Exemplo 6.6 de College Physics, 6.5 Newton’s Universal Law of Gravitation

Earth’s Gravitational Force Is the Centripetal Force Making the Moon Move in a Curved Path

Problema e solução em inglês, como no livro original.

(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.

Resolva no papel primeiro. Depois abra a solução um passo de cada vez e pare assim que conseguir terminar sozinho.

Estratégia para (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 $3 . 84 \times 10^{8} m$ .
Solução para (a)
Substituting known values into the expression for $g$ found above, remembering that $M$ is the mass of Earth not the Moon, yields
$\begin{array}{l} g & = & G \frac{M}{r^{2}} = (6 . 67 \times 10^{- 11} \frac{N \cdot m^{2}}{{kg}^{2}}) \times \frac{5 . 98 \times 10^{24} kg}{(3 . 84 \times 10^{8} m)^{2}} \\ = & 2 . 70 \times 10^{- 3} {m/s.}^{2} \end{array}$
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Estratégia para (b)
Centripetal acceleration can be calculated using either form of

$\begin{matrix} a_{c} = \frac{v^{2}}{r} \\ a_{c} = {rω}^{2} \end{matrix}} .$
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We choose to use the second form:

$a_{c} = {rω}^{2},$
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where $ω$ is the angular velocity of the Moon about Earth.
Solução para (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
$1 d \times 24 \frac{hr}{d} \times 60 \frac{min}{hr} \times 60 \frac{s}{min} = 86,400 s$
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we see that
$ω = \frac{Δ θ}{Δ t} = \frac{2π rad}{(27 . 3 d) (86,400 s/d)} = 2 . 66 \times 10^{- 6} \frac{rad}{s} .$
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The centripetal acceleration is

$\begin{array}{l} a_{c} & = & {rω}^{2} = (3 . 84 \times 10^{8} m) (2 . 66 \times 10^{- 6} rad/s)^{2} \\ = & 2.72 \times 10^{- 3} {m/s.}^{2} \end{array}$
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The direction of the acceleration is toward the center of the Earth.
Discussão
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.

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