# 8.4 Newton’s universal law of gravitation  (Page 3/11)

 Page 3 / 11

## Making connections

Attempts are still being made to understand the gravitational force. As we shall see in Particle Physics , modern physics is exploring the connections of gravity to other forces, space, and time. General relativity alters our view of gravitation, leading us to think of gravitation as bending space and time.

In the following example, we make a comparison similar to one made by Newton himself. He noted that if the gravitational force caused the Moon to orbit Earth, then the acceleration due to gravity should equal the centripetal acceleration of the Moon in its orbit. Newton found that the two accelerations agreed “pretty nearly.”

## Earth’s gravitational force is the centripetal force making the moon move in a curved path

(a) Find the acceleration due to Earth’s gravity at the distance of the Moon.

(b) Calculate the centripetal acceleration needed to keep the Moon in its orbit (assuming a circular orbit about a fixed Earth), and compare it with the value of the acceleration due to Earth’s gravity that you have just found.

Strategy for (a)

This calculation is the same as the one finding the acceleration due to gravity at Earth’s surface, except that $r$ is the distance from the center of Earth to the center of the Moon. The radius of the Moon’s nearly circular orbit is $3\text{.}\text{84}×{\text{10}}^{8}\phantom{\rule{0.25em}{0ex}}\text{m}$ .

Solution for (a)

Substituting known values into the expression for $g$ found above, remembering that $M$ is the mass of Earth not the Moon, yields

$\begin{array}{lll}g& =& G\frac{M}{{r}^{2}}=\left(6\text{.}\text{67}×{\text{10}}^{-\text{11}}\frac{\text{N}\cdot {\text{m}}^{2}}{{\text{kg}}^{2}}\right)×\frac{5\text{.}\text{98}×{\text{10}}^{\text{24}}\phantom{\rule{0.25em}{0ex}}\text{kg}}{\left(3\text{.}\text{84}×{\text{10}}^{8}\phantom{\rule{0.25em}{0ex}}\text{m}{\right)}^{2}}\\ & =& 2\text{.}\text{70}×{\text{10}}^{-3}\phantom{\rule{0.25em}{0ex}}{\text{m/s.}}^{2}\end{array}$

Strategy for (b)

Centripetal acceleration can be calculated using either form of

$\begin{array}{c}{a}_{c}=\frac{{v}^{2}}{r}\\ {a}_{c}={\mathrm{r\omega }}^{2}\end{array}\right\}\text{.}$

We choose to use the second form:

${a}_{c}={\mathrm{r\omega }}^{2}\text{,}$

where $\omega$ is the angular velocity of the Moon about Earth.

Solution for (b)

Given that the period (the time it takes to make one complete rotation) of the Moon’s orbit is 27.3 days, (d) and using

$1 d×24\frac{\text{hr}}{\text{d}}×60\frac{min}{\text{hr}}×60\frac{s}{\text{min}}=\text{86,400 s}$

we see that

$\omega =\frac{\text{Δ}\theta }{\text{Δ}t}=\frac{2\pi \phantom{\rule{0.25em}{0ex}}\text{rad}}{\left(\text{27}\text{.}\text{3 d}\right)\left(\text{86,400 s/d}\right)}=2\text{.}\text{66}×{\text{10}}^{-6}\frac{\text{rad}}{\text{s}}.$

The centripetal acceleration is

$\begin{array}{lll}{a}_{c}& =& {\mathrm{r\omega }}^{2}=\left(3\text{.}\text{84}×{\text{10}}^{8}\phantom{\rule{0.25em}{0ex}}\text{m}\right)\left(2\text{.}\text{66}×{\text{10}}^{-6}\phantom{\rule{0.25em}{0ex}}\text{rad/s}{\right)}^{2}\\ & =& \text{2.72}×{\text{10}}^{-3}\phantom{\rule{0.25em}{0ex}}{\text{m/s.}}^{2}\end{array}$

The direction of the acceleration is toward the center of the Earth.

Discussion

The centripetal acceleration of the Moon found in (b) differs by less than 1% from the acceleration due to Earth’s gravity found in (a). This agreement is approximate because the Moon’s orbit is slightly elliptical, and Earth is not stationary (rather the Earth-Moon system rotates about its center of mass, which is located some 1700 km below Earth’s surface). The clear implication is that Earth’s gravitational force causes the Moon to orbit Earth.

Why does Earth not remain stationary as the Moon orbits it? This is because, as expected from Newton’s third law, if Earth exerts a force on the Moon, then the Moon should exert an equal and opposite force on Earth (see [link] ). We do not sense the Moon’s effect on Earth’s motion, because the Moon’s gravity moves our bodies right along with Earth but there are other signs on Earth that clearly show the effect of the Moon’s gravitational force as discussed in Satellites and Kepler's Laws: An Argument for Simplicity .

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Source:  OpenStax, Physics 110 at une. OpenStax CNX. Aug 29, 2013 Download for free at http://legacy.cnx.org/content/col11566/1.1
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