Calculating gravity using radius and avg. density?

[NoXion]

Is it possible to calculate a planet's surface gravity this way? Ideally I'd like a result in Gs, but I have no idea what formulas to use, hence my Google-fu is weak. Any help would be greatly appreciated.

[Captain Seafort]

I don't know about the equations, but try using this, and guesstimate surface gravity until you get the required density.

[Terralthra]

Is it possible to calculate a planet's surface gravity this way? Ideally I'd like a result in Gs, but I have no idea what formulas to use, hence my Google-fu is weak. Any help would be greatly appreciated.

Yes, it is.

Volume of a sphere = 4/3 * pi*r^3
Mass of a sphere = volume * average density
Gravity = G*(m1*m2)/r^2

That Gravity is a force, which is divided by the mass it's acting on (m2) to get a constant acceleration, F=ma.

Have fun.

[NoXion]

Yes, it is.

Volume of a sphere = 4/3 * pi*r^3
Mass of a sphere = volume * average density
Gravity = G*(m1*m2)/r^2

That Gravity is a force, which is divided by the mass it's acting on (m2) to get a constant acceleration, F=ma.

Have fun.

Forgive my ignorance, but what do m1 and m2 stand for? I also take it G stands for the gravitational constant?

[Eternal_Freedom]

m1 is the mass that's pulling, m2 is the mass that's being pulled

That isnt set in stone, they are just the two masses, but in the case of planets thats the easiest way to think of it I find

[Terralthra]

Yes, it is.

Volume of a sphere = 4/3 * pi*r^3
Mass of a sphere = volume * average density
Gravity = G*(m1*m2)/r^2

That Gravity is a force, which is divided by the mass it's acting on (m2) to get a constant acceleration, F=ma.

Have fun.

Forgive my ignorance, but what do m1 and m2 stand for? I also take it G stands for the gravitational constant?

m1 and m2 are the two masses. You only need m1 to calculate the acceleration of gravity at a planet's surface, since m2 factors out when converting force to acceleration.

m1 is the mass that's pulling, m2 is the mass that's being pulled

That isnt set in stone, they are just the two masses, but in the case of planets thats the easiest way to think of it I find

Eh....both masses pull equally on each other. In the case of a planet and an object on that planet's surface, the force the object exerts on the planet is negligible.

[Dave]

So, working out a problem just for fun:
The average density of the earth is 5.519 g/cm3[1]
The radius of the earth is 6 378.1 kilometers[2]

Thus the radius of the earth is 637,810,000 cm. (km -> cm conversion)

Volume of a sphere being: 4/3 * pi*r^3
The volume of the earth is thus 4/3 * pi*637,810,000^3 or 1.08683241 × 10^27 cm^3
Multiplying by density gives us a mass of 5.519 g/cm^3 * 1.08683241 × 10^27 cm^3 or 5.99822808 × 10^27 grams
Converting grams to kg gives us, of course, 5.99822808 × 10^24 kg

The ISS weighs 375,727 kg and the lowest point of its orbit is 347 km above mean sea level. [3]

The gravitational force exerted on the ISS is given by G*(m1*m2)/r^2. In our terms, that's G * (mass of earth in kg * mass of ISS in kg) / (earth's radius in meters + height of orbit in meters)^2. (remember that the masses are treated as points, and r is the distance between the points.)

G * ( 5.99822808 × 10^24 kg * 375727 kg ) / (6378100 meters + 347000 meters)^2 gives us a force of 3 325 209.5 newtons. Handling significant digits, that 3,320,000 newtons

Re-plugging in the mass of the ISS to do an idiot check against the acceleration, we get 8.85 m/s^2 (on earth at sea level it's 9.81m/s^2), so we're in the right ballpark.

[1] http://hypertextbook.com/facts/2000/Kat ... ucci.shtml (note that this was actually calculated using the mass and radius of the earth, and we're just converting back...)
[2] http://www.google.com/search?q=radius+of+the+earth
[3] http://en.wikipedia.org/wiki/ISS

[Dave]

And if you wanted your acceleration in G's, just take the resulting acceleration and divide it by Earth's acceleration, 9.81 m/s^2.

[Steel]

I'm not sure if it is worth mentioning it, but I'll try anyway.

You can treat an object as a point mass with all its mass concentrated at its centre of mass only if the objects you are dealing with are 'outside'* each other. That is, you can treat an object as a point mass if you place a bounding sphere centred on the objects centre of mass which contains all of the object and that sphere does not intersect with the bounding sphere of the other object you are considering. Otherwise you'll need to do some thinking.

As an example, imagine we bore a (thin) tunnel through the centre of the earth from one side to the other. If we treat the earth as a point mass at its centre, then we would expect that the force (and therefore acceleration) of a rock we drop down the hole would increase to infinity as it falls down the hole towards the centre of mass: r is tending to 0 and gravitational force goes as 1/r2. However actually the force on the rock will go as r (no, not 1/r either), that is: the force will decrease linearly to zero. Now we can be pretty sure that zero is not infinity, so something is wrong with the method.

There is a theorem which shows that if you have a spherical shell of matter (ie the surface of a hollow sphere) then that shell exerts no force on an object inside the shell, no matter where the object is. If we apply this to the earth tunnel problem, then we see that we can treat the portion of the earth above the rock as it drops as a series of spherical shells that we are inside and hence there is no contribution from any of the mass of the earth above us, only the remaining spherical core below. We know that mass goes as r3, and gravity goes as r-2 and hence overall gravity will go as r while we are inside the earth, and then once we move outside it gravity will drop off with r-2 as normal.

*Note: the reason I put 'outside' in quotes earlier is because it is not sufficient to just be 'not inside' a solid body, as if we consider a doughnut or torus shape the centre of mass of the torus is in the middle of the hole, and again as you approach the centre of mass the force the torus exerts on you drops to zero rather than increasing to infinity -you can see this from symmetry rather than because of the shell argument as in the other case.

[Simon_Jester]

[cracks knuckles]

To answer the OP, yes, this can be done, as a fairly straightforward exercise in algebra; it's the sort of thing I'd expect to see as an exam question in a freshman physics class. For an answer, I'm going to try and use notation as readable as possible.

To get magnitude of gravitational force, Newton's Law of Gravitation is

F = (G*M*m)/(r2)

As in, gravitational constant G (you can look this up), mass of the two bodies (M and m), and the distance between them (r).

But you want acceleration due to gravity, which is effectively force-per-unit-mass: meters per second squared, or if you prefer g's. In that case, your mass cancels out (since you feel the same acceleration due to the planet's gravity no matter how light or heavy you are). So for acceleration, we plug in Newton's Second Law, F = ma, and get:

a = (G*M)/(r2)

That's acceleration due to an object of mass M, at distance r.

Now, suppose we are on a planet of radius R and density D (we'd normally use the Greek letter rho, but that's awkward in a forum post). We can get the mass from the volume and density. And since the planet is a sphere, we can get the volume from the radius easily:

M = D*V = D*[(4/3)*pi*(R3)]

Standing on the surface of the planet, you plug this into a:

a = [G*D*(4/3)*pi*(R3)]/(R2)

a = (4/3)*pi*G*D*R

That will give you your answer in metric units, if you look up G and know D and R.

Divide said answer by 9.8 to get the answer in g's.