Astronomy

Physicists count four fundamental forces between matter particles, but none impacts the large-scale structure of the universe more than gravity. Unlike the other forces, gravity is felt by every body, large or small; it links every last atom in the universe.

Gravity was also the first force to be understood through a modern scientific model. Inaugural scientists like Galileo and Newton pieced together a theory of gravity in the 1600's that is still used in applications today. The techniques of modern astronomy arose from their desire to understand and predict the motions of planets under gravity's action.

To get a feel for how gravity impacts the structure of the visible universe, let's start by calculating and comparing some typical gravitational forces in our solar system.

Gravity

                   

The force of gravity can be calculated from a simple quantitative law: gravity between two masses m1m_1 and m2m_2 separated by a distance rr is Fg=Gm1m2r2,F_g=\frac{Gm_1 m_2}{r^2}, where G=6.×671011 m3/(kgs2).G=\SI[per-mode=symbol]{6.67\times 10^{-11}}{\meter\cubed\per\kilo\gram\per\second\squared}. The distance rr is measured from the center of one object to the center of the other.

How large is the gravitational force of the Sun on Earth? Apply the gravitational force law above to the Sun-Earth system.


Details and Assumptions:

  • Earth's mass is 5.97×1024 kg.\SI{5.97e24}{\kilo\gram}.
  • The Sun's mass is 1.99×1030 kg.\SI{1.99e30}{\kilo\gram}.
  • In the next chapter, we will see how the ancient Greeks measured the Earth-Sun distance: dEarth=1.5×1011 m.d_\text{Earth}=\SI{1.5e11}{\meter}.

Gravity

                   

The Sun's gravity is responsible for shepherding the planets around nearly circular orbits in our solar system. The mass of Saturn, the ringed gas giant in our solar system, is 95.1695.16 times Earth's. In orbit, Saturn's distance from the Sun is 9.579.57 times the distance from Earth to the Sun.

Between the Sun and which planet, Earth or Saturn, is the gravitational force larger?


Details and Assumptions:

  • Rather than writing out dEarthd_\text{Earth} in meters, as in the previous question, astronomers define this distance to be 11 astronomical unit or 1 au.\SI{1}{au}. Thus, Saturn's orbital distance from the Sun is dSaturn=9.57 au.d_\text{Saturn}=\SI{9.57}{au}.

Gravity

                   

Calculating gravitational forces in our solar system requires knowing MSun,M_\text{Sun}, so you might wonder how astronomers first measured the Sun's mass. Putting it on a scale just wasn't an option, so they used Newton's theory of gravity, and measurements of Earth's orbital motion. Let's see how this works.


If we assume that Earth's orbit is circular (a pretty good approximation), then its acceleration a=v2/dEarth,a = v^2/d_\text{Earth}, where vv is Earth's speed in its orbit and dEarthd_\text{Earth} is its distance from the Sun.

Newton's law of motion relates acceleration to the net force on the Earth, which is the gravitational force. It says Fg=MEartha=MEarthv2dEarth,F_g=M_\textrm{Earth}a=\frac{M_\textrm{Earth} v^2}{d_\text{Earth}}, so Newton's law of motion gives us an alternate way to compute the gravitational force on the Earth, without using Newton's gravitational law1^1.

We first need to estimate Earth's orbital speed v.v. If Earth orbits the Sun once per 365365 days on a circular orbit with radius dEarth=1.5×108 km,d_\text{Earth}=\SI{1.5e8}{\kilo\meter}, what is its speed vv in km/s?\si[per-mode=symbol]{\kilo\meter\per\second}?


Details and Hint:

  • The distance Earth travels around its orbit is 2πdEarth.2\pi d_\text{Earth}.
  • The units of the answer will be km/s\si[per-mode=symbol]{\kilo\meter\per\second} if you first convert 365365 days into s\si{\second} and express the distance in km.\si{\kilo\meter}.

1^1 With his name on several laws, Newton was a busy guy in the 1600's.

Gravity

                   

Now that we have estimated the Earth's orbital velocity, a little algebra will lead us straight to the mass of the Sun.

We know that gravity is the force that pulls on Earth as it orbits the Sun, so in Newton's law of motion, Fg=MEartha,F_g = M_\textrm{Earth}a, we can substitute Fg=GMEarthMSun/dEarth2:F_g=GM_\textrm{Earth}M_\textrm{Sun}/d_\textrm{Earth}^2:

Fg=MEarthv2dEarthGMSunMEarthdEarth2=MEarthv2dEarth.\begin{aligned} F_g &= M_\textrm{Earth} \frac{v^2}{d_\textrm{Earth}} \\ \frac{GM_\textrm{Sun} M_\textrm{Earth}}{d_\textrm{Earth}^2} &= M_\textrm{Earth} \frac{v^2}{d_\textrm{Earth}}. \end{aligned}

Although this relationship may appear complicated, we can solve it for MSun,M_\text{Sun}, and calculate it, without any more information than we've already used.

Solve the above equation for MSun.M_\textrm{Sun}.

Gravity

                   

Astronomers routinely use orbital motion to deduce the masses of stars like the Sun, as well as much larger objects like galaxies; however, in this course, we are more interested in gravity's role energizing stars and other celestial objects, rather than the fine details of motion under gravity's influence.

In the remainder of this quiz, we will examine the birth of our solar system from a diffuse cloud of gas and debris, leftover from a star whose existence ended much earlier in a cataclysmic supernova explosion. We will work out an important relationship between gravitational potential energy and kinetic energy—sometimes called the virial theorem—that we will employ later in this course when we examine the life cycle of a star.

Gravity

                   

Several billion years ago, all of the mass that makes up the Sun and planets in our solar system was spread out as a low-density gas mixed with dust called a nebula.

Which stage of the solar system has lower gravitational potential energy—the spread-out gas, or our solar system today?


Details and Assumptions:

  • The Sun has a radius less than 1 au,\SI{1}{au}, whereas the region of gas that formed our solar system is roughly spherical with radius 100 au.\SI{100}{au}.
  • At the edge of the nebula or on the surface of the Sun, a gas molecule feels a gravitational force toward the center of the system. As a force like gravity does work it loses potential energy.

Gravity

                   

Let's consider a 1 kg\SI{1}{\kilo\gram} lump of metal as it is pulled from the outer fringe of the early solar system to a radius where it ultimately ends up in Earth's crust.

The metal lump is pulled from 100 au\SI{100}{au} to 1 au\SI{1}{au} as the nebula collapses. How much potential energy (( in MJ)\text{MJ}) does it lose, ΔUg?\lvert \Delta U_g \rvert ?



Details and Assumptions:

  • The gravitational potential energy of the lump with mass mm that is a distance of rr from the center of another object with mass MM is Ug=GmMr,U_g=-\frac{Gm M}{r}, where G=6.67×1011 m3/(kgs2).G=\SI[per-mode=symbol]{6.67e-11}{\meter\cubed\per\kilo\gram\per\second\squared}.
  • Take M=1.99×1030 kg,M=\SI{1.99e30}{\kilo\gram}, the mass of the solar system.
  • 1 au=1.5×1011 m.\SI{1}{au}=\SI{1.5e11}{\meter}.
  • Enter your answer (a positive number) in mega-Joules: 1 MJ=106 J\SI{1}{\mega\joule}=\SI{10^6}{\joule}.

Gravity

                   

Today, the lump of metal that we've tracked is now moving along with Earth with the kinetic energy

K=12Mv2,K=\frac12 M v^2,

where MM is the mass of the lump, and vv is the orbital speed of Earth that you calculated earlier in this quiz.

If you calculate the kinetic energy, you will find that it is less than ΔUg,\lvert \Delta U_g\rvert, which you calculated in the previous question. Because energy is conserved, we can assume the rest of the energy it lost was dissipated and reabsorbed by Earth as heat:

ΔUg=K+Heat.\lvert \Delta U_g \rvert = K + \text{Heat}.

About what percent of the potential energy lost by the metal lump was released as heat?


Details and Assumptions:

  • Neglect Earth's spin around its axis.

Gravity

                   

Earth's core, which is made of a mixture of dense molten metals, including iron and nickel, is very hot. Heating initially occurred as bits of dust and gas molecules lost potential energy during gravitational collapse of the cloud. In this quiz, we've calculated just 1 kg\SI{1}{\kilo\gram} of this material can release hundreds of MJ\si{\mega\joule} of energy.

We also discovered an important result that we will use several times in this Exploration. Specifically, the kinetic energy KK of the metal lump was about half of the potential energy Ug\lvert U_g \rvert it lost. In fact, if we had been more careful about including the lump's initial kinetic energy, we would have found that exactly half of the potential energy was lost as heat. This statement is formally called the virial theorem. We will see in a later chapter that stars always run the risk of gravitational collapse, and the virial theorem provides us a way to understand what happens to a star's temperature during collapse.

Gravity

                   
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