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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 size 12{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 × 10 8 m size 12{3 "." "84" times "10" rSup { size 8{8} } `m} {} .

Solution for (a)

Substituting known values into the expression for g size 12{M} {} found above, remembering that M size 12{M} {} is the mass of Earth not the Moon, yields

g = G M r 2 = 6 . 67 × 10 − 11 N ⋅ m 2 kg 2 × 5 . 98 × 10 24 kg ( 3 . 84 × 10 8 m ) 2 = 2 . 70 × 10 − 3 m/s. 2

Strategy for (b)

Centripetal acceleration can be calculated using either form of

a c = v 2 r a c = rω 2 } . size 12{ left none matrix { a rSub { size 8{c} } = { {v rSup { size 8{2} } } over {r} } {} ##a rSub { size 8{c} } =rω rSup { size 8{2} } } right rbrace "." } {}

We choose to use the second form:

a c = rω 2 , size 12{a rSub { size 8{c} } =rω rSup { size 8{2} } } {}

where ω size 12{ω} {} 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 hr d × 60 min hr × 60 s min = 86,400 s size 12{ω= { {Δθ} over {Δt} } = { {2π" rad"} over { \( "27" "." "3 d" \) \( "86,400 s/d" \) } } =2 "." "66" times "10" rSup { size 8{ - 6} } { {"rad"} over {s} } } {}

we see that

ω = Δ θ Δ t = 2π rad ( 27 . 3 d ) ( 86,400 s/d ) = 2 . 66 × 10 − 6 rad s . size 12{ω= { {Δθ} over {Δt} } = { {2π" rad"} over { \( "27" "." "3 d" \) \( "86,400 s/d" \) } } =2 "." "66" times "10" rSup { size 8{ - 6} } { {"rad"} over {s} } } {}

The centripetal acceleration is

a c = rω 2 = ( 3 . 84 × 10 8 m ) ( 2 . 66 × 10 − 6 rad/s ) 2 = 2.72 × 10 − 3 m/s. 2 alignl { stack { size 12{a rSub { size 8{c} } =rω rSup { size 8{2} } = \( 3 "." "84" times "10" rSup { size 8{8} } " m" \) \( 2 "." "66" times "10" rSup { size 8{ - 6} } " rad/s" \) rSup { size 8{2} } } {} #" "=2 "." "72" times "10" rSup { size 8{ - 3} } " m/s" rSup { size 8{2} } {} } } {}

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 .

Figure a shows the Earth and the Moon around it orbiting in a circular path shown here as a circle around the Earth with an arrow over it showing the counterclockwise direction of the Moon. The center of mass of the circle is shown here with a point on the Earth that is not the Earth’s center but just right to its center. Figure b shows the Sun and the counterclockwise rotation of the Earth around it, in an elliptical path, which has wiggles. Along this path the center of mass of the Earth-Moon is also shown; it follows non-wiggled elliptical path.
(a) Earth and the Moon rotate approximately once a month around their common center of mass. (b) Their center of mass orbits the Sun in an elliptical orbit, but Earth’s path around the Sun has “wiggles” in it. Similar wiggles in the paths of stars have been observed and are considered direct evidence of planets orbiting those stars. This is important because the planets’ reflected light is often too dim to be observed.

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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