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# Gravitational Physics

Here we lay out Newton's law of gravity and crack open the universe of consequences that spring from it.

In this quiz, you are tasked with testing Newton's universal theory of gravity by making a precision measurement of the force between two non-astronomical objects.

Naturally, you expect it will be easiest to make your measurement away from the influence of other massive bodies, so imagine you are in deep space, far away from all mass except two \(\SI{1}{\kilo\gram}\) metal spheres.

If the spheres are \(\SI{1}{\meter}\) apart, what is the order of magnitude of the gravitational force the spheres apply on each other?

How far away do you have to be from the spheres so that your body mass does not interfere with the experiment?

**Details & Assumptions**

- You move far enough away that the gravitational force you apply on one of the spheres is less than \(1/10\) of the force the spheres apply on each other.
- Your mass is \(\SI{50}{\kilo\gram},\) and assume you can treat yourself approximately as a point mass.

About how far away does the sphere experiment have to be **from the Earth** so that the Earth's gravitational pull is less than \(1/10\) of the force the spheres apply on each other?

**Details**

- The mass of the Earth is \(\SI{5.97\times 10^{24}}{\kilo\gram}\)

The idea that we can measure a gravitational force away from the influence of other massive bodies is futile. The mass of Earth is simply too large to remove the experiment from its gravitational influence.

Enter, Henry Cavendish, English experimentalist extraordinaire, peering into a hole in his woodshed, scribbling furiously. Inside is a

Cavendish's measurements of the gravitational force provided an estimate of Newton's constant, \(G,\) for the first time. We will follow his footsteps through this Earth-shattering measurement through the end of this quiz.

To make a precision measurement of gravity, Cavendish attached two small wooden masses, \(m=\SI{1}{\kilo\gram},\) on the ends of a horizontal rod. He balanced the rod around its center, hanging it on a 32-gauge copper wire, so that the rod would rotate as the wire twists, forming a torsion pendulum.

Finally he hauled into his shed two massive lead spheres, \(M=\SI{150}{\kilo\gram},\) which he placed in the plane of rotation of the rod.

This setup is ideal for measuring gravitational forces because the rod deflects predictably in response to minute forces within its plane of rotation.

Which of the following masses **does not** contribute to the size of the rod's deflection?

**Assume** the rod can only be deflected in the plane of rotation, perpendicular to the wire.

When the rod of the torsion pendulum is displaced by an angle \(\theta\) from its equilibrium orientation, the twisted wire applies a torque to the rod that is proportional to the angular displacement: \[N=\mu\theta\] where \(\theta\) is in radians and \(\mu\) is called the torsion constant of the wire.

What is the magnitude of a force \(F\) that would deflect the bar from its equilibrium orientation by \(\theta=\SI{0.1}{\radian}\) (about \(\ang{5.7}\))?

**Details & Assumptions**

- Cavendish's contraption was built from a \(D=\SI{1.75}{\meter}\) rod with negligible mass
- \(F\) holds the pendulum in
**static equilibrium**; it is perpendicular to both the rod and the wire. - The torsion constant is \(\mu=\SI{3.5\times 10^{-6}}{\newton\meter}.\)
- Express your answer in \(\si{\micro\newton}\) \((\SI{10^{-6}}{\newton}).\)

The lead mass that is closer to each wooden sphere exerts the most influence on the rod's deflection.

When the angular separation is \(\Delta \theta\) what is the magnitude of the gravitational force \(F_g\) between one of the small pendulum masses \(m\) and the closer of the large lead masses \(M\)?

In the last question, we considered the gravitational forces acting at the centers of the wooden and lead masses. To be really precise, we should instead calculate how the gravitational force **distributes** over the shape of each mass. Would this produce a significantly different prediction for the force? Newton was preoccupied with this **problem of spherical masses,** a question we will take up in the next chapter. Here is his important conclusion:

The force of gravity between two spherical masses is directed along the line connecting their centers, and is equal in magnitude to the force between two equivalent point masses located on their centers.

Note that we have recast Newton's force law in vector form, where it appears to have a \(1/r^3\) dependence. However, if we take the magnitude of this expression: \[\lvert \frac{GmM}{\lvert \mathbf{r} \rvert^3} \mathbf{r} \rvert= \frac{GmM}{\lvert \mathbf{r} \rvert^2}\] we see that it reduces to the inverse-square law.

Only the component of a force perpendicular to the rod in the plane of rotation causes the wire to twist.

If \(F_g\) is the magnitude of the gravitational force we found in the last question, which expression picks out the tangential component of \(F_g?\)

**Details & Assumptions**

- The centers of both spheres are in the plane of rotation of pendulum.
- Assume the centers of both spheres are on the circle with radius \(D/2\) centered on the wire.

So far, we have focused on the effect of \(F_\text{near},\) the gravitational force applied to the pendulum from the nearer of the two lead masses. Now we need to add in the effect of the farther mass, \(F_\text{far}.\)

Repeat the analysis of the previous two questions to calculate \(F_\text{far}^\parallel,\) the tangential component of \(F_\text{far}.\)

**Details**

- Express your answer as the ratio \(F_\text{far}^\parallel/F_\text{near}^\parallel,\) the ratio of the tangential components of \(F_\text{near}\) and \(F_\text{far}.\)

Cavendish did not know how much deflection to expect in his experiment because, at the time, the magnitude of \(G\) was unknown.

Finally, after years of minimizing sources of error and testing various positions of the lead masses, Cavendish is convinced he is observing a purely gravitational deflection of the pendulum in the setup below. He records the following data when the rod reaches static equilibrium: \[\begin{array}{c|c} \hline \theta_0 & \SI{0.355}{\radian} \\ \hline \theta & \SI{0.1}{\radian}\\ \hline \end{array}\]

By imposing the static equilibrium condition on the rod, derive an estimate of \(G\) to within a few percent.

**Details & Assumptions:**

- The counter-torque due to the twisting wire is \(N_t=\mu \theta.\)
- In the choices, \(\Delta \theta = \theta_0-\theta.\)

Cavendish made his measurements from outside his shed in order to minimize disturbances to the experiment, which limits the precision of his measurement of \(\theta=\SI{0.1}{\radian}\).

Suppose Cavendish was certain that the actual deflection of the pendulum was in a range \(\left[\SI{0.08}{\radian},\SI{0.12}{\radian}\right].\) Thus, we can write the range as \(\theta \pm \delta \theta\) where the **measurement uncertainty** \(\delta \theta = \SI{0.02}{\radian}\) (slightly more than \(\ang{1}\)).

We can use the estimating function \(G_\text{est}(\theta)\) from the previous question to estimate the corresponding uncertainty of \(G:\) \[G \pm \delta G\] where \(\delta G \approx G_\text{est}(\theta+\delta \theta)-G_\text{est}(\theta).\)

What is the approximate fractional error \(\delta G/G\) for our previous estimate of \(G\)?

Cavendish's experiment using a basic torsion balance provided an estimate of \(G\) within \(1\%\) of the current accepted value in the last years of the 18th century. Moreover, we have showed in this quiz that the uncertainty of his measurement was less than \(5\%,\) which is pretty remarkable considering the myriad sources of measurement error Cavendish battled.

As we analyzed Cavendish's setup, we gained some perspective on the scale of gravity:

- On one hand, the force of gravity is very weak. Gravitational forces between human-scale objects are almost undetectably small. Specialized devices are needed to detect them.
- On the other hand, astronomical masses are so large that it is impossible (with current technology) for us to escape to a region absent of gravitational field.
- By exploiting the radial configuration of a gravitational field around Earth, Cavendish was able to directly measure the gravitational interaction between two relatively small objects.

With a firm historical perspective, we move from Cavendish's triumph of measurement to a problem that baffled many great philosophers: the ocean tides. Like Cavendish's approach, the solution to the problem of tides is a consequence of gravity's vector nature.

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