Time of free fall does depend on mass

  • Thread starter parshyaa
  • Start date
  • #26
307
19
Yah he didn't said that , but he said that scientist are working on this field(it means scientists may thinks that time does/does not depends on mass). I just want to know what made them to do a research on it , you said that it is for varification of physics formula, then can you give me a link where it is written that experimental verification showed that time of free fall does or does not depends on mass.
 
  • #27
A.T.
Science Advisor
11,333
2,717
if any budy proved this then he will get a nobel prize
I will tell my buddies then.
 
  • #28
Vanadium 50
Staff Emeritus
Science Advisor
Education Advisor
27,578
11,776
he said that time of free fall does depends on mass
He said no such thing.
Yah he didn't said that

There is no point in discussing anything whatever with someone who will write things he knows not to be true.
 
  • #29
895
98
That is a unique rendering of Kepler's Law for free fall I've not seen. However, if I'm not mistaken Kepler eqn. for period of circular orbital motion should be equivalent to the oscillation period of m through a frictionless hole in the center of large Mass, M, if it has uniform mass density. In that case the free fall period would place m at center of mass, M in 1/4th the period, making the numerical constant merely pi/2 in your above eqn. Where does the extra factor of sq.rt. 2 come from?
Here's the Wikipedia page for that equation:
https://en.wikipedia.org/wiki/Free-fall_time
 
  • #30
27
1
Here's the Wikipedia page for that equation:
https://en.wikipedia.org/wiki/Free-fall_time
Thank you TurtleM for the derivation....
The usual (standard) Kepler eqn. that I was trying to reconstruct by squaring your equation is; T^2 / R^3 = 4(pi)^2 / G(M+m) ... but apparently the factor 1/32 kept appearing..
I think this clears it up here:(from your article)......
"Note that T (orbit) in the above equation, is the time for the mass to fall in a highly eccentric orbit, make a "hairpin" turn at the central mass at nearly zero radius distance, and then returns to R when it repeats the very sharp turn. This orbit corresponds to nearly linear motion back and from distance R to distance 0. As noted above, this orbit has only half as long a semimajor axis (R/2) as a circular orbit with radius R (where the semimajor axis is R), and thus the period for the shorter high-eccentricity "orbit" is that for one with an axis of R/2 and a total orbital pathlength of only twice the infall distance. Thus, by Kepler's third law, with half the semimajor axis radius it thus takes only (1/2)3/2 = (1/8)1/2 as long a time period, as the "corresponding" circular orbit that has a constant radius the same as the maximal radius of the eccentric orbit (which goes to essentially zero radius from the primary at its other extreme).

The time to traverse half the distance R, which is the infall time from R along an eccentric orbit, is the Kepler time for a circular orbit of R/2 (not R), which is (1/32)1/2 times the period P of the circular orbit at R. For example, the time for an object in the orbit of the Earth around the Sun, to fall into the Sun if it were suddenly stopped in orbit, would be P / (sq.rt.32) where P is one year...."

Since the first equation (Point masses) is somewhat useless, I included the free fall oscillating in a hole of a Large Mass and concluded it would be equivalent to Kepler's Orbital eqn. for uniform mass density....
They derived a similar result (there is an equivalent equation as the pt. mass eqn.) for extended mass of uniform mass distribution.... if M >> m.

Quite interesting...thanks for the link.
Pet Scan
 
Last edited:
  • #31
27
1
Sorry for this thread ,but he said that time of free fall does depends on mass , if any budy proved this then he will get a nobel prize and scientists are buisy to crack this. Suppose they have proved that yes it does depends on mass , then what will happen to the formula for time t, what is wrong in arguing that inertial mass and gravitational mass can be different[then this argument will raise a question how can a object have two mass] , oh god I am confused and making others too

@parshyaa (and Sophiecentaur) and whoever else....Maybe I can clarify a bit...Hopefully. It seems that the confusion is arising because there seems to be some confusion about which mass, (central mass or orbiting mass) you guys are referring to when commenting (it needs to be clarified) AND also because someone (Turtlemeister) used a "free -fall" equation based upon Kepler's 3rd law of ORBITAL motion....(which, BTW, is mass dependent) .. Let me explain>..and maybe it will help....maybe..LOL...It will require you to look at the equation briefly. OK, don't stop reading yet, this will be simple, not like my previous post.

Kepler's Law states (as mentioned by sophiecentaur) that an orbital planet sweeps out equal areas in equal times. What this means is that the orbital period, T, squared divided by the orbital Radius (or semi-major axis) CUBED is equal to a CONSTANT for ALL planets revolving around the CENTRAL MASS. Yes, amazingly the ratio of the orbital period squared to the orbital radius cubed is THE SAME value for ANY planet or asteroid or anything orbiting the sun................. Now what Sopie didn't mention is that this CONSTANT is equal to 4 (pi)^2 / GM....where M is the large Mass of the central body......(and G is the "gravitational constant".) This "Kepler" constant is dependent only on the central mass, say the sun for example, as long as the orbiting planets (or whatever) have "negligible" mass compared to the sun. As soon as a planet's mass becomes appreciable, then its mass, m, must be included in the calculation.

This Kepler orbital equation of motion can be manipulated to arrive at free-fall motion, (to which Turtlemeister alluded)...but this is not typical.... This no doubt caused some confusion...

The amazing genius of Kepler's equation is that it is applicable for all bodies orbiting around the same central mass. For example,,,ALL satellites orbiting earth must follow the same Kepler equation (except now using mass of earth as big M) . IOW, all satellites with the same orbital radius must have the same orbital period....(and even satellites in elliptical orbits have the same orbital period provided we substitute its semi-major axis for radius, R.) This is because the formula depends only upon the mass of earth and the satellites have "negligible" mass compared to that of earth.
Pres. Eisenhower in 1957, being greatly troubled, for example, when Russia sent the first orbiting satellite (Sputnik) over America, asked US scientists," How can we determine how big (how massive) this Russian satellite is?" The response was , "We can't .... even if we know its orbital period and orbital radius".
Why ? Because the mass of the orbiting satellite does not enter into the equation of motion...only that of the central mass, (earth) ...providing that the satellite isn't a big bugger like that on the movie "Independence Day" !! (LOL).

NOW, having said all that, the above stuff really says almost nothing about "gravitational" verses "inertial" mass. To really enter into that discussion you must start with NEWTON's equations...which deal with INERTIAL mass via F= ma.....and combine it with gravitational Force equations like F= GMm / R^2.... which gives the force between two masses.
I think I've said enough for now. Clear as mud, right?
Pet Scan
 
Last edited:
  • #32
307
19
During free fall : ma = GMm'/r^2 , I think newtons gravitational formula is formed for rest masses of a object (I am not shure )
When object falls down, its velocity increases , therefore according to this formula[ m = m'/√ {1-(v^2/c^2)}] , as velocity increases mass of a object also increases (by a tiny amount) , therefore m in LHS is different from m' of RHS therefore a = GMm'/mr^2 , ∴ accelaration depends on mass and it tells that time also depends on mass(for a very very very very small fraction but it does depends on mass) , this is just my thinking , so somebudy please tell me what is wrong in this .
 
  • #33
sophiecentaur
Science Advisor
Gold Member
2020 Award
26,542
5,573
Classical: Using Kepler's Law, which is the result of and inverse square law relationship is a bit dodgy, imo, when describing the behaviour of a 'through the Earth' pendulum, which involves a very different bit of Physics which gives a SHM.
I know that the resulting period is the same but, without some further detailed proof about the equivalence of the two processes (neither of which were discussed by Kepler, of course) it doesn't get us very far.
GR: In any case, the thread continues to hover between Classical and GR and Lewin's presentation is based on Classical and his Dimensional Analysis assumes things about the relationship that don't follow into GR. For a start, his time and distance variables will be changing in a GR way that he is not including in his lecture. This is all proving nothing.
 
  • #34
895
98
I just now had a chance to take a look at the section of Walter Lewin's lecture that the op is questioning. I think that the "very fancy experiments by very prestigious physicist" that he refers to are actually tests of the equivalence principle. And to be more precise, I think he is referring to tests of the equivalence principle as it relates to inertial and gravitational mass. Even though many of these tests have been performed by simply dropping objects from a height, this can only work when M>>m, such as M=Earth and m=any ordinary object. If you had an instrument sensitive enough (doesn't exist with current technology), or your test mass was massive enough, you could indeed detect a change in free fall time when the mass of the falling object is changed. You can see that in the equation that I posted.

Here's what the equivalence principle says:

Einstein's statement of the equality of inertial and gravitational mass:
A little reflection will show that the law of the equality of the inertial and gravitational mass is equivalent to the assertion that the acceleration imparted to a body by a gravitational field is independent of the nature of the body.

Notice that it says the acceleration imparted to a body by a gravitational field is independent of the nature of the body. So, if we were to change the mass of free falling body m, then it should have no effect on the acceleration of free falling body m. And that is true. However, changing the mass of free falling body m would have an effect on the acceleration of body M. And that would cause a change in the free fall time.

Keep in mind that the accelerations mentioned above are relative to the center of mass of the two bodies, NOT the relative acceleration. This is sometimes a source of misunderstandings in this type of discussion.
 
Last edited:
  • Like
Likes sophiecentaur and parshyaa
  • #35
sophiecentaur
Science Advisor
Gold Member
2020 Award
26,542
5,573
Handy test objects could be Planets? Any discrepancy between the two masses would / could be revealed in the details of their orbits, perhaps.
 

Related Threads on Time of free fall does depend on mass

  • Last Post
Replies
5
Views
9K
Replies
6
Views
3K
Replies
13
Views
874
Replies
11
Views
18K
  • Last Post
Replies
2
Views
891
  • Last Post
Replies
19
Views
6K
  • Last Post
Replies
3
Views
3K
Replies
2
Views
559
  • Last Post
Replies
2
Views
695
Top