# Problem involving gravitational constants

1. Aug 27, 2006

### Xkaliber

Hi all,

I was just working on a problem involving gravitational constants like $$\frac{\Delta(g)}{g}$$. Anyway, its been awhile since I have been in my intro physics classes and forgot how to make approximations that allow the adding of large numbers with small numbers (in my case 9.80665 + 2.94x10^-10) and still retain the significant figures. I can post the problem if needed.

Thanks

Last edited: Aug 27, 2006
2. Aug 28, 2006

### Xkaliber

I'll just go ahead and post the problem.

According the the Dicke experiment, the maximum fractional difference in the acceleration of gravity $$\frac{\Delta(g)}{g}$$ between objects of varying composition is not greater than 3x10^-11. Assume that the fraction has this maximum value. How far behind the first ball will the second ball be when the first ball reaches the ground if they are dropped from the top of a 46-meter vacuum chamber?

Answer: g1 = 9.80665 m/s^2 g2 = g1 - (3x10^-11 x g1) t=3.062905 s (t came from a previous problem)

g2 = 2d/t^2
d = (g2 x t^2)/2

Basically, I just need to plug in the numbers and solve for d (which will be a very small number). However, in the above equation in solving for g2, how can I get a calculation that still maintains the extremely small difference in g1 and g2 that can be plugged into the equation solving for d. When using a calulator, it just rounds g2 to g1, which undermines the basis of the entire problem. Any suggestions would be greatly appreciated.

3. Aug 28, 2006

### Bystander

Remember the binomial expansion?

4. Aug 28, 2006

### quasar987

The way I interpret this sentence is that for some object 'o', the acceleration of gravity is $a_o=g\pm \Delta g$, where $\Delta g \leq (3\times 10^{-11})g$.

If you let two object (1) and (2) go from the same height, their separation when one of them hits the ground will be maximum if $\Delta g = (3\times 10^{-11})g$ and $a_1=g+\Delta g$, $a_2=g- \Delta g$ (i.e. when their difference in acceleration is as large as the Dicke experiment's result let us go.

Is this how you interpreted what the fractional difference thing means too?

Last edited: Aug 28, 2006
5. Aug 28, 2006

### Xkaliber

hmm, I actually did not see it that way when working through the problem but now see that your explanation is what they are looking for. I was taking g to be the acceleration of one ball and using that to calculate the acceleration of the other ball. This being the case, my previous problem may be incorrect, and now appears to be difficult without a known acceleration.

It asks me to find the maximum $$\frac{\Delta(g)}{g}$$ if, when two balls were dropped from 100m, one hit the ground while the other was 7cm above the ground. (I used the equation t = sqrt(2d/g), but now that I do not know t or g, that would not work)

6. Aug 29, 2006

### quasar987

I've tought a lot about this and here's my best shot..

First we write the equations of kinematics for the two objects (object (1) being the one hitting the ground first):

$$x_1(t)=100-\frac{a_1}{2}t^2$$
$$x_2(t)=100-\frac{a_2}{2}t^2$$

We can easily solve for the time at which (1) hit the ground by setting x_1=0. We find

$$t_1^2=\frac{200}{a_1}$$

We are told that at that time, x_2=0.07 m. In equation, this writes

$$0.07=100-\frac{a_2}{2}\frac{200}{a_1}$$

We can then solve for the ratio of the accelerations:

$$\frac{a_2}{a_1}=1-\frac{0.07}{100}$$

And here we must think. Clearly the acceleration are not equal. We know from theory that they should both be equal to g. In the "worst" case, one of the acceleration is precisely g, and the other is off by a number $\Delta g$. So let's write $a_1=g$ and $a_2=g-\Delta g$ We get

$$\frac{\Delta g}{g}=7\times 10^{-4}$$

7. Aug 29, 2006

### quasar987

This is obviously in contradiction with the Dicke experiment, but then again, the Dicke experiment would only allow a maximum distance of $6\times 10^{-9}$ m between the balls as one of them hits the ground. So it is not our answer that is flawled, it is the hypothesis of the problem that is completely imaginary.

edit: Does anyone know what the Dicke experiment is? What did the guy do to get this result?

Last edited: Aug 29, 2006
8. Aug 29, 2006

### Xkaliber

Wow, quasar. I really did not mean for you to spend much of your time on this problem, but I greatly appreciate it.

Before proceding, please note that the 100m measurement in my last post should have been 46m, which I will use in my following calculations. (The first problem matches up with yours when I used 100m, but am uncertain about the second one since you did not post your steps.)

First Calculation:

My answer for $$\frac{\Delta g} {g}$$ was the same as yours (after compensating for our different initial heights)

{Gravity Acceleration for Gold} = g_gold
{Gravity Acceleration for Copper} = g_copper

{Distance Fallen by Gold in Time "T"} = D_gold = (1/2)*(g_gold)*T2
{Distance Fallen by Copper in Time "T"} = D_copper = (1/2)*(g_copper)*T2

Dividing Eq #1 by Eq #2, and then subtracting "1" from both sides:

{D_gold/D_copper} = {g_gold/g_copper}
{D_gold/D_copper} - (D_copper/D_copper) = {g_gold/g_copper} - (g_copper/g_copper)
{D_gold - D_copper}/D_copper = {g_gold - g_copper}/g_copper

We are given {D_gold = 46 m} & {D_copper = (46 m - 0.07 m) = 45.93 m},
so that placing these values in last equation:

{g_gold - g_copper}/g_copper = (0.07)/(45.93) = 1.524E-3

Second Calculation:

$$\frac{\Delta g} {g_1}$$=3x10^(-11)
(g_1 - g_2)/g_1 = 3x10^(-11)
1 - 3x10^(-11) = g_2 / g_1
[1 - 3x10^(-11)]g_1 = g_2

$$x_1(t)=46-\frac{g_1}{2}t^2$$
$$x_2(t)=46-\frac{g_2}{2}t^2$$

$$t_1^2=\frac{92}{g_1}$$

Substitute time in terms of g_1 in and also substitue g_2 in terms of g_1 that we solved above:

$$x_2(t)=46-\frac{1 - 3x10^{-11}g_1}{2}\frac{92}{g_1}$$
$$x_2(t)=46-46(1 - 3x10^{-11})$$
$$x_2(t)=1.38x10^{-9}m$$

Regarding the Dicke Experiment, it appears that these values were determined by examining the affects of the sun's gravitational pull on a ball of gold and a ball of aluminum. Each ball was attached to the end of a rod and the rod was suspended by a quartz fiber. The rod would rotate very slightly as the sun's position changed throughout the day. Any difference in the gravitational acceleration of Sun for gold and aluminum resulted in opposite sense of net torque on the torsion pendulum in the evening compared with the morning.

I will be learning more about it as I continue working on the problems throughout the corresponding set.

Last edited: Aug 29, 2006