13.4 Little g and Big G

The familiar equation $F_{G} = m g$ works well when an object is on the surface of a planet, but mg will not help us find the force exerted on the same object if it is in orbit around the planet. Neither can we use mg to find the force of attraction between the earth and the moon. Newton’s law of gravity provides a more fundamental starting point because it describes a universal force that exists between all objects.

To illustrate the connection between Newton’s law of gravity and the familiar $F_{G} = m g$ , FIGURE 13.6 shows an object of mass m on the surface of Planet X. Planet X inhabitant Mr. Xhzt, standing on the surface, finds that the downward gravitational force is $F_{G} = m g_{X}$ , where $g_{X}$ is the free-fall acceleration on Planet X.

The figure illustrates a planet in the form of a solid sphere labeled Planet X with a mass uppercase M subscript X.

Full Alternative Text

We, taking a more cosmic perspective, reply, “Yes, that is the force because of a universal force of attraction between your planet and the object. The size of the force is determined by Newton’s law of gravity.”

We and Mr. Xhzt are both correct. Whether you think locally or globally, we and Mr. Xhzt must arrive at the same numerical value for the magnitude of the force. Suppose an object of mass m is on the surface of a planet of mass M and radius R. The local gravitational force is

F_{G} = m g_{surface}

(13.5)

where $g_{surface}$ is the free-fall acceleration at the planet’s surface. The force of gravitational attraction for an object on the surface $(r = R)$ , as given by Newton’s law of gravity, is

F_{M on m} = \frac{G M m}{R^{2}}

(13.6)

Because these are two names and two expressions for the same force, we can equate the right-hand sides to find that

g_{surface} = \frac{G M}{R^{2}}

(13.7)

We have used Newton’s law of gravity to predict the value of g at the surface of a planet. The value depends on the mass and radius of the planet as well as on the value of G, which establishes the overall strength of the gravitational force.

The expression for $g_{surface}$ in Equation 13.7 is valid for any planet or star. Using the mass and radius of Mars, we can predict the Martian value of g:

g_{Mars} = \frac{G M_{Mars}}{R_{Mars}^{2}} = \frac{(6.67 \times 10^{- 11} {N m}^{2} {/kg}^{2}) (6.42 \times 10^{23} kg)}{{(3.39 \times 10^{6} m)}^{2}} = 3.8 {m/s}^{2}

NOTE We noted in Chapters 6 and 8 that measured values of g are very slightly smaller on a rotating planet. We’ll ignore rotation in this chapter.