How would one calculate the gravity of a planet?

Submitted by LoreCriticizer t3_z53aqr in askscience

I always assumed that doing so would simply involve somehow getting the overall mass of the planet, then taking (mass of other planet) divided by (mass of Earth) multiplied by our own Earth gravity. But today I was browsing Youtube when somebody in the comments section of a Star Wars video said that it wasn't so simple and it involved taking into account other nearby planets and the planet's moon(s).

Is this true? If so what are the factors that you need to calculate a planet's gravity?

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Humble_Indication798 t1_ixuejf8 wrote

They were probably talking about the net gravitational force felt from the spaceship or some object from multiple planetary bodies.

The equation for force from gravity is (Gm1m2)/r^2 G is a constant 6.67*10^-11 M1 is the planet M2 is the object you are interested in R is the distance BETWEEN THE TWO CENTERS OF THE OBJECTS

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sawkonmaicok t1_ixw2u4m wrote

Actually not the centers of the objects, but their centers of mass. Usually this is around the center but not always since things can have an uneven disteibution of density. Also that formula is only an approximation.

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Kered13 t1_ixy5uc7 wrote

For a planet sized object you can safely assume that the center of mass will effectively be it's geometric center.

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_Jaquen_Hgar_ t1_ixwrspc wrote

The question asked was about the strength of gravity exerted by one planet, so a slightly more apposite formula is the one for the gravitational acceleration felt by an object of any mass in free fall:

g=G.M/r²

i.e. it’s independent of the mass of the other object.

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Humble_Indication798 t1_ixwub3p wrote

Yeah but if you consider force=massacceleration, m2A=gm1m2/r^2 then A=g*m1/r^2 and A is acceleration from gravity which is the same as the formula you posted. Which is independent of the mass of the second object

I assumed the person who posed the question initially could get to this point algebraically.

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lacgibra t1_ixuglp0 wrote

Yeah right, the only change would depends whether spaceship or the object is bound to the system or not ( planet's atmosphere then the distance between them would be radius of the planet ), if it's extra planetary then the r would be distance between them, pretty much the same.

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[deleted] t1_ixwaqzr wrote

[removed]

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danceswithtree t1_ixwpoei wrote

I don't think that means what you think it means. The equation above gives only the first three digits for the gravitational constant-- because who wants to see it to the millionth decimal point? You can use as many decimal places as you want for the masses but the result (force in newtons) will only be good to 2 decimal places.

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critterfluffy t1_ixwqome wrote

The gravitational constant has proven very difficult to measure. The recommended max significant figure is actually a bit more, at 6. More than what they had but it is still a strong limit to modern measurement. The memory I had of this was wrong on length but my statement still holds. Anything using G will be stuck to 6 digits.

Info: https://en.wikipedia.org/wiki/Gravitational_constant#Value_and_uncertainty

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OlympusMons94 t1_ixxukz1 wrote

G and mass M can be measured with less precision individually, but the product G*M = mu, the standard gravitational parameter, can be measured with higher precision, and is more useful anyway, since most applications require or can use GM, not (necessarily) G and/or M separately.

But unless the distance from the mass is great enough to treat it as a single point mass, then well below even 6 digits of precision, the actual force of gravity at a given location cannot be treated as simply GM/r^2 . So a more precise value of G, M, or even GM, will not, by themselves, give a more accurate answer. Planets and other "spherical" celestial bodies are not perfectly spherical, and have topography, and an internal mass distribution that is not radially symmetric. Depending on the location and reference frame, you also have to consider centrifugal acceleration.

Because of the equatorial bulge and centrifugal acceleration, the effective force felt on Earth's surface is on average 1% lower at Earth's equator than at the poles. The non-uniform gravity field of Earth, and even more so the Moon, must be considered for spacecraft and artificial satellites orbiting these bodies.

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danceswithtree t1_ixwrdpp wrote

So what does your statement even mean then? Your answer is only good to however many decimal places your least precise measurement/constant is. But what do you mean that mass is only good to three decimal places? You can have as accurate a mass as you want but the answer, a force in Newtons, will only be good to 3 or 6 or however many decimal places in your least precise constant or measurement.

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critterfluffy t1_ixwrlcy wrote

The most accurate that the gravitational constant has been measured is 6 digits. After that the measurements drift. So anything being done using G will have a result no greater than 6 significant digits.

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danceswithtree t1_ixwrytd wrote

This response has drifted at least 6 significant digits from your first response.

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lacgibra t1_ixue9eg wrote

Are you speaking about Newton's law of gravitation? It's used to obtain force of attraction between two planets or it's satellites, which is deduced from f_g = G m_1 m_2/r²

f_g is force of attraction,

G is universal gravitational constant

m_1 mass of the planet

m_2 mass of the other planet

r is distance between them.

The above problem is otherwise called two body problem as well which turns into complicate when you apply classical mechanics

The formula for gravity of the planet obtained from the above equation as folloing as

We know that f = mg and also f = GmM/r²

Here g is gravity due to acceleration G is gravitational constant. m is mass of the object in the earth. M is mass of the earth.

Since the object is bound to the planet the r distance between them would be radius of the planet itself Thus r is radius of the earth

mg = GmM/r²

And you get g = GM/r²

I would suggest you to look into two body problem and barycentre you might find it relevant.

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phrankjones t1_ixvb757 wrote

Their question is more asking the line of: how do you get the mass of the earth?

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lacgibra t1_ixvevls wrote

The comment above still fits, You can obtain mass of the earth by doing the simple pendulum method to obtain g from the equation √g = 2π√L/T

g = earth's gravity L = length of the pendulum T = time period

Note that simple pendulum give adequate answers for small angles only, if you swing the pendulum to wide you won't br arriving at the earth's gravity.

then equate it with g = GM/r²

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Kered13 t1_ixy67ha wrote

It's actually very difficult, because we have to derive it from gravity, and measuring gravity is very difficult. Actually, measuring the acceleration due to gravity is quite easy, that's what /u/lacgibra described. The trouble is measuring the gravitational constant. Of all the fundamental physical constants, the gravitational constant is the one that is known to the least accuracy. The classic experiment for this is Cavendish's experiment, and indeed this was originally conducted in order to estimate the mass of the Earth. This is still basically the technique used, but with much more precise equipment.

Fun fact: We know the mass of the other planets as a ratio to the mass of the Earth much more precisely than we know their absolute mass. This is because the uncertainty in the gravitational constant is greater than the uncertainty in the acceleration due to the planets that we can measure.

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eldude2879 t1_ixvxcai wrote

using newton is not good enuff for precise calculations of planets

we have to use einstein, by that we have to stop thinking of gravety as a force and think of curved space time

Einstein's general relativity states that, in a Schwarzschild spacetime (which approximates our solar system), gravity is slightly different from the inverse square law by an extra fourth power term. It is this extra small term that causes the perihelion motion of Mercury's orbit. If we take this extra term into account and follow the same logic of classical considerations, we find that planets' orbits aren't closed, but are precessing consistently. Mercury's orbit has an apprant precession about 43 arcs per century at its perihelion position which is exactly the amount of precession observed.

https://physicsworld.com/a/correcting-einsteins-calculation-of-the-orbit-of-mercury/

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lacgibra t1_ixw0h05 wrote

Well yeah Newton's doesn't explain Mercury's precession. I were sticking the context of the questioner and I had mentioned defining it classically, you need to think what it is when required. You can't just tell all the concepts that neglected friction, air resistance are wrong, for the classical assumption gravity has to be assumed as force. Relativity got better explanation apsidal precession and all at the end of the day approximately g = 9.81 m/s². Not lower than that or higher than that.

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eldude2879 t1_ixw1mi9 wrote

light behind a galaxy that can be seen because of the mass of the galaxy infront is seen not because of gravity force

as we know photons have no mass so gravity has no effect if you think of it as a force

the space time is bend so the photons get bend with it

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lacgibra t1_ixw20m9 wrote

Okay do me a favour, workout a classical problem considering gravity as curved space time. Simple pendulum, compound pendulum, linear Harmonic oscillator two body problem anything.

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lacgibra t1_ixw2ey4 wrote

I'm simply saying, I apply gravity as space-time curve when I apply the Einstein's relativity in action. If had explained the question in the first scenario using relativity, I would've mentioned gravity as space time curve

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eldude2879 t1_ixw2f7o wrote

well, I am just a simple electrician with some knowledge of Einstein

I admit using Newton is perfectly fine if you dont need super precise

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lacgibra t1_iybna3g wrote

Yeah yeah quantum mechanics is wrong too, because it's foundation was laid by classical mechanics.

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Alfred_The_Sartan t1_ixwbl0l wrote

Is there any reasonable way to use equations to measure the mass of a rogue planet? Like let’s say there’s a brown dwarf that got ejected from its system and is rolling through interstellar space without anything really being able to measurably affect it. Would we have any idea of the mass?

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AstroBryGuy t1_ixxb99l wrote

Sure, via gravitational micro lensing. When the rogue planet or brown dwarf (which is not a planet) passes directly in front of a distant star (as viewed from Earth), the gravity of the rogue planet/brown dwarf will bend the starlight, focusing it towards Earth, causing the distant star to briefly increase in brightness. The amount of brightening depends on the mass of the body.

https://roman.gsfc.nasa.gov/exoplanets_microlensing.html

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JensAypa t1_ixufxao wrote

Gravity depends on the mass of the planet, but also on your distance to the center of the planet. If you're on the ground, that means it depends on the planet's radius. What you say would only work if the planet was the same size as the Earth, which is generally not the case.

Everything attracts everything with gravity, so in theory you would have to calculate the gravity pull from every planet, star, moon, everything. But generally, most of these objects are too far away to really have a significant effect. If you're on the surface of a planet for example, gravity of other planets and moons are completely negligible. We are attracted by the Moon when we're on Earth, but the gravity pull of the Moon is approximately 0.0006% that of the Earth.

But in space, if you're somewhere between Earth and Moon, you'll be attracted by both of them and you'll have to calculate both values to know towards which direction you'll be attracted.

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Sub2PewDiePie8173 t1_ixv889q wrote

So we’re 0.0006% lighter when the Moon is overhead?

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wildgurularry t1_ixvlhrd wrote

Yes, unfortunately not enough to affect the scale when you are weighing yourself.

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Sub2PewDiePie8173 t1_ixvr9d4 wrote

Do you think if Sun was close enough, but it somehow didn’t affect Earth itself, would we be a lot lighter?

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wildgurularry t1_ixvs4ao wrote

Sure, if you brought the sun close enough, you would start to float up towards it instead of being held down to the earth.

Of course, the entire earth would be ripped apart as a result of this, but you would definitely feel light on your feet as everything around you was being destroyed.

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Fushba t1_ixw05xf wrote

You’d also feel light on the rest of your body. A burning amount of it.

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uh-okay-I-guess t1_ixx3zrp wrote

Just to put this in perspective, you are also 0.0006% lighter if you climb 20 meters further from the center of the earth.

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Kered13 t1_ixy6fcj wrote

You're also 0.5% lighter by moving from one of the poles to the equator, because of centrifugal forces and because the Earth bulges at the equator.

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Towelyban t1_ixv9qcx wrote

There are a lot of great detailed explanations here. I'll just provide the simplest one feasible.

In low drag environments, the relationship between gravitational acceleration, time and distance are pretty clear: D = 0.5at^2.

All you need is an object to drop from rest, an object of known length (ie yardstick), and a stopwatch. Once you have time and distance, solving for gravitational acceleration becomes pretty straightforward.

Edit (in case you cant travel to or close to said planet): Alternatively, if one is unable to test within the gravitational field, the centripetal acceleration relationship also works: a = v^2 / r.

One would simply identify an orbiting satellite, mark it's orbiting velocity v and the orbiting radius, r. Gravitational acceleration can be calculated by solving for a. If orbiting radius is unknown, one could simply use the alternate formula: a = 2piv/T, where v is orbiting velocity and T is the amount of time needed to complete one full revolution.

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drhunny t1_ixun1vl wrote

Step 1 is to measure the mass of the planet. This is surprisingly easy. Just need to catch a view of some object in orbit around the planet (like a moon) or passing near the planet (like a comet). If you can a couple of good observations, preferable over several weeks, you can calculate the mass of the planet using newton's laws of motion.

Step 2 is to get a measurement of the diameter of the planet. This is surprisingly hard. I don't recall the methods used. One problem is "what's the definition of the diameter when you're looking at a gas giant? It basically goes from a solid core to a vet thick liquid transitioning to extreme pressure gas. Then the gas pressure, density drop off slowly to zero.

If you know the mass and diameter, it's a simple calculation.

But if you want the gravity of a gas giant at some point down inside the thick atmosphere, you do have to also figure out how much mass is inside your chosen depth.

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mfb- t1_ixxbrz9 wrote

Measuring the mass is almost always done by observing its gravitational influence, so we can skip the mass calculation and just use ratios of distances. We know the gravitational attraction of Earth better than 1 part in a billion, while the mass has an uncertainty of more than 1 in 100,000 because the gravitational constant isn't known that precisely.

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Kered13 t1_ixy6khj wrote

> Step 1 is to measure the mass of the planet. This is surprisingly easy. Just need to catch a view of some object in orbit around the planet (like a moon) or passing near the planet (like a comet). If you can a couple of good observations, preferable over several weeks, you can calculate the mass of the planet using newton's laws of motion.

Measuring the mass of a planet is actually very hard, but observing a moon tells you the gravitational acceleration, which is what we actually want here. So we just cut out mass altogether and calculate surface gravity based on the gravity that we observe pulling on the moon.

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fliguana t1_ixurxma wrote

"gravity" (field) of the planed is determined by its mass, nothing else.

The weight of an object is determined by all gravity fields and acceleration of its motion.

Solar gravity can be detected on earth, because tides are higher during new/full moon events. But acceleration plays even greater role, objects in equator weigh less than those on poles.

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haysoos2 t1_ixwhb16 wrote

Not actually just the mass. The distance between the center of mass to your center of mass has an effect as well. Therefore a dense planet will have a different surface gravity than a planet of the same mass that is built of lower density material.

If you know the diameter of the planet, and its average density it's fairly simple to calculate its surface gravity.

Take the planet's diameter (in miles), multiply by the average density (in g/cm3 or kg/l), and multiply that by 0.0000229

This will give you the surface gravity in G (where Earth's gravity = 1)

Earth has a diameter of about 8000 miles and density of 5.5. Thus, 1 G.

A planet the same size, but made of solid iron (density of 7.9) comes to 1.45 G

An Earth sized planet made entirely of water comes to 0.18 G

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fliguana t1_ixwnv90 wrote

>Not actually just the mass.

Gravity field outside the planet is determined by the mass, nothing else.

Field being the key word. 🧐

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haysoos2 t1_ixwpnks wrote

And the strength of that field is determined by your distance from the center of that mass.

Y'all are making this far more complicated than it needs to be. For pretty much all important considerations, the only number you need to know is a planet's surface gravity. This will govern such things as the escape velocity, and anything actually relevant to living or adventuring on that planet, like how far you can throw a grenade, how high you can jump, or how many puppies you can carry.

My formula will give you a close enough measure of that surface gravity for any planet.

Edit: changed "calculation" to "formula", which is a more accurate description

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fliguana t1_ixwq0dy wrote

>For pretty much all important considerations, the only number you need to know is a planet's surface gravity. This will govern such things as the escape velocity

Incorrect. Two planets with identical surface gravity can have different escape velocities.

Simplification is good to a point, dumbing down leads to errors.

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haysoos2 t1_ixwrzrc wrote

Ok, admittedly escape velocity will be a little different. In my examples above the escape velocity on the iron planet will be 1.7 times higher than on Earth, and on the water planet it will only be about 20% that of Earth's escape velocity, but it's close enough that for say, a fiction based on the Star Wars universe it will be much closer than anything actually used in the series.

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Quantum_Patricide t1_ixx2zxl wrote

If you are talking about surface gravity, and wanted to calculate the surface gravity of a planet in terms of Earth's gravity (g=(9.81ms^-2), then you couldn't just use the masses of the planets, you'd also need the radii. For example if you wanted to calculate the surface gravity of Mars, 0.38g.

The surface gravity of a planet is calculated as (4/3)*pi*G*density*radius, where G is the universal gravitational constant. If you know the density (mass/volume) and radius of Earth and the planet you're standing on, then the ratio of your planet's surface gravity to Earth's is the ratio of the densities multiplied by the ratio of the radii

or: g_planet/g_earth = (p_planet/p_earth) * (r_planet/r_earth)

More generally, the classical gravitational field at a point due to a distribution of masses is

Div(g) = -4*pi*G*density

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Future-Data-8076 t1_ixvpnf9 wrote

Take a rock push it off a table. Measure how long and how far till it hits. Earth gravity is equivalent to 9.8m/s Recreate on other planet.
Any other way would still require you to get the composition or at least be on it.

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Gnidlaps-94 t1_ixvpv0i wrote

Render both the mass and and surface area in multiples of Earth then divide mass by surface area, this will give you surface gravity in multiples of Earth

For example a planet of 8 Earth masses and 5 times the surface area

8/5 = 1.6

Will have a surface gravity of 1.6 Earths (or 1.6 g’s)

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pintofgeodesy t1_ixvxrs2 wrote

Approach 1: create a density model of the planet and integrate(sum) this model over all volume elements assuming every elements contributes as a point mass. For the Earth, the largest non-spherical anomaly comes from its flattened shape which cause satellite to deviate from a pure Kepler orbit. The further away, the more the planet looks like a point mass, and the more satellite orbits approach a Kepler orbit (perfect ellipse).

Approach 2: use satellite orbit or astronomy observations in combination with our knowledge that the Earth's (or other planet) external gravity field obeys the Laplace equation. This allows a gravitational model to be estimated for the external field. However, knowing the external field is not enough to know the internal mass distribution (infinite possibilities still generate the same external gravity field)

Additional contributions from other far objects (moons, other planets) can generally added as additional point masses, and are considered tidal accelerations. This requires knowledge of the position of those objects relative to the planet, which can be taken from a so-called planetary ephemeris.

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Kold91 t1_ixw9hoi wrote

As has already been said to calculate the gravitational attraction of two objects you can use:

GM1M2/r^2

But this only works for two objects you cant add a third object to this equation which means AFAIK we can not predict the attraction of a 3-or-more-body system (also called n-body problem)

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64vintage t1_ixwgfr4 wrote

Theoretically, the gravitational forces that you experience are a contribution of the gravity from every piece of mass in the universe.

This is why, for example, the tides rise and fall with the relative directions of the moon and sun.

But their contribution is tiny compared to that from the earth.

We are in orbit around the sun, so we don’t personally experience it’s gravity as such. Like, if you are in orbit around the earth, you experience zero-G because you are effectively falling freely.

However, it does still have a value, which I believe is about 0.006 m/s^2, compared to the surface gravity of the earth, 9.81 m/s^2. The influence of the earth is 1600 times stronger.

Basically you can calculate the surface gravity of a planet if you know it’s mass and diameter.

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SirNanigans t1_ixx39jr wrote

There's lots of interesting details about planetary mass and radius that others have mentioned but I have to wonder... if the question is simply "how strong is the gravity of this planet" then shouldn't that be measurable by simply examining a single orbiting body's velocity and distance, or just dropping something in vacuum and observing the acceleration, or even just taking your kitchen scale to the planet and putting a calibration weight on it?

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jbeech- t1_ixx8j9t wrote

And I suppose landing on said planet and weighing a known 1Kg bar is out of the question, eh? I can see it now, aliens visit the Earth to purchase 1kg bars expressly for the purpose of having a quick and dirty calculation of G for any planet they visit.

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drillpress42 t1_iy0ake2 wrote

Or, we discover a 1kg bar here with an alien inscription and we then know they know Earth's mass.

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Kered13 t1_ixy6ri4 wrote

Getting the mass of a planet is very difficult and can basically only be done by observing gravity, so it would be circular to use the planet's mass to calculate gravity. Instead you you can observe the orbit of something like a moon around the planet, based on the distance from the planet and time of the orbit you can easily calculate gravity.

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Electronic_Health22 t1_iy271v7 wrote

I think you're referring to the gravitational force exerted on an object by a planet. And in a Newtonian rather than relativistic sense. You can calculate the gravitational force between any 2 objects by F = G*m1*m2/r^2, where G is the universal gravitational constant. This is Newton's formulation. When you're talking about a 2-body system, this is simple.

But when you're talking about something floating in space where position is defined by its relationship to more than 1 object, this becomes more difficult. For example, the moon experiences the gravitational pull of the Earth. But it also experiences the gravitational pull of other massive objects nearby, like other planets. These are other vectors that all act on the object. Fortunately, other than things that are pretty close (astronomically speaking) together, the effect of r^2 is to dilute the effect of gravity out to near zero for things that are relatively far apart, to the point where it's negligible.

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Penis_Bees t1_ixvxuv3 wrote

This is complicated.

The reason we use g=9.81m/s^2 on earth is because for all practical purposes it's good enough. In most practical purposes you can even use g=10 and be fine.

"Gravity" is the sum of all quantities of mass interacting with each other based on their masses and the distance between those masses.

You can also have perceived gravity, where the rotation of a body such as the earth has centripetal forces that counter some gravity. The gravity didn't change compared to if the earth were stationary but it "feels" lower to some degree because of this.

No planet produces a set amount of gravity, it all depends on the mass of the "test object" (yourself) that's you're using to observe the gravity, and the relative distance between the bits of that planets mass and the bits of your mass.

Likewise the gravity acting up on you because of that planet does not necessarily change when a moon is introduced, instead the moon has a separate effect on you, and may effect the planet in a way that effects you.

Technically a car driving down the street has some gravitational effect on you, but it's so small that it's not worth considering. The interaction between you and the planet you stand on is so much stronger than any other heavenly bodies you see in the sky that unless you're planning on orbiting the planet, it's not worth considering.

If you want to know what you would feel on another planet, first decide on what degree of accuracy/precision you care about. If it's not very accurate, just estimate an average planet radius and the over all mass and use the law of gravitation.

If you need it to be very very very very accurate and preciseyou'll need to figure out the planets precise center of mass, the exact radius between your center of mass and the planets at the point you care about, then do the same for every other massive object nearby until the forces are small enough to be outside your desired degree of accuracy. This should still happen pretty quickly.

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eldude2879 t1_ixw7obk wrote

Greenland glacier has so much mass that it drags the sea to it, when its gone, the sea level around Greenland will go down 100 meters

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Your_Agenda_Sucks t1_ixwi7lg wrote

The only statistically relevant numbers you need are the mass of the planet and the radius to the surface.

Take 2 planets with the same mass, one made of light fluffy marshmallow and another made of solid metal. They can both have the same amount of mass but because you're so much closer to the center of gravity on the lead planet, the experienced surface gravitation acceleration will be higher on the lead planet.

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