Waste less time on Facebook — follow Brilliant.
×
Back to all chapters

Gravitational Physics

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

Newton's law of gravity

                           

The legend of Issac Newton usually begins in an idyllic English countryside and features a falling apple.

Newton's most significant accomplishment was inventing a single model, which we now call Newtonian gravity, that gives rise to the dynamics of both:

  • falling objects near Earth's surface;
  • and planets in orbit.

These two types of motion were thought to be entirely distinct until Newton unified them within the framework of his theory.

In this quiz, we will explore the insights that motivated Newton to propose his theory of gravity and we will derive its form with orbital data from Jupiter's moons.

Newton studies an apple in his hand. He estimates that the mass of the apple is about 10% of the mass of his head.

How do the following gravitational forces compare?

  • \(F_{E\to H} = \) the force the Earth applies on Newton's head
  • \(F_{E\to A} =\) the force the Earth applies on the apple

Newton understood that forces come in pairs, which he expressed in his third law.

How do the following forces compare?

  • \(F_{H\to E}=\) the force Newton's head applies on the Earth
  • \(F_{A\to E}=\) the force the apple applies on the Earth

Here's the idea that set Newton apart from his contemporaries:

Newton speculated that any pair of massive objects applies a gravitational force on each other. In this sense, he characterized gravity as a universal force.


How do the following forces compare?

  • \(F_{H\to A}=\) the force Newton's head exerts on the apple
  • \(F_{A\to H}=\) the force the apple exerts on Newton's head

The last series of questions show us that the form of Newton's universal force law \(F(m_1, m_2)\) needs to be

  • linear in each mass;
  • symmetric in the masses.

Which of these choices is a possible mass dependence of the law \(F(m_1, m_2)\)?

This brand of model building through intuitive reasoning can only take us so far; at some point, any physical theory needs to appeal to observations and experiments.

Newton suspected that the force of gravity between two objects would depend not only on their masses, but also on their separation \(r.\)

Of course, Newton didn't have the means to leave Earth's surface to test how gravity's strength varies away from the surface, so he turned his head skyward, using observations of celestial bodies to justify his theory.

In the remainder of the quiz, we are going to make a string of deductions in the vein of Newton's own reasoning.

The Earth's gravitational pull on the moon is the cause of its orbital motion.

Estimate the acceleration (in \(\si[per-mode=symbol]{\meter\per\second\squared}\)) of the moon, \(a_\text{moon},\) in its orbit around Earth.

Details & Assumptions

  • Assume the moon's motion is uniform on a circle.
  • The distance between Earth's center and the moon's center is \(R=\SI{3.8\times 10^5}{\kilo\meter}.\)
  • The period of the moon's orbit is about 27.3 days.

Since we do not know gravity's dependence on separation \(r,\) let's wrap it in an unknown function \(f(r),\) which we will try to deduce. The gravitational force on a mass \(m\) exerted by the Earth, which has mass \(M_E,\) is \[F_g=G m M_E f(r).\] \(G\) is a proportionality constant that encodes the strength of gravity in our system of units.


The figure below depicts three examples of how \(f(r)\) might vary over the distance between the Earth and moon.

In light of the last calculation of the moon's acceleration, only one of the three options is viable. Which one?

We cannot solve for \(f(r)\) directly from \(F_g=GM_Emf(r)\) because in Newton's time, Earth's mass \(M_E\) was unknown.

However, because Newton was quite clever, he could calculate the ratio \[Z=\frac{f(R_E)}{f(R_M)}.\]

Can you? What is the numerical value of \(Z\)?

Details

  • Use \(g=\SI[per-mode=symbol]{9.8}{\meter\per\second\squared}\) and the moon's orbital acceleration.

Scientific model building usually relies on fitting observations of an unknown function to one of several common functional forms. A common fit for scientific models is a power law: \[f(r)=r^p,\] where \(p\neq 0.\)


By assuming \[Z=\frac{f(R_E)}{f(R_M)} = \frac{R_E^p}{R_M^p}\] find a prediction for unknown power \(p.\)

Details

  • Use \(R_E=\SI{6.4\times 10^3}{\kilo\meter}\) and \(R_M=\SI{3.8\times 10^5}{\kilo\meter}.\) Both we known fairly accurately in Newton's time.
  • Round your answer to the nearest integer.

Newton devoted much of his life to deriving the effects of a power-law force \[F_g\propto \frac{1}{r^2},\] which we now call an inverse-square law. By the time he had performed the previous calculation, he had already proved the observed orbits of planets can be derived from it. (We will do these proofs explicitly in a later chapter.)

In a universal theory of gravity, orbital data reproduced the inverse-square dependence of the gravitational force law. So far, we have checked that \(F_g\propto 1/r^2\) for the moon's orbit, so now we'll see if orbital data from Jupiter's moons also validate this conclusion .

We can check if \(F_g\propto 1/r^2\) on a set of multiple data points by performing a linear regression. A regression returns parameters of the function that would most likely produce a set of data points. In a linear regression, the calculated parameters are the slope and intercept of a line.

Here is the orbital period \((T)\) and average orbital radius \((R)\) data for Jupiter's four largest moons, along with a plot of \(\ln T\) vs. \(\ln R\) and a best-fit line. \[\begin{array}{c|c} \text{Moon} & \text{Period (T) } (10^6 \si{\second}) & \text{Orbital R } (10^6 \si{\kilo\meter})\\ \hline \text{Io}& 0.153 & 0.422 \\ \text{Europa}& 0.307& 0.671 \\ \text{Ganymede}& 0.618& 1.070\\ \text{Callisto}& 1.442& 1.883 \\ \end{array}\]

Without making any assumptions about \(p\) in the power-law form of the gravitational force \(F_g=Gm_1 m_2 R^p,\) show that there is a linear relationship between \(\ln T\) and \(\ln R.\)

Using the regression coefficients on the plot, what is order of magnitude of \(\epsilon = \lvert -2-p\rvert,\) the deviation from \(p=-2\)?

Details

  • Assume the moons' orbits are circular and their motion is uniform.

Given that Jupiter's mass is \(\SI{1.898\times10^{27}}{\kilo\gram}\), find a prediction for proportionality constant \(G\) from this data.

The accepted value of the proportionality constant in Newtonian gravity is \[G=\SI{6.67408\times 10^{-11}}{\meter\cubed\per\kilo\gram\per\second\squared}.\] By what fraction does the predicted value deviate from the accepted value?

Calculate \(\frac{\lvert G_\text{measured}-G \rvert}{G}.\)

Orbital data has been used over the last several centuries to validate Newton's universal law of gravity between massive bodies.

Newtonian gravity is defined by a force law \[F = G \frac{m_1 m_2}{r^2},\] where \(r\) is the magnitude of the displacement of the centers of two masses \(m_1\) and \(m_2,\) with \(G=\SI{6.67\times 10^{-11}}{\meter\cubed\per\kilo\gram\per\second\squared}.\) The force between the masses is always attractive, directed between the centers of \(m_1\) and \(m_2.\)

Newton's work unearthed the connection between falling objects and orbital motion. His theory formed the foundation of orbital mechanics, more than 250 years before the first man-made object was launched into orbit in 1957. Today, with more than 2,000 man-made satellites in orbit around Earth (and even more space junk), Newton's theory remains more relevant today than ever.

Newton's theory withstands the test of time because it exemplifies the bedrock principle of modern science: that the laws of nature are systematic and universal, which was not a widely held belief in his time.

×

Problem Loading...

Note Loading...

Set Loading...