Newton’s Universal Law of Gravitation

\[F = G\frac{M m}{r^2}\quad\text{Eq. (5.7)}\]

Note: If you feel uncomfortable with working with fractions, exponents, or variables you should review the following tutorials before working through this tutorial:

Sections

Introduction

In Newton’s Universal Law of Gravitation, \(F\) is the gravitational force acting on one object by another object, the two masses, \(M\) and \(m\) are the masses of the two objects (say the Sun and the Earth), and \(r\) is the distance between the centers of those objects. \(G\) in this equation is the Universal Gravitational Constant, another of the fundamental constants of the universe. (We believe that the fundamental constants maintain their same values everywhere and for all time.) Note that there is nothing special about designating one object as \(M\) and the other as \(m\).

There is something different about this equation when compared to most of the rest of the equations you will encounter in the text; there is a variable in the denominator of the equation that is raised to an integer exponent. The only complication this introduces is that instead of multiplying by the quantity raised to the exponent, you need to divide instead. That’s it! Fully expanding the equation looks like \[F = G\frac{M \times m}{r\times r}\] which is the same as \[F = (G \times M \times m) \div (r \times r).\] Remember that for variables written immediately next to one another, the multiplication is implied so that \(Mm = M \times m\). You will always see the equation written as it is at the top of the page; you will never see it written in one of the forms just shown.

Numerator of the Equation (The Effect of Masses)

To understand how to work with Newton’s Universal Law of Gravitation, first consider the case where there are two stars orbiting one another, and both stars have the same mass, namely the mass of our Sun. In other words, \(M = 1~{\rm M}_{\rm Sun}\) and \(m = 1~{\rm M}_{\rm Sun}\), where \({\rm M}_{\rm Sun}\) symbolizes the mass of the Sun. (You can think of \({\rm M}_{\rm Sun}\) as representing a specific amount of mass just like kg represents a specific amount of mass, the kilogram.) How do you work with the \(M\) and \(m\)? Since they are in the numerator, you simply multiply them together. For example, if the distance \(r\) between the two stars is unchanged, and one has the mass of the Sun, but the other star somehow magically doubles in mass so that it now has twice the mass of the Sun, then the force of gravity between the two is just twice the value it would be if both stars had the same mass as the Sun. Rather than \(Mm = (1~{\rm M}_{\rm Sun}) \times (1~{\rm M}_{\rm Sun})\) you would now have \(Mm = (1~{\rm M}_{\rm Sun}) \times (2~{\rm M}_{\rm Sun})\); \((1\times 2 = 2)\).

What if both stars are twice the mass of the Sun and the distance is unchanged? In this case the force would be \(Mm = (2~{\rm M}_{\rm Sun}) \times (2~{\rm M}_{\rm Sun})\) or 4 times larger \((2\times 2 = 4)\).

Denominator of the Equation (The Effect of Distance of Separation)

If the distance between the two stars somehow magically doubled but the masses of the two stars remain unchanged, what impact would that have on the force acting between them? The force would become one-fourth as strong. To see this, note that if \(r\) doubles, then \(r^2\) will change by a factor of \(2^2 = 2 \times 2 = 4\). However, since \(r^2\) is in the denominator, that means that \(F\) must decrease by a factor of 4. To say it another way, \(F\) is now \(\frac{1}{2^2} = \frac{1}{4}\) times as strong.

Rather than doubling, let’s assume that \(r\) increases by a factor of 5. In this case \(r^2 = 5^2 = 5\times 5 = 25\) so the force between the two stars must decrease by a factor of 25; in other words, \(F\) is now \(\frac{1}{25}\) times as strong.

What happens if the two stars get closer together rather than farther apart? Clearly the force must get stronger! Suppose the distance between the two stars magically decreased by a factor of 2, meaning that \(r\) is now 1/2 the value it was before. First recall that when you divide by large numbers the answer is smaller, and conversely when you divide by smaller numbers the answer is larger. So, if the distance between the two stars is now 1/2 as large as it was, the force between the two stars is now 4 times greater. To understand this mathematically, note that for the case where \(r\) is now 1/2 of the value it had before, \[\frac{1}{r^2} = \frac{1}{\left(\frac{1}{2}\right)^2} = \frac{1}{\frac{1}{2}\times \frac{1}{2}} = \frac{1}{\frac{1}{4}} = 4.\]
(Note: If you didn’t understand the last step you should review the fractions tutorial.)

In a similar way, if the distance between the two stars magically decreased by a factor of 5, the force of gravity between them would suddenly increase by a factor of \[\frac{1}{r^2} = \frac{1}{\left(\frac{1}{5}\right)^2} = \frac{1}{\frac{1}{5}\times \frac{1}{5}} = \frac{1}{\frac{1}{25}} = 25.\]

Combining Changes in the Numerator (Masses) and the Denominator (Distance)

Suppose that both stars magically doubled in mass at the same time that the distance increased by a factor of two. This just means that \[\frac{2\times2}{2^2} = \frac{4}{4} = 1,\] overall the force is exactly the same as before all of the magic happened!

If one mass, say \(M\) increased by a factor of 3 while the other mass decreased by a factor of 5, and at the same time the distance of separation decreased by a factor of 4, by what factor would the gravitational force change? In this case the equation dictates that \[\frac{3\times \frac{1}{5}}{\left(\frac{1}{4}\right)^2} = \frac{\frac{3}{5}}{\frac{1}{4}\times \frac{1}{4}} = \frac{\frac{3}{5}}{\frac{1}{16}} = \frac{3}{5}\times \frac{16}{1} = \frac{3\times 16}{5\times 1} = \frac{48}{5} = 9.6.\]

Calculating the Force in Newtons (Advanced Topic)

If your instructor wants you to actually calculate the force of gravity between two objects you just have to put the values into the equation and carry out the operations, remembering the order of operations. If you put your values in using the SI system of unites (kg, m, and s), then the force unit becomes newtons (N). (You may want to review scientific notation for this example.)

As one example, what is the force of gravity between the Earth and the Sun? In this case let \(M\) be the Sun and \(m\) be the Earth. These values are: \[M= 1~{\rm M}_{\rm Sun} = 1.99 \times 10^{30}~{\rm kg},\] \[m = 1~{\rm M}_{\rm Earth} = 5.97 \times 10^{24}~{\rm kg}.\] Note that it doesn’t matter which object you assign to \(M\) and to \(m\). Mathematically this is because \(Mm = mM\) (the commutative property of multiplication); physically it is because of Newton’s third law. Physical law is represented in the associated mathematics. Mathematics can also lead to discoveries in the physical universe (as was the case with Einstein’s relativity theory; see Sections 8.2 and 8.3).

The distance between the Earth and the Sun is one astronomical unit, or \[r = 1~{\rm au} = 1.50 \times 10^{11}~{\rm m}.\] In order to get a numerical result for the gravitational force you will also need to substitute the value for the Universal Gravitational Constant (discussed in Section 6.3): \[G = 6.67 \times 10^{-11}~{\rm N}\,{\rm m}^2/{\rm kg}^2\].
Making all of the substitutions: \[F = \left(6.67 \times 10^{-11}\frac{{\rm N}\,{\rm m}^2}{{\rm kg}^2}\right) \times \left[\frac{\left(1.99\times 10^{30}~{\rm kg}\right) \times \left(5.97 \times 10^{24}~{\rm kg}\right)}{\left(1.50\times 10^{11}~{\rm m}\right)^2}\right]\].
The final answer is \[3.52 \times 10^{22}~{\rm N}.\]

Quiz Yourself

Try the following exercises to make sure that you really do understand the material just presented. By the way, computers and many calculators write scientific notation in “computer-speak,” where \(10\) raised to an integer (either positive or negative) replaces \(\times 10^\text{exponent}\) with \({\rm E(exponent)}\) or \({\rm e(exponent)}\); in other words, \(3.14159 \times 10^9\) is represented by \(3.14159\text{E}9\). If your calculator has the \(\Large{\hat{ }}\) symbol, then you would enter the number as \(3.14159 \times 10{\Large{\hat{ }}}9\). If your calculator has a \(10^x\) button then you would enter the number as \(3.14159\times9\,\,10^x\). If you have having problems with entering numbers in scientific notation, read the manual or help screens, or ask your instructor.

(Answers are available below.)

For the first eight problems, you will just take a look at by what factor the force of gravity would increase or decrease relative to the case of two stars, each with masses equal to the mass of our Sun, separated by a distance of 1 au.

  1. If one of the two stars has a mass three times greater than the mass of the Sun while the distance of separation remained 1 au, how many times greater would the force of gravity between the two be?
  2. If one of the stars is 10 times more massive than the Sun, with the other mass and the separation being unchanged, how many times larger would the force of gravity be?
  3. If one star is three times the mass of the Sun and the other star is five times the mass of the Sun, but the distance between them remained unchanged, how many times greater would the force of gravity be?
  4. If both stars are ten times the mass of the Sun, how many times greater would the force of gravity be assuming that the distance between them is 1 au?
  5. If the masses of the two stars are equal to the mass of the Sun, but the distance between them increased by a factor of four, how many times weaker would the force of gravity be?
  6. If the masses of the two stars remained equal to the mass of the Sun, but the distance between them increased by a factor of 13, how many times weaker would the force of gravity be?
  7. If star 1 is twice the mass of our Sun, star 2 is 6 times the mass of our Sun, and the separation between them is 7 au, how many times weaker would the force of gravity be?
  8. If star 1 is one-half the mass of our Sun, star 2 is one-fourth the mass of our Sun, and the distance between them is 0.5 au, how many times weaker would the force of gravity be?
  9. For the previous problem, what is the force of gravity between the two stars (in N)?
  10. Star 1 is \(3~{\rm M}_{\rm Sun}\), star 2 is \(1.5~{M}_{\rm Sun}\), and \(r = 3.4~{\rm au}\). What is the force of gravity between the two stars?

Answers

  1. \(3\)
  2. \(10\)
  3. \(15\)
  4. \(100\)
  5. \(16\) (The force would be \(1/16\) the value if they were separated by 1 au.)
  6. \(169\) (The force would be \(1/169\) the value if they were separated by 1 au.)
  7. \(2\times6/7^2 = 12/49 = 0.245\)
  8. \((1/2)\times(1/4)/(1/2)^2 = (1/8)/(1/4) = 1/2 = 0.5\) or (0.5\times0.25\)/(0.5^2) = 0.125/0.25 = 0.5\)
  9. We can first find the force of gravity between two stars both having the mass of the Sun and separated by 1 au, then simply multiply by the answer in Exercise 8.
    \[6.67 \times 10^{-11}~{\rm N}\,{\rm m}^2/{\rm kg}^2 \times (1.99 \times 10^{30} {\rm kg})^2/(1.5\times 10^{11} {\rm m})^2)=1.17\times 10^{28} {\rm N}\]
    This means that the force between the two stars is \(0.5*1.17\times 10^{28} {\rm N} = 5.85\times 10^{27} {\rm N}.\)
  10. Using the same “trick” as in Problem 9, first find the force of gravity relative to two stars having the mass of the Sun and separated by 1 au then multiply by \(1.17\times 10^{28} {\rm N}\). The relative force is \(3 \times 1.5/3.4^2 = 4.5/11.56 = 0.389\). Finally, the force between the two stars in newtons is \(0.389 \times 1.17\times 10^{28} {\rm N} = 4.55\times 10^{27} {\rm N}.\)
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