Gravity Calculation

peacekeeper · Jan 27, 2005

How would you calculate the gravity at the uppermost levels of a Jovian atmosphere? I take it the surface gravity of a rocky world is calculated in very much the same way, right? If not, how do you calculate that gravity?

bbrock · Jan 27, 2005

Peacekeeper The force of gravity is calculated the same everywhere. You need to know the mass of the two bodies and the distance between them. Then it's a simple calculation. Using (slugs) as a unit of mass, slugs is the weight in pounds devided by the acceleration of gravity. I weight 180 lbs ( yea right! ). slugs = 180/32.1725 = 5.5948 slugs. I don't know how much a Jovian Moon weights, but lets say it is 1/10th the weight of the earth ( 1.319x10 ^ 25 lbs ) / 10 = 1.319x10^24 lbs The mass of the moon in slugs is 1.319^24/32.1725 = 4.099775x10^22 slugs. The gravitational constant G = 3.436x10^-8 The Force of gravity now only depends on the distance between the center of the moon and the 180 lb man. The distance needs to be in feet. So if the man is 4000 miles above the center of the moon, the distance in feet is 4000(mi) x 5280(ft/mi) = 2.112x10^7 feet. The force of Moon gravity pulling equally on the man and on the moon will be. F= G x ( Moon Slugs x Man Slugs )/ distance^2 F = 3.436x10^-8 x (4.09977x10^22 x 5.5948 ) / 4.460544x10^14 F= 17.669 lbs This example is general. The force of gravity only depends on the mass of two objects and the distance separating them. It doesn't matter if the object is at the uppermost level of the atmospher or on the survace. What matters is how far the separation between the two centers of mass. Bill

siarad · Jan 27, 2005

As the speed of gravity has recently been shown to move only at light-speed I don't see how it can act at the centre of gravity, normally taken as the centre of mass. Gravity, like light, comes in at an angle missing the centre of mass. Has theory been changed to accommodate this new difference or have I got it wrong.

bobvanx · Jan 27, 2005

Bill! Slugs!? Did you learn your physics from a Hungrian?

peacekeeper · Jan 27, 2005

BB: Thank you very much for that answer. It helped me alot.

bbrock · Jan 27, 2005

Bobvanx For the units of Force and Mass, in the Mks system, the unit of force is the newton and mass is the kilogram. In the Gaussian System ( Cgs ) the unit of force is the dyne and the unit of mass is the gram. In the English Engineering System the unit of force is the pound and the unit of mass is the slug. I learned this is high school and college. I thought the name was pretty cool. As for the Gravitation/Speed of Light problem and comment. The calculation for gravitational force is the Newton Calculation. The results differ from Einstine,s relativety results by so little for low gravity systems, you have to carry the results out to about the 7th or 8th decimal place to see any difference. I can't help it if you disagree with this. I suggest you take the argument up with someone more knowledgable then yourself. Bill

bobvanx · Jan 27, 2005

I agree, mass in slugs is cool. Just so non-typical around here. Most posts are in newtons and kilograms. BTW you of course know the noise a dyne-centimeter makes? It goes, "erg."

peacekeeper · Jan 27, 2005

<blockquote>In reply to:<hr />Most posts are in newtons and kilograms.<hr /></blockquote> Yes, it took me some time to convert the answer to units I was familiar with. It wasn't too hard though, so there was no problem with that. Besides, "slugs" do sound quite cool <img src="/images/icons/wink.gif" />

tony873004 · Jan 27, 2005

To calculate gravity at Jupiter's cloudtops, you need to know the mass of Jupiter, and the altitude of the cloudtops. Then use this formula: g = GM/r^2 where G is the gravitational constant, M is the mass of Jupiter, and r is the altitude of Jupiter's cloudtops G = 6.67 e-11 Mass of Jupiter is 1.9e27 kilograms Altitude of Jupiter's cloudtops is 71,500,000 meters g = 6.67e-11 * 1.9e27 / 71500000^2 g = 24.79 meters / second^2 g on Earth's surface is 9.81 meters / second^2, so at Jupiter's cloudtops you would experience gravity that was about 2.5 times stronger than at Earth's surface.

Saiph · Jan 27, 2005

if you wish to get even more technical, you must also consider how much of jupiters mass you have above you, as that provides no net contribution to it's gravity. Only the mass enclosed by a sphere, with radius equal to your altitude, is considered in the gravitational calculations. <div class="Discussion_UserSignature"> --------------------SaiphMOD@gmail.com -------------------"This is my Timey Wimey Detector. Goes "bing" when there's stuff. It also fries eggs at 30 paces, wether you want it to or not actually. I've learned to stay away from hens: It's not pretty when they blow" -- The Tenth Doctor, "Blink" </div>

tony873004 · Jan 27, 2005

I was assuming that ALL of Jupiter mass is below you when you are at the cloudtops. There will still be some clear atmosphere above you. But it's mass is so negligable compared to what is below you that it may be ignored. For example, the same problem applies to Earth. When you are at the surface, there's still about 60 kilometers of atmosphere above you. But the formula g = GM/r^2 using 5.976e24 for Earth's mass and 6371000 for Earth's radius to the surface, not top of atmosphere gives a value of 9.8 which is in agreement with experiments done on the surface. A factor which would create a much larger uncertainty on Jupiter is its shape. It has a noticable equatorial buldge. So any answer done using the formula I posted would only be an approximation anyway, and would vary depending on your exact location.

nexium · Jan 27, 2005

I suspect the equations assume the density is uniform if the distance to the center of mass is close to the surface. Decending below sea level the gravity decreases measurably as the equasions suggests, but it also decreses as you fly higher over the ocean, even though we could reasonably reguard the surface of Earth as 100 miles above sea level; the density being very low near the 4100 mile radius of the Earth. Neil

peacekeeper · Jan 27, 2005

Thank you very much, Tony.

siarad · Jan 28, 2005

Wow Slugs, you must be about my age <img src="/images/icons/wink.gif" /> <blockquote>In reply to:<hr />BTW you of course know the noise a dyne-centimeter makes? It goes, "erg."<hr /></blockquote> <img src="/images/icons/laugh.gif" /> <img src="/images/icons/laugh.gif" />

Saiph · Feb 21, 2005

pun alert! Btw, what brings you to the boards shila? I noticed you registered (or appear to have registered) on: 04/06/03 01:01 PM but only have 140 posts. And yet they are very detailed posts (you grabbed my attention with them anyway!). I hope you stick around. <div class="Discussion_UserSignature"> --------------------SaiphMOD@gmail.com -------------------"This is my Timey Wimey Detector. Goes "bing" when there's stuff. It also fries eggs at 30 paces, wether you want it to or not actually. I've learned to stay away from hens: It's not pretty when they blow" -- The Tenth Doctor, "Blink" </div>

thalion · Feb 22, 2005

Check out the Jupiter fact sheet: http://nssdc.gsfc.nasa.gov/planetary/factsheet/jupiterfact.html It actually gives two different values for the gravity at Jupiter's equator. For the gravity, the figure is 24.79 m/s^2, but for the acceleration, it's 23.12 m/s^2, about 93% of the previous value. This apparent discrepancy mystified me for a while, but I think it's finally clicked. "Gravity" must refer to the actual gravitational acceleration exerted by Jupiter's mass at its equator. However, "acceleration" probably refers to the force that someone would *experience* at the equator, due to the considerable centripetal force caused by its rapid rotation. In short, we can't leave rotation out of the equation, at least not for bodies that spin as quickly as Jupiter and Saturn. Let's not forget that we may also have to take oblateness into account. Gravity at Jupiter's poles would be stronger than at its equator because of both the lesser centripetal acceleration, and because its oblateness brings its poles closer to its center than its equator.

Saiph · Feb 22, 2005

Absolutely correct. On earth these same effects give IIRC from my previous calculations (years ago) of ~.02 m/s^2 discrepancy at the pole and equator from the quoted 9.8 m/s^2 average. <div class="Discussion_UserSignature"> --------------------SaiphMOD@gmail.com -------------------"This is my Timey Wimey Detector. Goes "bing" when there's stuff. It also fries eggs at 30 paces, wether you want it to or not actually. I've learned to stay away from hens: It's not pretty when they blow" -- The Tenth Doctor, "Blink" </div>

xmo1 · Feb 27, 2005

What are the chances that on a very large body, of any form that has mass, that the force of gravity may vary from one point to the next? An example would be a planet the size of the sun which has a preponderance of mass in one location, for whatever reason, but say the density on the top of the planet was greater by a factor of x than the density of matter at the bottom of the planet. The x here thinks that there must be actual numbers due to the properties of elements. I could talk about the center of gravity and the center of mass, but could a single planet have various (not variable) measures of gravity at different points around it? <div class="Discussion_UserSignature"> DenniSys.com </div>

yevaud · Feb 27, 2005

Try this site as a modest resource on Earth's Gravitational anomalies: http://rst.gsfc.nasa.gov/Intro/Part2_1b.html <div class="Discussion_UserSignature"> Differential Diagnosis: "I am both amused and annoyed that you think I should be less stubborn than you are." </div>

vogon13 · Feb 27, 2005

I believe the famous masscons of earth's moon (discovered in the sixties by NASA craft) is precisely what you are asking about. Probes orbiting moon speed up and slow down (very slightly, of course) as they pass over various regions of moon. May be denser rock formation, or buried remains of an asteroid with differing density from surrounding mediacause the effect. <div class="Discussion_UserSignature"> TPTB went to Dallas and all I got was Plucked !!So many people, so few recipes !!Let's clean up this stinkhole !! </div>

Saiph · Feb 27, 2005

sure can. earth for instance, is a pear. It's an oblate, sure, but it's bottom heavy as well. The gravity is measurably non-uniform. The problem is, it can't be to non-uniform, otherwise the mass shifts and evens it out a bit. <div class="Discussion_UserSignature"> --------------------SaiphMOD@gmail.com -------------------"This is my Timey Wimey Detector. Goes "bing" when there's stuff. It also fries eggs at 30 paces, wether you want it to or not actually. I've learned to stay away from hens: It's not pretty when they blow" -- The Tenth Doctor, "Blink" </div>

xmo1 · Feb 28, 2005

Related, but slightly off topic: I was thinking there might be a nodule of mass denser, and rotating slower than the fluid it is floating in at the Earth's core. When it arrives at some position in that rotation it causes the pole reversals. Isn't there some tension in the flux of the magnetic fields just before such a reversal (thinking now of common magnets), and then it breaks and the fields reverse almost instantly. That is what I expect to happen when the Earth's poles reverse. Then my interest would go to what might happen to the ionosphere and other layers of the atmosphere when that happens, and the affects on business objects (electrical, navagational, and so on). Think I'll do an archive search. <div class="Discussion_UserSignature"> DenniSys.com </div>

xmo1 · Feb 28, 2005

If this is true then it is possible to differentiate between elements and find specific mineral resources in any topology. Probably already been done. ... hunting for gold deposits by satellite ... <div class="Discussion_UserSignature"> DenniSys.com </div>

Saiph · Feb 28, 2005

problem with the rotation alignment thing, is magnetic fields are created by the rotation. They don't care about how it all fits together. If the two things are rotatin gin the same direction, the fields will add together and strengthen. It's all very puzzling. <div class="Discussion_UserSignature"> --------------------SaiphMOD@gmail.com -------------------"This is my Timey Wimey Detector. Goes "bing" when there's stuff. It also fries eggs at 30 paces, wether you want it to or not actually. I've learned to stay away from hens: It's not pretty when they blow" -- The Tenth Doctor, "Blink" </div>

Gravity Calculation

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