Acceleration of Gravity for large objects

[Guest Drondo]

I have done a lot of searching on this topic and haven't found any answers. I may be overlooking something, but here is what I'm wondering about:

 

From memory the formula is G*M / (r + h)^2. For example, at Earth's surface it is 9.8 m/s^2. It seems that the acceleration due to gravity only takes into account the "planet" or "base" object. I'm guessing this is because the object that is in free fall is always assumed to be so small (relative to something like Earth)that it's own gravitational pull is unimportant. However, suppose that the object in freefall is of similar size to the earth. Or if you had two earths that started out at x distance from eachother. How would you calculate their resulting acceleration toward eachother? Would you simply calculate it for each object independent of the other and then add them together? (that doesn't sound quite right). Or is there a separate formula specifically for large objects that each have significant gravity. Any help would be much appreciated

[swansont]

What you need to do to be accurate is to be in a frame that won't accelerate, and often the center-of-mass frame is useful for this, though any inertial frame will do. The person on the earth is almost in that frame. As you guessed, it's because a person's mass is small, and the implication of that is we can assume the earth is an inertial frame of reference and not get into too much trouble.

 

The force will end up being F = GMm/r² on each object, and the acceleration will be the force divided by the appropriate mass, and toward the other object.

[Klaynos]

Some little formulae that I know from memory...

 

(m and M are masses, F is the force applied on m due to gravity.)

 

g (acceleration due to gravity) = F/m

 

g=-GM/r^2

so:

 

F/m = -GM/r^2

 

F = -GMm/r^2

 

The force is related to the mass of both objects.

[Guest Drondo]

Ok I think I'm starting to understand...I've been playing with the formulas. So it seems no matter what the mass of the 2nd object, the acceleration is always constant (Earth always 9.8m/s^2 etc.)

 

Is it somewhat accurate to say then, that two Earths spaced 1 meter apart will both accelerate toward the other at 9.8m/s^2 and therefore if you were standing on either one it would appear that the other was accelerating at 19.6m/s^2? That's what I'm really trying to get at. Thanks for the replies

[swansont]

As long as the motion doesn't become relativistic, yes.

[Johnny5]

Ok I think I'm starting to understand...I've been playing with the formulas. So it seems no matter what the mass of the 2nd object' date=' the acceleration is always constant (Earth always 9.8m/s^2 etc.)

 

Is it somewhat accurate to say then, that two Earths spaced 1 meter apart will both accelerate toward the other at 9.8m/s^2 and therefore if you were standing on either one it would appear that the other was accelerating at 19.6m/s^2? That's what I'm really trying to get at. Thanks for the replies [/quote']

 

 

The magnitude gravitational force between two objects is

 

[math] G \frac{M_1M_2}{r^2} [/math]

 

M1 is the mass of one object,

M2 is the mass of the other object

G is the gravitational constant

And r is the distance between their centers of inertia.

 

Suppose that both objects have the mass of earth.

 

[math] M_e = 6 \times 10^{24} Kg [/math]

 

And suppose that both objects are spherical in shape, and have the radius of the earth.

 

[math] R_e = 6,000,000 m [/math]

 

Now, you want to drop one earth, from one meter above the other earth, and measure the acceleration of the one towards the other.

 

The center to center distance R is:

 

[math] R = (6,000,000)+1+(6,000,000) = 12,000,001 [/math]

 

[math] M1 g = G \frac{M_1M_2}{(12,000,001)^2} [/math]

 

[math] g = G \frac{M_2}{(12,000,001)^2} [/math]

 

[math] g = G \frac{6 \times 10^{24} Kg}{(12,000,001)^2} [/math]

 

G = 6.672 x 10^-11 m^3/Kgs^2

 

[math] g = (6.672 \times 10^{-11}) \frac{6 \times 10^{24} Kg}{(12,000,001)^2} [/math]

 

Using a calculator this is 2.77 meters per second squared, which is much less than 9.8. The reason is, that even though the surface of the second earth is one meter away from the surface of the first earth, the center to center distance between the two earths is enormous, and you can treat this situation as to particles of very large mass, but enormously far apart.

 

Instead, what I think you wanted, was to drop two earths at each other, leave one of them the size of the real earth, but squash the other into the size of a walnut, and then drop the walnut one meter above the surface of the earth. This changes things a great deal.

 

Because then, the center to center distance is given by:

 

6,000,000 +1 = 6,000,001 meters

 

So that in this case we have:

 

[math] g = (6.672 \times 10^{-11}) \frac{6 \times 10^{24} Kg}{(6,000,001)^2} [/math]

 

And using a calculator we have:

 

11.11 meters per second

 

If we use a more realistic earth radius of 6,378,000 meters we get:

 

[math] g = (6.672 \times 10^{-11}) \frac{6 \times 10^{24} Kg}{(6,378,001)^2} [/math]

 

And using a calculator we find:

 

g = 9.84 meters per second squared

 

The two earths do not fall together at 2g=9.8+9.8 = 19.6 meters per second squared.

[Janus]

 

The two earths do not fall together at 2g=9.8+9.8 = 19.6 meters per second squared.

 

Yes, they do. Your calculations only took into account the acceleration of M1 due to its attraction to M2, but neglected the acceleration of M2 due it's attraction to M1.