# 0.4 4.5 newton’s universal law of gravitation  (Page 3/9)

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## 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 .

## Tides

Ocean tides are one very observable result of the Moon’s gravity acting on Earth. [link] is a simplified drawing of the Moon’s position relative to the tides. Because water easily flows on Earth’s surface, a high tide is created on the side of Earth nearest to the Moon, where the Moon’s gravitational pull is strongest. Why is there also a high tide on the opposite side of Earth? The answer is that Earth is pulled toward the Moon more than the water on the far side, because Earth is closer to the Moon. So the water on the side of Earth closest to the Moon is pulled away from Earth, and Earth is pulled away from water on the far side. As Earth rotates, the tidal bulge (an effect of the tidal forces between an orbiting natural satellite and the primary planet that it orbits) keeps its orientation with the Moon. Thus there are two tides per day (the actual tidal period is about 12 hours and 25.2 minutes), because the Moon moves in its orbit each day as well).

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Source:  OpenStax, Unit 4 - uniform circular motion and universal law of gravity. OpenStax CNX. Nov 23, 2015 Download for free at https://legacy.cnx.org/content/col11905/1.1
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 By Eric Crawford By By Subramanian Divya