13 Gravitation
13.2 Gravitation Near Earth’s Surface
Learning Objectives
By the end of this section, you will be able to:
- Explain the connection between the constants [latex]G[/latex] and [latex]g[/latex]
- Determine the mass of an astronomical body from free-fall acceleration at its surface
- Describe how the value of [latex]g[/latex] varies due to location and Earth’s rotation
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 [latex]g=9.80\,{\text{m/s}}^{2}[/latex]. 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 [latex]{\mathbf{\overset{\to }{F}}}_{12}[/latex] in Newton’s law of universal gravitation, m for [latex]{m}_{1}[/latex], and [latex]{M}_{\text{E}}[/latex] for [latex]{m}_{2}[/latex], 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 [latex]r={R}_{\text{E}}[/latex] (Figure). 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.
Example
Masses of Earth and Moon
Have you ever wondered how we know the mass of Earth? We certainly can’t place it on a scale. The values of g and the radius of Earth were measured with reasonable accuracy centuries ago.
- Use the standard values of g, [latex]{R}_{\text{E}}[/latex], and Figure to find the mass of Earth.
- 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, [latex]5500\,{\text{kg/m}}^{3}[/latex].
Strategy
With the known values of g and [latex]{R}_{\text{E}}[/latex], we can use Figure to find [latex]{M}_{\text{E}}[/latex]. For the Moon, we use the assumption of equal average density to determine the mass from a ratio of the volumes of Earth and the Moon.
Solution
- Rearranging Figure, we have
[latex]{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}^{-11}\,\text{N}\cdot {\text{m}}^{2}{\text{/kg}}^{2}}=5.95\times {10}^{24}\,\text{kg.}[/latex]
- The volume of a sphere is proportional to the radius cubed, so a simple ratio gives us
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