How do you calculate gravity

[Everten P.]

Of a celestial body how do you calculate the amount of gravity?

[Nibb31]

This link:

http://lmgtfy.com/?q=how+to+calculate+gravity

[shynung]

You need to find out the mass of the body in question; that's the only thing that matters.

After that, if you know your own mass, and your current position from the object's center-of-mass, plug it into this equation:

Where:

F = gravitational force currently acting on you (shaping your orbit/trajectory), in Newtons.

G = gravitational constant (6.67×10^{−11} N·(m/kg)²).

m1 & m2 = object mass & your mass (swappable), in kilogram.

r = distance between you and center-of-mass of object (core-to-surface radius if landed, radius+orbital height if in orbit), in meters.

[K^2]

Note, these equations assume a roughly spherical body of reasonable density. Essentially, larger moons, planets, stars. That sort of thing. If you need to compute surface gravity on an asteroid of irregular shape or surface of a neutron star, you have to get creative.

[YNM]

If you need to compute surface gravity on an asteroid of irregular shape or surface of a neutron star, you have to get creative.

For asteroids, can one assume something of the same mass at the average radius ? And why would it be hard for neutron stars, would it be the same for things like white dwarfs and blackhole (at schwarzschild radius) ? I mean, there's like no limit on how many acceleration you can have (unlike velocity), right ?

[xmaslightguy]

So does this equation mean that heavier objects technically do fall faster, since the gravitational pull would be greater?

[K^2]

So does this equation mean that heavier objects technically do fall faster, since the gravitational pull would be greater?

No. The pull is proportional to the mass, but so is the amount of force required to accelerate a heavier object. So the acceleration of a falling object is exactly the same regardless of the mass.

For asteroids, can one assume something of the same mass at the average radius ? And why would it be hard for neutron stars, would it be the same for things like white dwarfs and blackhole (at schwarzschild radius) ? I mean, there's like no limit on how many acceleration you can have (unlike velocity), right ?

No, unfortunately, you can't just assume an average radius. When you are far enough away, from the mass distribution, it behaves a lot like a point object + some corrections. But on the surface, any changes in elevation will cause differences. The gravity's pull doesn't even have to be "straight down" everywhere, either. Center of the rock can be right bellow your feet, yet you might be pulled sideways because there is a larger lump of mass to the side. You really have to sit down and carry out the integration to compute the field at each point.

At the surface of the neutron star, the gravity is strong enough for Newtonian estimates to be far off. You have to use formulas from General Relativity. So long as the source is spherical, the formulas aren't that much more complicated, but there are some additional factors. A rotating neutron star will have a different gravitational pull than a static one. Magnetic field also plays a huge difference. In fact, magnetic field is the only reason some massive neutron stars don't collapse into black holes. So that should give you some idea of how much difference that makes.

[YNM]

No, unfortunately, you can't just assume an average radius. When you are far enough away, from the mass distribution, it behaves a lot like a point object + some corrections. But on the surface, any changes in elevation will cause differences. The gravity's pull doesn't even have to be "straight down" everywhere, either. Center of the rock can be right bellow your feet, yet you might be pulled sideways because there is a larger lump of mass to the side. You really have to sit down and carry out the integration to compute the field at each point.

Hmm, yeah, I thought I read somewhere that making a probe orbiting an irregular object is hard, becausr of the varying gravitational pull. This is also the reason why an oblate Earth will have some special sun-synchronous orbit, no ?

At the surface of the neutron star, the gravity is strong enough for Newtonian estimates to be far off. You have to use formulas from General Relativity. So long as the source is spherical, the formulas aren't that much more complicated, but there are some additional factors. A rotating neutron star will have a different gravitational pull than a static one. Magnetic field also plays a huge difference. In fact, magnetic field is the only reason some massive neutron stars don't collapse into black holes. So that should give you some idea of how much difference that makes.

Is it fully because of general relativity, or because of the total sum of accelerations (ie. gravitational pull + centrifugal force + magnetic force) ? Maybe you can post some links ?

[K^2]

This is also the reason why an oblate Earth will have some special sun-synchronous orbit, no ?

Indeed. The low order correction to orbital motion around spheroid is orbit precession. With the right choice of orbit, you can match rate of precession to the Earth's motion around the Sun, resulting in a Sun-synchronous orbit. But this still primarily works because Earth is a fairly symmetrical object.

Is it fully because of general relativity, or because of the total sum of accelerations (ie. gravitational pull + centrifugal force + magnetic force) ? Maybe you can post some links ?

One of the cool things about General Relativity is that it completely covers effects of non-inertial frames of reference. That includes centrifugal force. The magnetic field's contribution in neutron star is primarily to the total energy. Yeah, these fields are strong. Ultimately, though, these are effects of General Relativity. Gravity just doesn't work the way you normally expect when you are that close to something that dense. All orbits end up wonky, and the way tidal forces and gravity vary with radius is a bit different. Worse, the pull of gravity depends on how you are moving. It takes an infinite amount of force to keep an object at event horizon. The tidal force, then, is also infinite. Yet, if you free-fall into a black hole of sufficient size, tidal forces can be quite manageable.

Unfortunately, almost any source I can point to assumes that you can work out gravity from the metric, and only list metrics. But just as an example, if you start with Schwarzschild Metric, which is valid for any spherical object of mass M with no significant charge, magnetic field, or angular momentum, you can derive surface acceleration due to gravity to be:

GM/(R^3/2 Sqrt(R - 2GM/c²))

The quantity R_S = 2GM/c² is the Schwarzschild radius. That's where event horizon of the black hole is going to be. It's easy to see that if R = R_S, surface acceleration is infinite. However, if R is much, much larger than R_S, then the quantity under the square root is essentially just Sqrt®, and formula reads GM/R², which is your typical Newtonian acceleration due to gravity formula. The interesting things happen when R > R_S, but not too much greater. Surface of a neutron star is such a place.

If you add angular momentum, electrical charge, or strong magnetic fields, things get even more interesting.

[YNM]

I was told by someone that teaches me orbital motion is that the keplerian and newtonian models are actually not right. The keplerian model fails at binary stars. She also wanted to tell me the relativity model (she said that Einstein didn't like the newtonian model because it oversimplifies a lot of things so there's relativity), but then the time was over. So, thank you for the explanation.

(additionally, could this be the reason for "dark matter"... ?)

[Creature]

No. The pull is proportional to the mass, but so is the amount of force required to accelerate a heavier object. So the acceleration of a falling object is exactly the same regardless of the mass.

Sorry if this is a silly question but the force acts both bodies separately, even though it's equal for both directions? If I'm regular sized me floatin in space, Earth pulls me at 9,81 m/s^2 and I pull the Earth at essentially zero force upwards. If I were the size of Earth, it would still pull me at 9,81 m/s^2 but I'd pull the Earth at the same acceleration towards me so while the acceleration caused by Earth on me is the same in both cases, in second case I would collide faster because the falling speed of Earth would be bigger due to my enormous mass? Or is this already taken into account somehow?

[YNM]

Sorry if this is a silly question but the force acts both bodies separately, even though it's equal for both directions?

The force on both bodies have the same magnitude. Just that because F = m*a you body receives more acceleration than Earth, to compensate for your small mass compared to Earth. And so, the Earth barely moves when you jump - you're coming back to it, not the Earth comes back to you.

What matters is mass and distance from CoM of both bodies. Of course it's not really like that if you're on an irregularly shaped bodies or your shape is a thin, very long rod (like the gravitational tethers some are proposing), but it's the same more or less.

Edited July 6, 2014 by YNM

[Creature]

Well I'm not sure if I'm expressing myself all too well (english not my first language).

What I mean is that let's say I'm teleported so that I'm floating in space next to Earth. I start to experience the gravitational pull F=G m(me)m(Earth)/r^2. For my acceleration F=m(me)a where F is the gravitational pull. So m(me)a = G m(me)m(Earth)/r^2. m(me) cancels out which leads to gravitational acceleration g (roughly 9,81 m/s^2 on sea level). So that's why the acceleration is independent of my mass because it cancels out.

But when I appear, I start to pull the Earth towards me with the equal force F=G m(me)m(Earth)/r^2. Now the m(Earth) cancels out and the Earth accelerates at a=G m(me)/r^2 which is pretty small. But if you substitute the regular old me with a super-sized me that's the size of Earth, then suddenly my surface acceleration is also 9,81 m/s^2. Now the Earth starts to "fall" towards me at a=9,81 m/s^2 (independent of Earth's mass, this is me pulling Earth).

So the point was that in first case I fall at 9,81 m/s^2 towards Earth and it falls at me with essentially zero acceleration.

In second case we both have the same acceleration towards each other, so if you're caught between the Earth and super-me, you'll have the planet rushing towards you at 9,81 m/s^2 acceleration and a huge scared-looking bearded guy from the other direction at 9,81 m/s^2.

So doesn't the collision happen twice as early in case 2 than in case 1? So the both masses do matter, it's just irrelevant unless the masses are in the same order of magnitude. Or am I misunderstanding something?

EDIT: Of course not taking into account the r^2 factor here when I'm tossing the 9,81 m/s^2 around but I don't think it's relevant to the actual question.

[YNM]

No, in the case that suddenly you have the mass of Earth, the acceleration is different. Just to give some figures, Earth mass is approx. 6*10^(24) kg, G approx. 6.67*10^(-11) N m^2 kg^(-2) , and I assume that the distance (yeah, it's actually distance, not radius) between your CoM and Earth's CoM is precisely 5000 km (even while that won't be possible actually).

F(grav) = (G*(m earth)^2)/(d^2) = 9.6048*10^(25) N

a = F/m = 16.008 m s^(-2)

So, both Earth and your Earth-massed body attracted pretty fast, with a relative acceleration of twice the value (observer have the same acceleration with the object).

Edited July 7, 2014 by YNM

[K^2]

Creature, you seem to be confusing force and acceleration.

If Earth is pulling me down with a force of 650N, then I'm pulling on the Earth with the force of 650N. That's Newton's 3rd Law in action. However, if I jump off a chair, these 650N will cause me to accelerate downwards at 9.8m/s². But Earth is really massive, so the same 650N aren't going to make much of a difference.

[Creature]

So I don't want to soud argumentative, but:

No, in the case that suddenly you have the mass of Earth, the acceleration is different. Just to give some figures, Earth mass is approx. 6*10^(24) kg, G approx. 6.67*10^(-11) N m^2 kg^(-2) , and I assume that the distance (yeah, it's actually distance, not radius) between your CoM and Earth's CoM is precisely 5000 km (even while that won't be possible actually).

F(grav) = (G*(m earth)^2)/(d^2) = 9.6048*10^(25) N

a = F/m = 16.008 m s^(-2)

So, both Earth and your Earth-massed body attracted pretty fast, with a relative acceleration of twice the value (observer have the same acceleration with the object).

This is all fine, except you're calculating at roughly 1400 kilometers below the surface. Plug in the correct 6 400 ~ish kilometers for Earth's radius and you arrive at a more familiar value. And no matter how much you change the second mass, it always cancels out. Like I already calculated in abstract form and this is also what K^2 said earlier about the falling body's mass not affecting the acceleration caused by gravity, only the mass of the second body affects it due to elimination.

Creature, you seem to be confusing force and acceleration.

If Earth is pulling me down with a force of 650N, then I'm pulling on the Earth with the force of 650N. That's Newton's 3rd Law in action. However, if I jump off a chair, these 650N will cause me to accelerate downwards at 9.8m/s². But Earth is really massive, so the same 650N aren't going to make much of a difference.

Again I don't want to sound argumentative but to quote myself with some added bolds for emphasis:

What I mean is that let's say I'm teleported so that I'm floating in space next to Earth. I start to experience the gravitational pull F=G m(me)m(Earth)/r^2. For my acceleration F=m(me)a where F is the gravitational pull. So m(me)a = G m(me)m(Earth)/r^2. m(me) cancels out which leads to gravitational acceleration g (roughly 9,81 m/s^2 on sea level). So that's why the acceleration is independent of my mass because it cancels out.

But when I appear, I start to pull the Earth towards me with the equal force F=G m(me)m(Earth)/r^2. Now the m(Earth) cancels out and the Earth accelerates at a=G m(me)/r^2 which is pretty small. But if you substitute the regular old me with a super-sized me that's the size of Earth, then suddenly my surface acceleration is also 9,81 m/s^2. Now the Earth starts to "fall" towards me at a=9,81 m/s^2 (independent of Earth's mass, this is me pulling Earth).

So the point was that in first case I fall at 9,81 m/s^2 towards Earth and it falls at me with essentially zero acceleration.

Which at least in my mind sounds exactly what you just said? Or could you be more precise as to what you think I'm confusing with what? As I said english isn't my first (or second) language so maybe I'm not expressing myself the way I think I am?

But my point was that if my mass is equal to Earth's, the gravitational force between me and Earth is increased, leading to Earth actually undergoing a significant acceleration towards me while still not affecting my own acceleration towards Earth in any way. So if you're watching from a stationary frame of reference at a third point (again ignoring the fact that the actual force depends on distance due to being irrelevant towards this problem):

Case 1, Earth and normal me: I accelerate towards Earth at 9,81 m/s^2, Earth stays essentially still. Impact at t= X seconds.

Case 2, Earth and Earth-sized me: I accelerate towards Earth at 9,81 m/s^2, Earth accelerates towards me at 9,81 m/s^2. Impact at t= X/2 seconds.

Right or wrong?

So to revisit the original question of

So does this equation mean that heavier objects technically do fall faster, since the gravitational pull would be greater?

If my above statement is right, then this is essentially correct too, although we need to be talking about bodies around the same order of magnitude for it to be of any practical meaning.

[dbmorpher]

The quick way is to drop a 1 kg ball and measure the acceleration.

[YNM]

But my point was that if my mass is equal to Earth's, the gravitational force between me and Earth is increased, leading to Earth actually undergoing a significant acceleration towards me while still not affecting my own acceleration towards Earth in any way. So if you're watching from a stationary frame of reference at a third point (again ignoring the fact that the actual force depends on distance due to being irrelevant towards this problem):

Case 1, Earth and normal me: I accelerate towards Earth at 9,81 m/s^2, Earth stays essentially still. Impact at t= X seconds.

Case 2, Earth and Earth-sized me: I accelerate towards Earth at 9,81 m/s^2, Earth accelerates towards me at 9,81 m/s^2. Impact at t= X/2 seconds.

I realized the problem : the true acceleration of an object due to Earth's gravity stays the same. What changes is the Earth's true acceleration due to the object's gravity.

Fgrav = Fgrav

(G*(m Earth)*(m Object))/(d^2) = m*a

You can set the mass into either the object's mass or the Earth mass to obtain each object's (either Earth or the object) true acceleration, but in reality, both object have acceleration towards each other counted from an observer at an infinite distance, and so, if you take the reference frame so that either the Earth or the object is stationary, the apparent acceleration of the other object is the sum of both object's "true" acceleration. Additionally, this is why the keplerian model of orbital motion fails when both object have nearly the same mass - look at binary stars.

It's not necessarily half the time, the relation between acceleration and time is an exponential function.

Edited July 7, 2014 by YNM

[Creature]

I realized the problem : the true acceleration of an object due to Earth's gravity stays the same. What changes is the Earth's true acceleration due to the object's gravity.

Fgrav = Fgrav

(G*(m Earth)*(m Object))/(d^2) = m*a

You can set the mass into either the object's mass or the Earth mass to obtain each object's (either Earth or the object) true acceleration, but in reality, both object have acceleration towards each other counted from an observer at an infinite distance, and so, if you take the reference frame so that either the Earth or the object is stationary, the apparent acceleration of the other object is the sum of both object's "true" acceleration. Additionally, this is why the keplerian model of orbital motion fails when both object have nearly the same mass - look at binary stars.

It's not necessarily half the time, the relation between acceleration and time is an exponential function.

Yes this is what I was after, thank you for taking the time to deal with my questions For some very odd reason orbital mechanics and gravitational forces haven't really been even looked at on any of the physics courses I've studied. I should probably get a proper text book on the subject so I don't have to figure out everything on my own