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# Jupiter’s Gravity

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### #1 Gschnettler

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Posted 16 September 2020 - 01:55 PM

I have read a lot about Jupiter and how it’s gravity is so powerful that it impacts the whole solar system and so on.  With 318 times the mass of Earth that is not surprising at all.

However, I was surprised to see that the gravity of Jupiter is only 2.5 times that of Earth.  Looks like they measure this on both planets where the atmospheric pressure is at 1 bar.

Seems hard to believe that it is so low, but I understand that gravity drops off rapidly as the surface diameter expands.

Is it best just to think of gravity in terms of mass when considering it’s impact?  So in this case, Jupiter’s gravity has 318 times the impact of Earth’s when it comes to things like comets and other objects flying through space?  And the 2.5 times number is pretty much irrelevant unless you are visiting near the surface?

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### #2 Sleep Deprived

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Posted 16 September 2020 - 02:42 PM

Remember, your weight on Jupiter (as quoted) is on its 'surface' which is approx. 45,000 miles from its center.  Your weight on Earth's surface is approx. 4,000 from its center.  If you were to measure your weight on a (non-moving) surface 45,000 miles from Earth's center, it would be much less than at 4,000 miles - I haven't done the math.  Also, Jupiter is a gaseous planet, while Earth is solid - there is a big density difference.  I guess what I am saying is that comparing weight on Jupiter and on Earth is almost like comparing apples and oranges - there are large differences in many of the basic aspects of the two planets, so comparing a single effect (weight, eg.) on the two may very well end up appearing odd.

In a sense, one's weight is a 'local' measurement.  On a more non-localized scale (scale of the solar system) overall mass is the ruler of gravity.

If you've done the math on how-much-would-I-weigh-45,000-miles-from-Earth, you might want to try how-much-would-I-weigh-5-MILLION-miles-from-Earth.  Do the same calc for Jupiter.  You should see a HUGE difference.  I am too lazy to do that math.

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### #3 TOMDEY

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Posted 16 September 2020 - 02:58 PM

Yep... you're right. Note that both obey Newton's Gm1m2/R2, with R being the distance between centers. So, it all makes sense. Anytime you interpret gravity strength text, be sure to note which they are referencing, surface or centers... or something else. They even often further confuse/camouflage things by measuring expressing distances between surfaces. The principle is so reliable, that you can even gauge gravity strength as only the ball ball below your feet, completely ignoring the shell above (derivation is easy, consequent of any R2 field).   Tom

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### #4 DaveC2042

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Posted 16 September 2020 - 06:24 PM

Interesting, thanks.

Here's something in return.

When a little object orbits a big one, the big one 'wobbles' due to the little one's gravitational attraction. They both actually orbit a point, the barycenter, a bit away from the big one's center.

Jupiter is so massive, that the barycenter of the Sun-Jupiter system (and indeed the whole solar system) is outside the surface of the Sun. This despite the Sun being a thousand times as massive.

I find that amazing and not at all intuitive.

### #5 Sleep Deprived

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Posted 16 September 2020 - 07:08 PM

Jupiter is so massive, that the barycenter of the Sun-Jupiter system (and indeed the whole solar system) is outside the surface of the Sun. This despite the Sun being a thousand times as massive.

There are a bunch of physical laws in play here.  In the end, it is mostly about Conservation of Energy.  One form of energy is angular momentum.  In a closed system, we can expect Conservation of Angular Momentum - add up all the bits of matter, take into account their mass, their distance from the center of rotation, and their speed, and that should remain constant.  So... the angular momentum we see in the Solar System today is close to that of when it was formed.  There will be differences because the Solar System isn't a 100% closed system - stars have passed nearby in the past, etc - but the general idea holds.  Also, some angular momentum can be converted into heat when orbiting bodies flex and stretch - this is why Io is such a volcanically active body.  Anyway, we should have never been surprised that exoplanets exist.  In the star formation process, clouds collapse into stars, but because the clouds are so big they will have a certain amount of angular momentum resulting in a relatively fast-spinning star.  It is the same idea that makes a spinning skater spin faster when he/she pulls his/her arms in tight.  Anyway, there is so much angular momentum in a dust-cloud a light-year across that if you were to scrunch it all into one body, it may very well spin so fast that the body would fly apart.  One way around this is that there are small bits of matter orbiting the star, holding a large amount of angular momentum, thus preventing the star from flying apart.  These bodies simply have too much orbital energy to have ever gotten close enough to the star to get gobbled up.  THAT'S why we have planets and exoplanets.

I found this little tidbit:

Jupiter has most of the rest of the mass of the Solar System and it is five times as far from the Sun as the Earth. As a result it actually has almost 60% of the Solar System's angular momentum.

Amazing that Jupiter is a small fraction of the Sun's mass, yet it holds most of the angular momentum of the Solar System.  Seems counter-intuitive that a small piece of star-schmaltz (Jupiter) holds so much energy in the Solar System.  We should expect similar circumstances at other stars.

FWIW, this is highly simplified and the complicated stuff is the devil-in-the-details, but the basic message holds pretty well.

Edited by Sleep Deprived, 16 September 2020 - 07:20 PM.

### #6 TOMDEY

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Posted 16 September 2020 - 08:28 PM

Kinetic energy can be converted into heat... but the net momentum of the system remains invariant, even in the presence of inelastic interactions... like tidal flexing.    Tom

### #7 llanitedave

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Posted 16 September 2020 - 08:41 PM

Remember, your weight on Jupiter (as quoted) is on its 'surface' which is approx. 45,000 miles from its center.  Your weight on Earth's surface is approx. 4,000 from its center.  If you were to measure your weight on a (non-moving) surface 45,000 miles from Earth's center, it would be much less than at 4,000 miles - I haven't done the math.  Also, Jupiter is a gaseous planet, while Earth is solid - there is a big density difference.  I guess what I am saying is that comparing weight on Jupiter and on Earth is almost like comparing apples and oranges - there are large differences in many of the basic aspects of the two planets, so comparing a single effect (weight, eg.) on the two may very well end up appearing odd.

In a sense, one's weight is a 'local' measurement.  On a more non-localized scale (scale of the solar system) overall mass is the ruler of gravity.

If you've done the math on how-much-would-I-weigh-45,000-miles-from-Earth, you might want to try how-much-would-I-weigh-5-MILLION-miles-from-Earth.  Do the same calc for Jupiter.  You should see a HUGE difference.  I am too lazy to do that math.

The concept of "on" Jupiter is itself meaningless.  You can either be outside Jupiter or you can be inside of it.  You can't be "on" it.

I know, I'm just being pedantic.  I'd rather the explanation be given for a point "at" a certain distance from the center, of which the 1 atmospheric pressure line is good, or maybe even the 273 K line.

### #8 Sleep Deprived

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Posted 16 September 2020 - 09:12 PM

Kinetic energy can be converted into heat... but the net momentum of the system remains invariant, even in the presence of inelastic interactions... like tidal flexing.    Tom

Yeah, Tom, I think I misspoke on that point.  I know that the moon was much closer to Earth billions of years ago, and that Earth's day was much shorter.  The moon was probably not tidally locked to Earth in its earliest stages.  The flexing/stretching/etc causes the rotational period (think 'day', not 'year') to change.  The moon, being smaller, tidally locked to Earth.  The flexing/stretching/etc on Earth has a lot to do with the fact that Earth's day is very different than the moon's orbital period.  Energy is being taken out of the Earth (flexing/etc) and is going into heat AND raising the moon's orbit (also slowing its orbital speed - THIS is conservation of angular momentum).  Ultimately, after a few billion years more, Earth and moon will be tidally locked to each other - Earth's day, moon's day, and moon's orbital period will all be the same.  This has already happened in the Pluto/Charon system.  Given enough time, it'll happen on Earth, too (unless there is some stable resonance - like 2 Earth days=1moon orbit - happens first.  I don't know if this is stable.).  Earth and Pluto each only have one large moon (Pluto has a few small moons, but Charon is MUCH bigger than any of those) so this mutual tidal locking is probable.

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