the fluid, so half the density, half the force due to drag. A 100mph wind would feel like a 50mph wind. I hope this helps.

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    $\begingroup$ I think your calculations are off. Half mass and half radius give 20 m/s^2. Also, if you re-read the question, I only stipulate that the planet has half the radius of Earth. I don't specify exactly in the question, but to get to half gravity, then the mass has to be 1/8th of Earth's. I could keep the radius of the planet the same as Earth's, and then half the mass, but the density of the planet would not be realistic and it would likely break apart or compress anyway $\endgroup$
    – Seth
    Oct 27, 2015 at 15:08
  • $\begingroup$ You are right. I was referencing the wrong cell. The actual answer is 19.6m/s^2. Interesting. Mars is still 3.71 m/s^2 $\endgroup$
    – ozone
    Oct 28, 2015 at 1:08
  • $\begingroup$ You are right. I was referencing the wrong cell. The actual answer is 19.6m/s^2. Interestingly Mars is still 3.71 m/s^2. But that means, unless I grabbed the wrong cell again, that the average density of your planet is 4 times that of Earth. Could have a larger core, which could change the magnetism of the planet. That's a different question though. In your scenario, the mass is still half so I think my answer still stands and it seems to have been strengthened. I believe the Coriolis effect would be weaker which would lessen the chances of a storm and the air would still be thinner. $\endgroup$
    – ozone
    Oct 28, 2015 at 1:42

Trying to address the actual question This related response may be of interest concerning the effect of low gravity on storms Storms on a low-gravity planet It would appear that low gravity tends to return wind speed so less severe weather would be expected. Also as already pointed out increased air density would also reduce the intensity of the wind so all in all the atmosphere would be a great deal calmer than that seen of Earth. So the answer is no storms on this world could not be as violent as those on Earth.


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