# physics query **URL:** \ **Category:** Factual Questions\ **Created:** [May 11, 2002, 7:55am UTC](https://boards.straightdope.com/t/physics-query/108463 "2002-05-11T07:55:07Z")\ **Posts on this page:** 12\ **Page:** 1
**Author:** ![sfwarlock](https://avatars.discourse-cdn.com/v4/letter/s/b19c9b/32.png) [@sfwarlock](https://boards.straightdope.com/u/sfwarlock)\ **Post date:** [May 11, 2002, 7:55am UTC](https://boards.straightdope.com/t/physics-query/108463/1 "2002-05-11T07:55:07Z")
I sent this to Cecil, but I figure I might as well post it here too. (since I figure the odds of that one being picked are roughly P(snowball surviving in hell)) We all know that Galileo proved (by dropping cannon balls from the leaning tower of Pisa) that objects of different weights fall at the same rate, and indeed, physics tells us that this rate is 9.8 m/sec². However, physics _also_ tells us that the gravitational pull between two objects is (g_m1_m2)/d² where g is the gravitational constant as expressed dependant on the units you’re working in. From where I’m sitting (that being California) that sure looks like the force pulling two objects together _does_ depend on their masses. So which is it? ---
**Author:** ![Schnitte](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/schnitte/32/9033_2.png) [@Schnitte](https://boards.straightdope.com/u/Schnitte)\ **Post date:** [May 11, 2002, 8:07am UTC](https://boards.straightdope.com/t/physics-query/108463/2 "2002-05-11T08:07:38Z")
g is not the force pulling, it’s the acceleration caused by the force. Newton tells us: F = ma, thus a = F/m The heavier the object is, the stronger is the force pulling, but the acceleration remains constant because with large m’s, F/m gets smaller. Since the mass of the Earth is more or less constant, the rresult of the formula you give is proportional to m1 (the objects’s mass), and if you divide it by m you’ll get a constant a, regardless of the mass. This a for falling objects is g; it equals 9.81 N/kg or m/sec². ---
**Author:** ![sailor](https://avatars.discourse-cdn.com/v4/letter/s/a587f6/32.png) [@sailor](https://boards.straightdope.com/u/sailor)\ **Post date:** [May 11, 2002, 8:08am UTC](https://boards.straightdope.com/t/physics-query/108463/3 "2002-05-11T08:08:26Z")
IIRC, this was discussed at length in a recent thread and the answer was that _yes_, acceleration does depend on mass but for all practical purposes it is unnoticeable because the earth is so massive. But if you were dropping the moon and a ball, yes, the moon would fall faster. ---
**Author:** ![Jim\_B](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/jim_b/32/4338_2.png) [@Jim\_B](https://boards.straightdope.com/u/Jim_B)\ **Post date:** [May 11, 2002, 8:08am UTC](https://boards.straightdope.com/t/physics-query/108463/4 "2002-05-11T08:08:47Z")
Well, it has been a while since I’ve taken HS physics–so my answer will be vague and might be wrong. As you probably know, 9.8m/sec^2 is the _acceleration due to gravity_. Force, as I’m sure you’re aware, is defined as mass times acceleration–acceleration being V/T [where V=veloc. & T=time]. But wait a minute, where is _mass_ in the velocity equation? If we turn the equation for acceleration inside out, we just find velocity equals acceleration times time. Does any of that make sense? Probably not, but I tried :). ---
**Author:** ![Achernar](https://avatars.discourse-cdn.com/v4/letter/a/e274bd/32.png) [@Achernar](https://boards.straightdope.com/u/Achernar)\ **Post date:** [May 11, 2002, 8:09am UTC](https://boards.straightdope.com/t/physics-query/108463/5 "2002-05-11T08:09:34Z")
**Schnitte** is right. The force between the Earth and the object _does_ depend on the object’s mass. However, the _acceleration_ does not, because of Newton’s Second Law, F = ma. So in this case: F = GM[sub]Earth[/sub]m[sub]object[/sub] / d[sup]2[/sup] = m[sub]object[/sub]a and so: GM[sub]Earth[/sub] / d[sup]2[/sup] = a As I understand it, there’s no underlying physical reason that the two m[sub]object[/sub]s should exactly cancel out. But it’s been measured to about nine decimal places, and they do. ---
**Author:** ![Achernar](https://avatars.discourse-cdn.com/v4/letter/a/e274bd/32.png) [@Achernar](https://boards.straightdope.com/u/Achernar)\ **Post date:** [May 11, 2002, 8:13am UTC](https://boards.straightdope.com/t/physics-query/108463/6 "2002-05-11T08:13:18Z")
**sailor** , I think you’re half right. The instantaneous acceleration does not depend on mass, not one iota. However, because the Moon will tend to pull the Earth a lot closer than the small mass will, then the amount of time it takes the object to hit the earth _will_ depend on mass. ---
**Author:** ![Cheesesteak](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/cheesesteak/32/3694_2.png) [@Cheesesteak](https://boards.straightdope.com/u/Cheesesteak)\ **Post date:** [May 11, 2002, 10:20am UTC](https://boards.straightdope.com/t/physics-query/108463/7 "2002-05-11T10:20:12Z")
**Achernar** has hit it on the head. The acceleration of the falling body will be 9.8m/s[sup]2[/sup] regardless of mass, for the reasons noted. If the object is sufficiently massive, however, there will be a noticable acceleration of the _Earth_ towards the body, reducing the time it takes for the body and Earth to come together. ---
**Author:** ![rsa](https://avatars.discourse-cdn.com/v4/letter/r/839c29/32.png) [@rsa](https://boards.straightdope.com/u/rsa)\ **Post date:** [May 11, 2002, 1:27pm UTC](https://boards.straightdope.com/t/physics-query/108463/8 "2002-05-11T13:27:14Z")
Putting it another way… **M** is the _gravitational_ mass. If one object has 1000 times the mass of another object, it will indeed exert 1000 times the gravitational force of the smaller object. **m** is the _inertial_ mass and the more massive object would have 1000 times the resistance to the gravitational force than the smaller object. So, Msub[/sub]/msub[/sub] = 1.000000000000 ---
**Author:** ![Dr\_Paprika](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/dr_paprika/32/3042_2.png) [@Dr\_Paprika](https://boards.straightdope.com/u/Dr_Paprika)\ **Post date:** [May 11, 2002, 1:38pm UTC](https://boards.straightdope.com/t/physics-query/108463/9 "2002-05-11T13:38:20Z")
The gravity of the distant moon acts on both the Earth and the ball. The moon is far, so this action is weak, but it acts on the Earth much more strongly than the ball because it is heavier. The Earth is close to the ball and its force of gravity acts much more strongly on the ball, but if the taller the tower is, the weaker this force would be (by very marginal amounts). Practically, if not dropped in a vacuum, there would also be an air resistance force proportional to the cross-sectional area of the ball and its velocity. So the moon pulls the Earth more than the ball. The mass matters, but not very much, maybe somewhere in the twentieth decimal place. The height of the tower matters about as much… if the Earth’s surface is a million miles from the Earth’s centre, adding a mile to your tower height won’t affect gravity much. But it will affect it; gravity may be 9.81 m/s\*s everywhere, but it ain’t the same everywhere. ---
**Author:** ![David\_Simmons](https://avatars.discourse-cdn.com/v4/letter/d/9de053/32.png) [@David\_Simmons](https://boards.straightdope.com/u/David_Simmons)\ **Post date:** [May 11, 2002, 2:00pm UTC](https://boards.straightdope.com/t/physics-query/108463/10 "2002-05-11T14:00:05Z")
The gravitational force is: F[sub]g[/sub] = G\*m[sub]1[/sub]\*m[sub]2[/sub]/r[sup]2[/sup] G is the gravitation constant, m[sub]1[/sub] is the mass of one object, in this case we’ll call that the earth. m[sub]2[/sub] is the mass of the falling object and r is the distance between the centers of mass of the two objects. The acceleration of the falling object caused by the mutual gravitation of the earth and the object is: a = F[sub]g[/sub]/m[sub]2[/sub] substituting the gravitation force from the first equation for the force in the second equation gives: a = G\*m[sub]1[/sub]/r[sup]2[/sup] showing that the acceleration is independent of the mass of the falling object. ---
**Author:** ![scm1001](https://avatars.discourse-cdn.com/v4/letter/s/c4cdca/32.png) [@scm1001](https://boards.straightdope.com/u/scm1001)\ **Post date:** [May 11, 2002, 3:16pm UTC](https://boards.straightdope.com/t/physics-query/108463/11 "2002-05-11T15:16:57Z")
I had thought that the story of galileo dropping the balls was a legend - any one know for sure? ---
**Author:** ![David\_Simmons](https://avatars.discourse-cdn.com/v4/letter/d/9de053/32.png) [@David\_Simmons](https://boards.straightdope.com/u/David_Simmons)\ **Post date:** [May 11, 2002, 3:29pm UTC](https://boards.straightdope.com/t/physics-query/108463/12 "2002-05-11T15:29:24Z")
> [@](#): > > \*Originally posted by scm1001 \* > \*\*I had thought that the story of galileo dropping the balls was a legend - any one know for sure? \*\* Galileo experimented with falling objects by rolling balls down inclined rails. His method was all carefully described in his write-up of the work in _Two New Sciences_.