13.2 Gravitation near earth's surface  (Page 30/6)

In this section, we observe how Newton’s law of gravitation applies at the surface of a planet and how it connects with what we learned earlier about free fall. We also examine the gravitational effects within spherical bodies.

Weight

Recall that the acceleration of a free-falling object near Earth’s surface is approximately $g=9.80{\text{m/s}}^2$ . The force causing this acceleration is called the weight of the object, and from Newton’s second law, it has the value mg . This weight is present regardless of whether the object is in free fall. We now know that this force is the gravitational force between the object and Earth. If we substitute mg for the magnitude of ${\overrightarrow{F}}_{12}$ in Newton’s law of universal gravitation, m for $m_1$ , and $M_{\text{E}}$ for $m_2$ , we obtain the scalar equation

where r is the distance between the centers of mass of the object and Earth. The average radius of Earth is about 6370 km. Hence, for objects within a few kilometers of Earth’s surface, we can take $r=R_{\text{E}}$ ( [link] ). The mass m of the object cancels, leaving

This explains why all masses free fall with the same acceleration. We have ignored the fact that Earth also accelerates toward the falling object, but that is acceptable as long as the mass of Earth is much larger than that of the object.

Masses of earth and moon

1. Use the standard values of g , $R_{\text{E}}$ , and [link] to find the mass of Earth.
2. Estimate the value of g on the Moon. Use the fact that the Moon has a radius of about 1700 km (a value of this accuracy was determined many centuries ago) and assume it has the same average density as Earth, $5500{\text{kg/m}}^3$ .

Strategy

Solution

1. Rearranging [link] , we have
   
   $M_{\text{E}}=\frac{g{R}_{\text{E}}^2}{G}=\frac{9.80{\text{m/s}}^2{(6.37\times {10}^6\text{m})}^2}{6.67\times {10}^{\mathrm{-11}}\text{N}·{\text{m}}^2{\text{/kg}}^2}=5.95\times {10}^{24}\text{kg.}$
2. The volume of a sphere is proportional to the radius cubed, so a simple ratio gives us
   
   $\frac{M_{\text{M}}}{M_{\text{E}}}=\frac{R_{\text{M}}^3}{R_{\text{E}}^3}\to M_{\text{M}}=\left(\frac{{(1.7\times {10}^6\text{m})}^3}{{(6.37\times {10}^6\text{m})}^3}\right)(5.95\times {10}^{24}\text{kg})=1.1\times {10}^{23}\text{kg.}$
   
   We now use [link] .
   
   $g_{\text{M}}=G\frac{M_{\text{M}}}{r_{\text{M}}^2}=(6.67\times {10}^{\mathrm{-11}}\text{N}·{\text{m}}^2{\text{/kg}}^2)\frac{(1.1\times {10}^{23}\text{kg})}{{(1.7\times {10}^6\text{m})}^2}={2.5}^{}{\text{m/s}}^2$

Significance