Can mass play a factor in the force of gravity on Earth?

[Pratto]

So I was thinking about the equation for the acceleration of gravity on earth, or g:

m*g=G*M(earth)*m/d^2

I guess it's universally accepted that different masses on Earth's gravitational acceleration is 9.8m/s^2, regardless of the mass of the object. This is because the two "m"s cancel out, no? But do they HAVE to cancel out? I mean, if mass is supposed to be a playing factor for gravity universally, wouldn't on Earth count too? That's like saying that me and another guy have a mass of 2*40kg=2*40kg, with my mass on the left and his on the right. Now, the two "2"s "can" cancel out, but if I did that, I would no longer have a mass of 80kg, I would only be 40kg, which simply isn't true. So doesn't canceling out the "m"s make the equation untrue as well? You don't have to necessarily cancel two variables out just because you can, because it could possibly make the equation "untrue", even if mathematically it's still correct, in reality it isn't "true".

 

[SteamKing]

It gets worse. Not only do the 2's cancel out, but the 40 kgs cancel as well. You and your friend weigh nothing at all!

 

[Bandersnatch]

When you cancel out one of the masses from the gravitation force equation you end up with the acceleration the mass you've just got rid of experiences in the gravitational field of the other one.
But that's just one half of the whole shebang, as you can do it the other way - calculating the acceleration of e.g. Earth due to your mass. This is the way in which your mass counts.
Note that you no longer have a force then - it's the acceleration. It tells you that while the force depends on both masses, the acceleration does on just one.

If there existed an equation in which cancelling out variables invalidated it, then that equation would be false.

Look at your 80kg=80kg example. You're saying in it that whatever is on the left is equal to what is on the right. That you can cancel the two masses tells you that this equality is independent of the actual value of the masses, as long as they're the same. You can't really pin any more significance to it.

 

[BruceW]

Hey pratto, welcome to physicsforums, buddy.
Like bandersnatch says, cancelling out the m's doesn't make the equation untrue. It then becomes an equation relating the acceleration of objects which are the same distance from the earth. And we find that the acceleration doesn't depend on the mass of the object, which is exactly what we expect to find (ignoring air resistance). Generally, when you rearrange an equation, you will end up with a different quantity as the subject of the equation. That is what is happening here.

 

[gmax137]

... I mean, if mass is supposed to be a playing factor for gravity universally, wouldn't on Earth count too? ...

When you use the resulting equation to calculate how long it takes the small mass to hit the ground, that's where you are ignoring the effect it's mass has on the answer. It is normally OK to do this, because the big M is so much bigger than small m; so that the acceleration of the Earth is entirely negligible. In "reality" the Earth is accelerating up towards the small m at the same time as the small m is accelerating down, so the distance closes faster than you calculate. You could figure out the magnitude of this error in the calc, you'd find it to be microscopically small.

 

[rcgldr]

The force between the two is

f = \frac {-G \ m_{object} \ m_{earth}} {r^2}

The acceleration of the object with respect to the center of mass of the object and Earth = f / m_object

a_{object} = \frac {-G \ m_{earth}} {r^2}

and the acceleration of the Earth would be:

a_{earth} = \frac {G \ m_{object}} {r^2}

The acceleration of closure between the two is a_object - a_earth :

\ddot{r} = \frac {-G (m_{object} + m_{earth} ) }{r^2}

Hmm, lost yet another edit. As pointed out in the next post, assuming the object isn't some other huge object like the moon, then since m_object << m_earth, then m_object + m_earth ~= m_earth. Wiki lists the mass of the Earth as 5.97219 × 10²⁴ kg, where the least significant digit is 9 x 10¹⁹ kg. The object would need to be close to 1 x 10¹⁹ kg to even effect the least significant digit of the sum of m_object + m_earth.

 

[Darwin123]

The acceleration of closure between the two is a_object - a_earth :

\ddot{r} = \frac {-G (m_{object} + m_{earth} ) }{r^2}

However, the mass of the Earth is much greater than the mass of the object. Therefore,

\ddot{r} = \frac {m_{object} &lt;&lt; m_{earth} }

The sum of a large positive quantity and a small positive quantity is almost the same as the large positive quantity. Therefore,

The acceleration of closure between the two is a_object - a_earth :

\ddot{r} = \frac {-G m_{object} }{r^2}

This is the same expression as the acceleration of an object on Earth with an Earth of infinite mass. So the measured acceleration is almost independent of the mass of the object.

Now here is a harder question. How does one copy a text equation in a post? Copied equations don't seem to come out right for me.