ANOTHER question about gravity

[shrumeo]

howdy,

so, we are told in school that because of d=1/2a*t^2

and a=F/m

that a will be constant, 9.8, g, no matter what the mass of the "falling" object

so that the moon and Earth would fall toward each other at 9.8 m/s/s just as a bowling ball toward the earth

Why is there only one mass in the above acceleration equation but 2 in
F=G(m1*m2)/R^2 ?

Based on the above assumption the Earth falls toward the moon at 9.8 m/s/s
but when you go to the moon, the value of a is not 9.8 m/s/s. If the above were totally true I could interchange either mass.

Is d=1/2g*t^2 a huge approximation that neglects the gravity of small objects? Just wondering because i was reading the sample chapter of
"The final theory" here: http://www.thefinaltheory.com/images/Final_Theory_--_Chapter_1.PDF

and the claim on page 56 is that this constant acceleration of gravity is a big mystery of science. I wanted to tell him he is wrong, but then I started to doubt myself. Could someone please clear this up for me?

 

[jcsd]

howdy,

so, we are told in school that because of d=1/2a*t^2

and a=F/m

that a will be constant, 9.8, g, no matter what the mass of the "falling" object

so that the moon and Earth would fall toward each other at 9.8 m/s/s just as a bowling ball toward the earth

Why is there only one mass in the above acceleration equation but 2 in
F=G(m1*m2)/R^2 ?

Based on the above assumption the Earth falls toward the moon at 9.8 m/s/s
but when you go to the moon, the value of a is not 9.8 m/s/s. If the above were totally true I could interchange either mass.

Is d=1/2g*t^2 a huge approximation that neglects the gravity of small objects? Just wondering because i was reading the sample chapter of
"The final theory" here: http://www.thefinaltheory.com/images/Final_Theory_--_Chapter_1.PDF

and the claim on page 56 is that this constant acceleration of gravity is a big mystery of science. I wanted to tell him he is wrong, but then I started to doubt myself. Could someone please clear this up for me?

9.8 ms^-2 is only rough value for the accelartion due to garvity at the Earth's surface. It is not constant.

If you wanted to find the accelration due to (the Earth's) gravity for some object of mass m you'd use the second formula:

$$F = \frac{GM_{Earth}m}{r^2}$$

substitute in a = F/m

$$a = \frac{GM_{Earth}}{r^2}$$

So we go to http://en.wikipedia.org/wiki/Earth and get the values for mass and mean radius of the Earth, we get an answer of a = 9.8199 ms^-2, which is in pretty good agreement of our value of 9.8.

If we now input r as the mean radius of the orbit of the moon (again from http://en.wikipedia.org/wiki/Moon) we get a = 0.0027 ms^-2 which considerably less than 9.8.

The reason why there is only one mass in the first equation is that the first equation describes the force of a single body, whereas the second describes a force between two bodies.

 

[Doc Al]

acceleration due to gravity

There are a few things to sort out here...

so, we are told in school that because of d=1/2a*t^2

and a=F/m

that a will be constant, 9.8, g, no matter what the mass of the "falling" object

Well, it's true that the acceleration due to gravity near the surface of the Earth is about 9.8 m/s^2 and is independent of the mass of the object (as long as the object is reasonably small compared to the earth--so that you can ignore the earth's acceleration). But not for the reason you gave.

That first equation is just a kinematic relationship for uniform acceleration, and the second is Newton's 2nd law.

To find the acceleration due to gravity, we must combine the force of gravity:
$$F = G\frac{M_{earth}m}{R^2}$$
with Newton's 2nd law:
$$a = F/m$$
This gives an acceleration due to gravity of:
$$a_{gravity} = G\frac{M_{earth}}{R^2}$$
Note that m (the mass of the object) cancels out: If you double the mass, you double the force, so the acceleration is constant. Also note that the acceleration depends on R, which is the distance from the center of the earth.

If you plug in the numbers (mass of earth, radius of earth, G) you will get a value of around 9.8 m/s^2 (that's what we call g).

so that the moon and Earth would fall toward each other at 9.8 m/s/s just as a bowling ball toward the earth

Careful. Firstly, the distance to the moon is about 60 Earth radii, thus the acceleration due to Earth's gravity at the moon's position is about 1/3600 times smaller than g at the Earth's surface!

Second, if the Earth pulls on the moon (or any object) then the moon (or object) pulls back on the Earth with exactly the same force. But the resulting acceleration is not equal! The Earth is pretty big, so its acceleration due to the force of gravity is usually quite small. For example, as huge as it is, the moon is still only 1/81 times the mass of the earth, so the acceleration of the Earth due to the moon is 1/81 times smaller that the acceleration of the moon due to the earth. That's pretty small. Can you imagine how much smaller is the acceleration of the Earth due to the attraction of a bowling ball? Undetectable!

I hope this helps a little.

Note: I see that jcsd has already answered the question. Oh well, now you'll hear it twice.

 

[shrumeo]

Ok, I really appreciate the detail that you two gave in your explanations. Thanks!

So, I guess that the idea of every object falling to Earth with the exact same rate of acceleration is an approximation of the actual case? I was just trying to make sure of this so that I could be sure that this "Final Theory" guy is smoking crack. (Oh, I'm looking at my question above and I should have said that g = 9.8 m/s/s is the approximation. It's relation to distance and time wouldn't be, or would it? O well.)

When you said "careful", Doc Al, I wasn't expressing my beliefs or understanding, I was expressing the way one would understand the subject based solely on what we are taught in junior high. O well, no biggie.

Thanks again!

 

[jcsd]

9.8 is only an approximation valid at or near the Earth's surface.

It's interesting that we can do a simlair thing for any object no matter it's mass. For example we could find how much I accelarte the Earth in free fall. If we approximate me to a point mass (which isn't a major approximation if you consider the difference between the Earth's radius and my maximum radius) and take my mass of ~90kg. We find that the accelartion due to my gravity of the Earth ( take r as R_Earth) is about 1.5e-22 ms^-2.