Mass and gravity question

[wehaveaproblem]

Yo,

What's the basic calculation for how much gravity is created by how much mass? I'm curious of this in regards to how big an asteroid or comet, or anything for that matter, has to be to have practical amounts of gravity with regards to human influence. Specifically, im wondering if a cosmological, or artificial, lump of something could be used as the core of a spaceship for the purposes of creating gravity on that ship.

And don't worry about giving me hard or advanced maths, I'm just after getting a general understanding for some sci-fi creative writing I'm working on. So giving me measures and scales in the way documentaries do will be fine.
"A lump of lead the size of a sports stadium would have 0.2g", or "a solid iron asteroid the size of Ohio would be like walking on the moon".... You know the sort of thing.

Cheers

 

[RisingFury]

Force between two bodies can be calculated using the F = m1 * m2 * G / r^2 equation, where m1 and m2 are the masses of both bodies, G is the gravitational constant and r is the distance between the centers of mass of both bodies.

Keep in mind that gravitational force always acts between two bodies with mass (at least in classical mechanics), so the question "how much gravity" is a bit ambiguous. You can ask yourself how much gravitational acceleration a body with mass causes and the answer is a = M * G / r^2, where M is the mass of the body and r the distance from its center of mass.

Some numbers:
A kilogram is attracted to Earth with a force of 9.81 N.
A 75 kg human on Earth is attracted to Earth with a force of 736 N and so on...

If you're trying to compare how much people would weight on other bodies, compare their gravitational acceleration at ground level with that of Earth. For example, the gravitational acceleration on the surface of the Moon is roughly 1.62 m/s^2, whereas on Earth it's 9.81 m/s^2. A human on the Moon would then feel an attractive force of only 17% that of Earth, weighing roughly as much as a medium size dog on Earth.

 

[orb]

a=G*M/R^2

G=6.67384e-11 [m^3*kg^(-1)*s^(-2)] - Gravitational constant
M - mass of the planet
R - radius of the planet, or radius of orbit where you want to get the gravitational acceleration

If you want the Earth's g ratio then:
g_ratio=a/g

 

[wehaveaproblem]

Cheers gents,

A little follow up before I play with the equations. What's the output unit of measure for these? And what units should I put into them. I'd like to get something out that I can compare to the 1g of earth's gravity. Like I say, I'm only looking to get some ball park figures. I am assuming simple metric is best?

Cheers

 

[ADSWNJ]

Cheers gents,

A little follow up before I play with the equations. What's the output unit of measure for these? And what units should I put into them. I'd like to get something out that I can compare to the 1g of earth's gravity. Like I say, I'm only looking to get some ball park figures. I am assuming simple metric is best?

Cheers

Well... gravity is a force, so the correct units are units for force. For the SI system, that would be Newtons.

If you want something more simple, then you could use kilograms-force (kgf) or pounds-force (lbf). 1 kgf = 9.80665 Newtons, or in simple terms, the force on a MASS of 1kg in standard Earth gravity near the surface is 1kgf of WEIGHT.

---------- Post added at 02:43 PM ---------- Previous post was at 02:11 PM ----------

Yo,

"A lump of lead the size of a sports stadium would have 0.2g", or "a solid iron asteroid the size of Ohio would be like walking on the moon".... You know the sort of thing.

Cheers

Here's a quick calc for your lead example. Per Wiki (the reference in the sky!), Lead's density is 11.34 g/cm3, which is 11.34 x 10^12 kg/km3. Planet Earth is 5.98x10^24kg of mass, and 1.083 x 10^12 km3 of volume. So the Earth's average density is approximately 5.52x10^12 kg/km3.

So good news ... your lead is about twice as dense as the Earth (about 2.054 times, but let's round to 2x).

Hmm - so we want to generate a 0.2g effect from a lump of lead. Well, the Earth does it from 20% of 1.08x10^12 km3 of "earthy stuff". Lead, being 2x as dense, would therefore need 10% x 1.08x10^12 km3 of pure lead to do it. (Nice ... 10% makes the simplification easier!)...

10% x 1.08 x 10^12km3 = 1.08 x 10^11km3 = 108 x 10^9 km3.

So, you will need to put 108 blocks, each 1000km x 1000km x 1000km next to your sports stadium to generate 0.2g.

Quick cross-check: 108 x 10^9km3 at 11.34x10*12 kg/lm3 density = 1.22x10^24 kg, or approximately 20% of Earth.

Isn't the Earth an awesome thing?

 

[RisingFury]

Cheers gents,

A little follow up before I play with the equations. What's the output unit of measure for these? And what units should I put into them. I'd like to get something out that I can compare to the 1g of earth's gravity. Like I say, I'm only looking to get some ball park figures. I am assuming simple metric is best?

Cheers

Mass in kilograms [kg], distance in meters [m], gravitational constant is a bit more complicated [m^3 / (kg * s^2)]. Feed in these numbers and you'll get force in Newtons [N = kg * m / s^2] and acceleration in meters per second squared [m / s^2].

Remember the equation F = m * a? Force equals mass times acceleration? Check the breakdown of the Newton and you'll see where it comes from...

 

[ADSWNJ]

One more ... to relate it back to the stadium size ... 108 blocks, 1000km cubed. Roughly (real dirty), North America is 4000km across by 3000km down (ugh - don't fact check too much). So we can put down 12 blocks and cover the USA up into Canada, then fill in 9 layers deep of that. A 9000km high lead weight across all of North America, to make a fifth the G-force of Planet Earth. Wow-wee.

 

[wehaveaproblem]

Earth. Wow-wee.

Wow-wee indeed. Thanks for that dumbed down answer, much appreciated. It certainly puts my thoughts into perspective!

 

[ADSWNJ]

What you need is some Super-High Unobtanium-like material from la-la land, with a density never seen in our universe before. Maybe with a way to activate and deactivate the mass by applying a ray gun to it. {shrug!}

That way, you could put it in your pocket, take it to the ball game, then zap it to create the effect you wanted.

That would make a cool Earth defense system too ... fire to just miss the Armageddon rock, then zap on the mass to deflect the course, then zap it off again to stop messing with the orbits of the rest of the Solar System!

 

[RisingFury]

What you need is some Super-High Unobtanium-like material from la-la land, with a density never seen in our universe before. Maybe with a way to activate and deactivate the mass by applying a ray gun to it. {shrug!}

I don't know. There are some pretty high densities in the universe. Water has a density of 1000 kg / m^3, (I won't be writing out units anymore) rock is around 3000, metals are between (roughly) 2000 and 20 000. That's what normal pressure and temperature can do. But that's not much.

Cores of stars can be in the 100 000 to (not quite) 1 000 000 range, white dwarfs are upwards of 1 000 000 = 10^6, neutron stars are in the 10^17. It's hard to talk about the density of black holes, because the larger they are, the less dense they are. That being said, the singular region is infinitely dense... derp.

That way, you could put it in your pocket, take it to the ball game, then zap it to create the effect you wanted.

That would make a cool Earth defense system too ... fire to just miss the Armageddon rock, then zap on the mass to deflect the course, then zap it off again to stop messing with the orbits of the rest of the Solar System!

Yes, zap it on, zap it off., deflect the asteroid... but what happens to all other asteroids? What happens to the orbit of the Moon and Earth? Don't screw around with gravity!

 

[wehaveaproblem]

My original thoughts were triggered by a Baxter novel set on a big old lump of space rock, which created some gravity for those on it. So I was curious as to the real maths and possibility of such a thing. But it does seem you basically need a small moon made of lead for it to have any real possible application. And attaching engines to such a thing for use as a ship seems somewhat like attaching an outboard motor to a continental land mass lol.

But if were talking high scifi, Perhaps some sort of unobtanium CERN collider space ship could create and maintain a tiny black hole in its core.

 

[Ares]

Hmm - so we want to generate a 0.2g effect from a lump of lead. Well, the Earth does it from 20% of 1.08x10^12 km3 of "earthy stuff".

It's a bit better than this in reality because taking 20% of the Earth also reduces it's radius. You can get closer to the centre so the gravitational acceleration is higher than just 0.2g, more like 0.6g (assuming the original Earth and the 20% chunk of Earth are the same density). Mars has 0.15 the volume of Earth but 0.36g at the surface.

Maybe the tiny black hole is the way to go

 

[RisingFury]

It's a bit better than this in reality because taking 20% of the Earth also reduces it's radius. You can get closer to the centre so the gravitational acceleration is higher than just 0.2g, more like 0.6g (assuming the original Earth and the 20% chunk of Earth are the same density). Mars has 0.15 the volume of Earth but 0.36g at the surface.

That's correct. Density matters more when calculating surface acceleration than mass of the object. As you said, a more massive object might also be a lot larger. The two effects could cancel each other out to produce roughly the same gravitational acceleration as Earth. Same with smaller objects.

 

[wehaveaproblem]

Cheers for that reminder Ares/RF, I think I may have to have a play with the maths before I give up on the idea completely.

 

[dgatsoulis]

Here you go WHAP.

A calculator to help you speed things up.

It assumes spherical objects and returns you the gravitational acceleration in m/s² on the object's surface and a comparison to Earth's g.

All you have to do is enter the density and radius of the object.

Next to it you will find a small table with the densities of some meterials/elements.

Remember that the decimal is the comma (,) not the dot (.)
Don't touch any of the numbers in the red area, that's where the calculation is made.

 

[wehaveaproblem]

Nice one dg, I'll be sure to have a play with that. Thanks a lot.