gravity stacking

[514void]

would something as massive as the earth fall at g or 2g towards the earth?

[Mordred]

Every point mass attracts every single other point mass by a force pointing along the line intersecting both points. The force is proportional to the product of the two masses and inversely proportional to the square of the distance between them

 

now think of your question and ask yourself at what distance? then use the formula below

 

[latex]F = G \frac{m_1 m_2}{r^2}[/latex]

 

F is the force between the masses,
G is the gravitational constant, 6.67×10⁻¹¹ N·(m/kg)² or if you prefer one of the other forms see http://en.wikipedia.org/wiki/Gravitational_constant
m1 is the first mass,
m2 is the second mass, and
r is the distance between the centers of the masses

 

Earths mass is

5.97219 × 10²⁴ kg don't forget to convert the units

1 g is 9.80665 m/s² (at the Earths surface) in newtons units its determined by [latex]f=mg[/latex] 1 newton=1 kg*m/s²

On Earth's surface, a mass of 1 kg exerts a force of approximately 9.81 N

 

(a rough calculation will show there is 19.82*10¹⁹ newtons of force between the Earth and the moon.)

 

I'll leave it up to you to calculate how far a body the size of the Earth would have to be from the Earth to equal 2 g

Edited June 9, 2014 by Mordred

[514void]

umm, like at sea level or something?

[Mordred]

a 1 kg object would fall at sea level to the center of mass of the Earth at 1 g. An object the mass of the Earth would fall towards the Earth at 2g much farther out than the radius of the moon.

 

1 g is 9.80665 m/s² in terms of acceleration

 

in terms of force

Fgrav = m * g or force=m*a
where g = 9.8 N/kg (on Earth) at sea level

and m = mass (in kg)

 

one of the fastest meteors on record hit the Earths atmosphere at 28.6 kilometers per second. this is far greater

than 2 g. with an estimated mass of roughly 40,000 kg.

 

http://blogs.scientificamerican.com/observations/2012/12/20/california-meteor-broke-speed-record-for-atmospheric-entry/

Edited June 9, 2014 by Mordred

[Spyman]

Speed is not equal to acceleration, heavier objects doesn't accelerate more towards the center of the Earth and meteors don't get their high speed from being pulled by Earth's gravity - normally they orbit the Sun and crosses path/collide with Earth.

 

Meteoroids travel around the Sun in a variety of orbits and at various velocities. The fastest ones move at about 42 kilometers per second through space in the vicinity of Earth's orbit. The Earth travels at about 29.6 kilometers per second. Thus, when meteoroids meet Earth's atmosphere head-on (which only occurs when meteors are in a retrograde orbit such as the Eta Aquarids, which are associated with the retrograde Halley's Comet), the combined speed may reach about 71 kilometers per second. Meteoroids moving through Earth's orbital space average about 20 km/s.

http://en.wikipedia.org/wiki/Meteoroid#Meteoroids_in_the_Solar_System

 

Nasa image of the orbit for 2008 TC3, in your article link:

Orbit Diagram of Asteroid 2008 TC3

http://neo.jpl.nasa.gov/news/news159.html

 

----------

 

If we would bring another Earth sized planet to sea level then they would already be touching each other and no longer be in free fall.

But if they are still slightly separated, then they would both accelerate towards their center of mass with 1/4 g.

 

If we would bring a Black Hole with the mass of Earth to sea level then it would accelerate towards Earth's center with 1 g.

However it would rip apart Earth since Earth's surface would acclerate towards the BH with 515 000 000 000 000 000 g.

[xyzt]

would something as massive as the earth fall at g or 2g towards the earth?

[math]a=\frac{GM}{d^2}[/math] where [math]d[/math] is the distance between the centers of the two attracting bodies.

 

1. For a test probe mass (negligible radius) situated on the surface of the Earth [math]d=R[/math] so:

 

[math]a_1=\frac{GM}{R^2}[/math]

 

2. For two "Earths" touching each other [math]d=2R[/math] so:

 

[math]a_2=\frac{GM}{(2R)^2}=\frac{a_1}{4}[/math]

 

Spyman has ninja'd me.

Edited June 9, 2014 by xyzt

[Mordred]

Yeah I must have been half asleep I was defining the force of gravity which isn't the same as g force
long work day wasn't thinking right lol (mind you the beer didn't help much)

Edited June 9, 2014 by Mordred

[Janus]

would something as massive as the earth fall at g or 2g towards the earth?

Okay, let's assume you have an object both as massive as the Earth and small enough that you could drop it from one meter above the Earth's surface. How fast will it fall?

 

It really depends on how you define its "fall".

 

The force of gravity is equal to

 

[math]\frac {GM_eM_2}{d^2}[/math]

 

where Me is the mass of the Earth and M2 the mass of the object.

 

Acceleration is found by F=ma, and in this case m= M2

 

If we substitute the first equation for F in the second, we find that M2 cancels out, and thus the acceleration of our object is independent of its mass. In this sense, the object falls at 1g.

 

However, this only half the scenario, the Earth itself is also subject to acceleration due to gravity, so just as the object falls towards the Earth at 1g, the Earth falls towards the object at one G. What is happening is that both are falling towards the barycenter between them. If you are standing on the Earth and using the Earth as your reference frame, it would appear the the object is falling at 2g with respect to the Earth.

In fact, every time you drop an object, the Earth also falls towards it. However, in real life, the difference in mass between earth and object is so great that the Earth's rate of fall is just to tiny to take notice of.

 

The same type of thing occurs when we consider orbits. As long as one of the bodies is very small compared to the other, we can ignore its mass when calculating its orbit, but once the masses become comparable to each other, we have to factor it in to get a proper result.

 

For example: If we find the orbital period of the Moon with:

 

[math]T= 2 \pi \sqrt{\frac{d^3}{GM_e}}[/math]

 

ignoring the mass of the Moon, we get 2364668 sec

 

However, in order include the mass of the moon, we need to use

 

[math]T= 2 \pi \sqrt{\frac{d^3}{G(M_e+M_m)}}[/math]

 

and get an answer of 2350205 sec, which is ~4 hrs shorter.

 

The same comparison made for the ISS only accounts for a 9e-18 sec difference.

 

4 hrs represents a difference of position of roughly 14.400km(some 4 times its diameter) for the Moon, while 9e-18 sec represents a position difference of less than 1/1,000,000 of a nanometer for the ISS.

 

So in the first case it might pay to consider the mass of the orbiting object, while in the second case, it doesn't.

[md65536]

[math]a=\frac{GM}{d^2}[/math] where [math]d[/math] is the distance between the centers of the two attracting bodies.

I think this answers the question as I interpret it. If you double the total mass, the relative acceleration will double, regardless of how the additional mass is distributed between the two bodies. Basically, a 2nd Earth mass would fall toward a "fixed location Earth" at the same rate as any other object would, but the Earth isn't fixed and would also fall toward the other equal mass, doubling the total acceleration.

 

A lot of the answers in this thread are assuming an Earth-sized object, even though only mass is mentioned, and I don't think the question intended to allow for the location of the masses to be moved to account for different sizes.

Edited June 10, 2014 by md65536

[ydoaPs]

Every point mass attracts every single other point mass by a force pointing along the line intersecting both points. The force is proportional to the product of the two masses and inversely proportional to the square of the distance between them

 

now think of your question and ask yourself at what distance? then use the formula below

 

[latex]F = G \frac{m_1 m_2}{r^2}[/latex]

 

F is the force between the masses,

G is the gravitational constant, 6.67×10⁻¹¹ N·(m/kg)² or if you prefer one of the other forms see http://en.wikipedia.org/wiki/Gravitational_constant

m1 is the first mass,

m2 is the second mass, and

r is the distance between the centers of the masses

 

Earths mass is

5.97219 × 10²⁴ kg don't forget to convert the units

1 g is 9.80665 m/s² (at the Earths surface) in newtons units its determined by [latex]f=mg[/latex] 1 newton=1 kg*m/s²

On Earth's surface, a mass of 1 kg exerts a force of approximately 9.81 N

 

(a rough calculation will show there is 19.82*10¹⁹ newtons of force between the Earth and the moon.)

 

I'll leave it up to you to calculate how far a body the size of the Earth would have to be from the Earth to equal 2 g

Note that here F=ma, so when calculating the acceleration, mass cancels out. So, each is accelerated in opposite directions at the same magnitude.

[Mordred]

I've already realized I did the wrong set of calculations, above lol. (forgot the misnomer)

I wish to just add the related G_force calculations. of a falling object impact. (older textbooks) now its properly called impact force

Even though the OP is referring to g -force in terms of acceleration so the above posts by others is correct. (this is just in terms of g-force of a falling object at impact)(or sudden deceleration), just for informational purposes. of the older formulas once used

 

(the above answers by others is correct for the OP)

 

funny part is this is what you would find in an engineering safety data sheet. However G-force is a misnomer the correct definition of g-force is

g-force A force acting on a body as a result of acceleration or gravity, informally described in units of acceleration equal to one g

One g is the acceleration due to gravity at the Earth's surface and is the standard gravity (symbol: gn), defined as 9.80665 metres per second squared, or equivalently 9.80665 newtons of force per kilogram of mass.

 

some older textbooks still have whats below for example my year published 1923 physics textbook. or sites as follows

 

http://hyperphysics.phy-astr.gsu.edu/hbase/carcr2.html Gforce (due to deceleration of a car in the example)

 

down below is from one of my older engineering books

 

velocity upon impact is give by

 

v = (v^{_o 2} +2gs)^1/2

v= velocity upon impact (ft/s)
v_o= initial velocity (ft/s)
g= acceleration due to gravity (32.2 ft/s²)
s= distance of the fall (ft)

 

rate of deceleration a

[latex]a=\frac{v^2}{2d}[/latex]

 

a= rate of deceleration (ft/s²)
v= the velocity at the point of impact (ft/s)
d= deceleration distance (ft)

 

G_force conversion of rate of deceleration to G_force

 

[latex]G_{force}=\frac{a}{g}[/latex]

 

force of impact f_i

 

[latex]f_i=\frac{Wa}{g}=Wg[/latex]

 

F_i= force of impact(pounds force)
W= object weight (lbs)
a= rate of deceleration (ft/s²)
g= acceleration due to gravity (32.2 ft/s²)
G= G_force

 

time to fall

[latex]t=\sqrt{\frac{2s}{g}}[/latex]

t= time (sec)
s= distance (ft)
g= 32.2 ft/s²

 

pressure of impact

 

[latex]p_i=\frac{f_i}{A_i}[/latex]

 

P_i= pressure of impact (force per unit area in lbs/in²)
F_i= force of impact (pounds force)
A_i= surface area of impact (in²)

Edited June 10, 2014 by Mordred

[Cosmobrain]

Instead of posting so many equations, can't you guys answer with a yes or no?

[md65536]

Instead of posting so many equations, can't you guys answer with a yes or no?

Yes (depending on the distance).

[Janus]

Instead of posting so many equations, can't you guys answer with a yes or no?

What good is a yes or no answer without the explanation as to why it is yes or no?