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DyerWolf
2006-Sep-12, 05:31 PM
I tried googling, but can't find the answer(s):

Is there a direct correlation between a rocky planet's size and its gravitational pull?

For instance - if we were to find a planet that was 1.2 Earth gravities, would the planet be 1.2 times larger or only fractionally larger (I guess presuming the same planetary make up, to the extent that matters)?

Does planetary rotation counter gravitation to any significant degree?

i.e. does the centrifugal force generated when our planet rotates at 1 revolution per 24 hours change the apparent weight of an object on the surface? If Earth had a 15 hour day or a 30 hour day without changing the diameter of the planet, would an object weigh less or more respectively?

Is a rocky planet's atmosphere determined by its size/gravity?

Would a 1.2 G planet have significantly more atmosphere? Is there a ratio of atmospheric density relative to planetary size / gravitation?

grant hutchison
2006-Sep-12, 06:11 PM
Is there a direct correlation between a rocky planet's size and its gravitational pull?Surface gravity is proportional to mass and inversely proportional to radius squared. Since mass varies with radius cubed for a given density, there's a linear relationship between surface gravity and radius for planets of equal density. Of course, in real life, a planet will become more dense as you add mass, so surface gravity will rise faster than radius.

Does planetary rotation counter gravitation to any significant degree?Certainly measurably. The equatorial bulge of the Earth moves you a little farther from the centre of mass, and the centrifugal pseudoforce decreases your weight as well. The details were discussed here (http://www.bautforum.com/showthread.php?t=42621).

Would a 1.2 G planet have significantly more atmosphere? Is there a ratio of atmospheric density relative to planetary size / gravitation?Not necessarily. Very low mass planets can't retain atmosphere for geologically significant time periods, because their escape velocity is too low: the gas leaks away. Once a planet can retain an atmosphere, however, there are other influences on the amount of gas it retains: look at the difference between Venus and Earth, which have roughly the same mass.

Grant Hutchison

antoniseb
2006-Sep-12, 06:14 PM
Is there a direct correlation between a rocky planet's size and its gravitational pull?
It depends on the density of the rocks. If that remains contant, then you can treat the force of gravity for the surface of a planet as though all the mass were at the center. The mass goes up as the cube of the radius, and the force (for constant mass) goes down as the square of the radius. Thus Surface Gravity goes up linearly with radius (average density being kept constant).

Does planetary rotation counter gravitation to any significant degree?
It is small (under one percent) for the Earth, but exaggerated by the equatorial bulge it causes raising the equatorial surface further from the center by a few miles.

Is a rocky planet's atmosphere determined by its size/gravity?
Knowing the gravity is necessary but not sufficient. Other factors include the composition and temperature of the atmosphere, as well as how well it is protected by the planet's magnetic field.

neilzero
2006-Sep-12, 06:16 PM
The formula adds not multiplies my mass to the mass of the planet and divides by the distance squared between mass centers, so the gravity force decreases as the square of the radius of the planet, but it increases as the cube of the radius assuming the larger planet has the same density. Typically, the density will be larger for the larger planet, because some elements and compounds are compressed to higher density. there is a finaglefactor in the formula called big G which, I think depends on weather you are thinking MKS = meter, kilogram, second or CGS = centimeter, gram, second or English units such as inch, ounce, hour. G also depends on if you want the answer in Earth gravities= g or newtons What ever newtons are, I think. Apparently newtons are equal to kilograms but force instead of mass or weight? Neil

grant hutchison
2006-Sep-12, 06:24 PM
... there is a finaglefactor in the formula called big G ...You can forget about G (the universal gravitational constant) if you work in "Earth units": Earth masses, Earth radii, Earth surface gravity.

Grant Hutchison

Jeff Root
2006-Sep-12, 07:58 PM
For any real planet, the composition and thus density would depend
on the composition of the protosolar nebula, the size, density and
distribution of angular momentum in the resulting protosolar disk,
and details of the formation of the sun and planets. In particular,
how strong the sun's solar winds become and at what point in the
planet formation process they become that strong. An early, strong
solar wind will blow away all the gases and dust from the inner part
of the solar system, while a late solar wind in a dense disk could
allow gas giant planets to form close in.

-- Jeff, in Minneapolis

neilzero
2006-Sep-12, 08:04 PM
Hi Grant: So if Earth's mass is one quintillion times 10E18 times my mass: F = 10E-18 times 1 over 1 squared = 10E-18 Earth force? How do I convert to pounds, poundals, newtons or something I can relate to? This post is good only as a horrible example of results when one has an errors in the formula. Thanks Grant for leading me down the correct path. Neil

grant hutchison
2006-Sep-12, 08:15 PM
Hi Grant: So if Earth's mass is one quintillion times 10E18 times my mass: F = 10E-18 times 1 over 1 squared = 10E-18 Earth force? How do I convert to pounds, poundals, newtons or something I can relate to? NeilI'm not sure what you mean.
DyerWolf is interested in the surface gravity, which is usually measured as an acceleration: 9.8 m.s-2 (or 9.8 N.kg-1) on Earth, which is equal to one g.
Surface gravity is given by GM/R2 and, as you say, the numerical value of G depends on the units you're using. If you use "Earth units", G becomes 1. So a planet with 3 Earth masses and 2 Earth radii would have a surface gravity of 3/22 = 0.75 g. If your mass is 70kg, then your weight would be 0.75 x 9.8 x 70 = 514.5 N.

Grant Hutchison

neilzero
2006-Sep-12, 08:45 PM
I think I know part of my problem: I should ADD my mass to the Earth's mass: not multiply. a = GM/radius squared = 1 times 1, all divided by 1 or the bigger planet with double the mass but 1.2 times the radius: a = 1 times 2 all divided by 1.2 = 1.666 g = 9.6 times 1.666 = 16 meters per second per second = the acelleration due to gravity of the bigger planet. Neil

DyerWolf
2006-Sep-12, 09:09 PM
I'm not sure what you mean.
DyerWolf is interested in the surface gravity, which is usually measured as an acceleration: 9.8 m.s-2 (or 9.8 N.kg-1) on Earth, which is equal to one g.
Surface gravity is given by GM/R2 and, as you say, the numerical value of G depends on the units you're using. If you use "Earth units", G becomes 1. So a planet with 3 Earth masses and 2 Earth radii would have a surface gravity of 3/22 = 0.75 g. If your mass is 70kg, then your weight would be 0.75 x 9.8 x 70 = 514.5 N.

Grant Hutchison

Grant - I need to try to break this down Barney style for a second: Your 3 Earth mass, 2 Earth radii planet - would the 70 kg person on Earth weigh 514kg (equivalent) or does the "N" above mean something different?

On this bigger planet, would the acceleration be greater such that a dropped brick is going to hit my foot faster there than it would here?

Saluki
2006-Sep-12, 09:10 PM
kg is a mass term. N is a force. If you want "weight", you need to measure in terms of N. The mass (as measured by kg) stays the same.

grant hutchison
2006-Sep-12, 09:14 PM
Grant - I need to try to break this down Barney style for a second: Your 3 Earth mass, 2 Earth radii planet - would the 70 kg person on Earth weigh 514kg (equivalent) or does the "N" above mean something different?See Saluki's answer.
You can really forget the conversion to newtons and just think about that 0.75g. It means you'd weight 0.75 times as much as you would on Earth, and that objects accelerate 0.75 times as rapidly. So: lighter and slower. My imaginary planet's surface gravity is only three quarters of Earth's.

Grant Hutchison

DyerWolf
2006-Sep-12, 09:40 PM
kg is a mass term. N is a force. If you want "weight", you need to measure in terms of N. The mass (as measured by kg) stays the same.

Thanks Saluki -

I understand that a 70 kg body in space will be weightless, and will have more weight on Grant's Planet. But how do I convert 514 N into a meaningful equivalent? If the healthy 70 kg Earthling landed on Grant's Planet, would she suddenly feel like she weighed 514 lbs?

--Whoops - GH answered as I typed. So it's the .75 number that communicates the equivalent. -- Thanks, GH.

Earlier you wrote "a planet will become more dense as you add mass, so surface gravity will rise faster than radius." So my theoretical 1.2 Earth gravity planet is not likely to be much bigger than Earth.

Very interesting stuff.

Saluki
2006-Sep-12, 09:54 PM
1 Newton = 0.224808943 pounds force. So, a 514 N object weighs about 115.5 pounds.

Celeste
2006-Sep-12, 10:22 PM
Thanks Saluki -

I understand that a 70 kg body in space will be weightless, and will have more weight on Grant's Planet. But how do I convert 514 N into a meaningful equivalent? If the healthy 70 kg Earthling landed on Grant's Planet, would she suddenly feel like she weighed 514 lbs?

--Whoops - GH answered as I typed. So it's the .75 number that communicates the equivalent. -- Thanks, GH.

Earlier you wrote "a planet will become more dense as you add mass, so surface gravity will rise faster than radius." So my theoretical 1.2 Earth gravity planet is not likely to be much bigger than Earth.

Very interesting stuff.

The International System (SI) introduced in 1960 is but the latest version of the old 18th century Metric System. The SI Newton adds to the internal coherence within the system but in old days another unit for forces was used: the pond and its big brother the kilopond (also called kilogram-force)

A pond is the force equivalent to the weight of a gram of mass under Earth&#180;s average surface gravity. A kilopond is the same for a kilogram of mass.

In your example, a mass of 70 Kg weights 70 Kp on Earth. With a 0.75g gravity, it would weight 70 * 0.75 = 52.5 Kp

You can easily guess that a Kp is equivalent to 9.80665 N

I remember that for the purpose of school exercises sometimes we were allowed to aproximate it as 10 N (a decaNewton)

Saluki
2006-Sep-13, 02:20 PM
The International System (SI) introduced in 1960 is but the latest version of the old 18th century Metric System. The SI Newton adds to the internal coherence within the system but in old days another unit for forces was used: the pond and its big brother the kilopond (also called kilogram-force)

A pond is the force equivalent to the weight of a gram of mass under Earth´s average surface gravity. A kilopond is the same for a kilogram of mass.

In your example, a mass of 70 Kg weights 70 Kp on Earth. With a 0.75g gravity, it would weight 70 * 0.75 = 52.5 Kp

You can easily guess that a Kp is equivalent to 9.80665 N

I remember that for the purpose of school exercises sometimes we were allowed to aproximate it as 10 N (a decaNewton)

It is interesting to note that the Metric System and SI use mass as a base (non-derived) unit, while the US Customary system uses force is a base (non-derived) unit.

In USCS, we have one force unit, the pound force; and two optional derived mass units, the pound mass and the slug. An object that weighs 100 pounds force on earth has a mass of 100 pounds force, or 3.1 slugs.

Whats the difference? The gravitational acceleration in USCS units, which is aproximately 32.2 feet per second per second.

DyerWolf
2006-Sep-17, 03:33 PM
Need some math help...

Tried the surface gravity = GM/R^2 equation, and think I'm getting the right numbers (apparently) but the decimal seems wrong. Any insight on what I need to change?

I've learned that:

"big G" is generally thought to be = 6.673x10^-11 (or .00000000006673)
Earth's mass is 5.98 x 10^24 kg (or 5,980,000,000,000,000,000,000,000 Kg)
the radius of the earth is 6376.5 Km (and r^2 is 40,659,752.25)

So I plug those numbers into the equation and get 9,814,260.49. That looks a lot like the surface gravity number of 9.8 m/s^2 GH supplied, but my decimal is way off. I tried multiplying by .001 to convert km to m, but that only makes it 9,814.2. Somehow I have to move the decimal over three more places, and can't figure where/how. (Is it to again multiply by .001 to convert kg to grams?)

Finally, I am starting to see how the equation works. But if I get a number of grams per meter^2, I am still missing something to understand the acceleration aspect.

I know (unrelated to this question-as a former parachuter) that a body falling on earth accelerates at 9.8 m/s^2 until it reaches terminal velocity due to air resistance. As I mentioned, I am starting to understand the equation. I need help making the next leap to connect both ideas and understand how the equation solves for acceleration (the time aspect, for one thing).

Can someone clear the mud?

grant hutchison
2006-Sep-17, 03:52 PM
So I plug those numbers into the equation and get 9,814,260.49.You need the radius in metres: multiply by a thousand before squaring.

But if I get a number of grams per meter^2, I am still missing something to understand the acceleration aspect.You're neglecting the units of G, which, with the value you quote, are N.m2.kg-2 (which is why you need the radius in metres, to match the units of G). These units cancel with the mass and radius squared to yield a final answer in N.kg-1, which is equivalent to acceleration, if you recall the equation that force equals mass times acceleration. (In fact, one could just as well express G in m3.kg-1.s-2, which makes the relationship to acceleration more explicit.)

I know (unrelated to this question-as a former parachuter) that a body falling on earth accelerates at 9.8 m/s^2 until it reaches terminal velocity due to air resistance.A quibble. That acceleration pertains during only the first instant of falling. Thereafter, air resistance decreases the acceleration steadily towards zero, at which point terminal velocity has been achieved.

Grant Hutchison

Jeff Root
2006-Sep-17, 06:18 PM
You need the radius in metres: multiply by a thousand before squaring.
I trip up on that one all the time.

That acceleration pertains during only the first instant of falling.
Thereafter, air resistance decreases the acceleration steadily
towards zero, at which point terminal velocity has been achieved.

That was argued here within the last year. Some thought that
acceleration would continue until terminal speed was reached.
I took your view. Dropping a ping-pong ball and steel ball
side-by-side a distance of about seven feet didn't make enough
of a difference for me to detect.

-- Jeff, in Minneapolis

grant hutchison
2006-Sep-17, 06:35 PM
An OT diversion into the physics underlying terminal velocity:
The force of gravity can be treated as constant, assuming the distance fallen is small compared to the radius of the planet.
As an object moves through air (or any fluid medium), it experiences a pressure drop between its leading and trailing surfaces. This exerts a force tending to oppose the object's motion. The force increases as velocity increases: approximately as the square of velocity, for velocities well short of the speed of sound.
In the case of a vetically-falling object, the fluid resistive force opposes the gravity force, and must be subtracted from it to give the resultant force which is acting to accelerate the object. At low velocities the resistive force is low, and so the net force accelerating the object is only slightly diminished in comparison to the force of gravity on an object at rest. As the object accelerates, however, it experiences progressively higher resistive forces, and therefore progressively lower accelerations. The resistive force eventually rises to match the gravitational force, at which point there is no net force, and therefore no further acceleration: the object is at terminal velocity.

Grant Hutchison

DyerWolf
2006-Sep-17, 08:37 PM
You need the radius in metres: multiply by a thousand before squaring.

You're neglecting the units of G, which, with the value you quote, are N.m2.kg-2 (which is why you need the radius in metres, to match the units of G). These units cancel with the mass and radius squared to yield a final answer in N.kg-1, which is equivalent to acceleration, if you recall the equation that force equals mass times acceleration.
Grant Hutchison

Thanks! Now to try and take this the next step...

If I take your planet, a 3 earth mass, 2 earth radii planet, and run the numbers (3xMass # above, 2x radii # above), I get:
g= (6.673x10^-11) x (17.94x10^24)/162.639x10^12 = 7.3607

Presuming no math errors there, please check me on the following:
If I take your 70 kg person, and multiply by the surface gravity of Grant's planet, 7.3607, I get the person's weight in Newtons (515.25). If I then divide that number by Earth's surface gravity, I get 52.5 kg - which I interpret to mean a 70 kg person on Earth would only weigh 52.5 kg on Grant's planet.

Is that correct?

hhEb09'1
2006-Sep-17, 08:56 PM
Thanks! Now to try and take this the next step...

If I take your planet, a 3 earth mass, 2 earth radii planet, and run the numbers (3xMass # above, 2x radii # above), I get:
g= (6.673x10^-11) x (17.94x10^24)/162.639x10^12 = 7.3607

Presuming no math errors there, please check me on the following:
If I take your 70 kg person, and multiply by the surface gravity of Grant's planet, 7.3607, I get the person's weight in Newtons (515.25). If I then divide that number by Earth's surface gravity, I get 52.5 kg - which I interpret to mean a 70 kg person on Earth would only weigh 52.5 kg on Grant's planet.

Is that correct?Another way to approach it, is that the 3 earth mass makes that planet's gravity 3 times stronger, but the twice radius makes it one fourth, so 3/4 of 70 is 52.5.

PS: Like grant did earlier (http://www.bautforum.com/showthread.php?p=824313#post824313)

grant hutchison
2006-Sep-17, 08:58 PM
... I get 52.5 kg - which I interpret to mean a 70 kg person on Earth would only weigh 52.5 kg on Grant's planet.
Is that correct?Yes. If a 70-kg person stood on a spring balance calibrated for Earth gravity, it would report a "weight" of 52.5 kg on my imaginary planet.
Notice this is consistent with the "Earth units" calculation I proposed earlier, which predicted my planet would have a surface gravity of 0.75g. 70 x 0.75 = 52.5.

Grant Hutchison

publius
2006-Sep-18, 02:42 AM
Grant,

After all that mess I went through with the Schwarzschild metric, I can't help but point out the local acceleration (for a spherically symmetric mass) is actually

GM/r^2 * 1/sqrt(1 - R/r),

where R is the Schwarzschild radius. This is what makes the force required to remain stationary go to infinty at r = R, rather than zero. For the earth and just about any conceivable body where we could walk on, that extra factor is so close to 1 as to be of no consequence. The earth's R is about the size of a ping-pong ball.

Amazingly, the local radial free-fall/escape velocity is exactly the same as Newton, v^2 = c^2*R/r = 2GM/r, and the local tidal forces felt by both stationary and radial free-fallers are the same as Newton, 2GM/r^3. But orbiting observers, those with non-zero L feel slightly different tidal forces. This is an amazing property of sphericall symmetric mass distributions that doesn't hold in general.

About air resistance and terminal velocity. You can generally model air resistance as some F = -bv, and that will give you a fairly good trajectory for most "heavy" objects (wouldn't work too good for a feather or an airplane with significant lift) If you need more accuracy, you can represent the "friction" by a power series in v, but that makes the equations of motion impossible to solve analytically. You can put a e^(-z/z0) on bv to account for the reduction in air resistance with height. IIRC, that can still be solved analytically. Anyway, you find things like the maximum range of a projectical will increase above a firing angle of 90 degrees if has enough velocity to quickly get high enough where the air resistance is less, and therefore gain horizontal range, even though the component of initial horizontal velocity is less than the max at 45 degrees.

-Richard

DyerWolf
2006-Sep-18, 03:19 PM
Of course, in real life, a planet will become more dense as you add mass, so surface gravity will rise faster than radius.

Grant Hutchison

Grant, I've been looking at this: are there formulae?
I graphed the mass, diameter and mean density of the inner rocky planets. My graph shows that mass and diameter scale similarly, so that I can see there is (probably) a ratio of diameter to mass.** However density does not scale with either number (the numbers I have show Mercury as denser than Venus). (In other words, the line charting planetary mass*10^-21 compared to the line charting planetary diameter tend to rise and fall almost identically, but the density lines don't change predictably.)

I haven't gone back and tried to do the numbers yet for surface gravity of the planets (based upon the work you've helped me with so far (that spreadsheet is at home...)) but if I do, will I see surface gravity rising faster than radius?