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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 = 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 = 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 = 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 = 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).

Questions & Answers

where we get a research paper on Nano chemistry....?
Maira Reply
what are the products of Nano chemistry?
Maira Reply
There are lots of products of nano chemistry... Like nano coatings.....carbon fiber.. And lots of others..
learn
Even nanotechnology is pretty much all about chemistry... Its the chemistry on quantum or atomic level
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da
no nanotechnology is also a part of physics and maths it requires angle formulas and some pressure regarding concepts
Bhagvanji
Preparation and Applications of Nanomaterial for Drug Delivery
Hafiz Reply
revolt
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Application of nanotechnology in medicine
what is variations in raman spectra for nanomaterials
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Crow Reply
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RAW Reply
please someone correct me if I'm wrong but I think one can use nanoparticles, specially silver nanoparticles for water treatment.
Damian
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Professor
I think
Professor
Nasa has use it in the 60's, copper as water purification in the moon travel.
Alexandre
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Alexandre
what is the stm
Brian Reply
is there industrial application of fullrenes. What is the method to prepare fullrene on large scale.?
Rafiq
industrial application...? mmm I think on the medical side as drug carrier, but you should go deeper on your research, I may be wrong
Damian
How we are making nano material?
LITNING Reply
what is a peer
LITNING Reply
What is meant by 'nano scale'?
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What is STMs full form?
LITNING
scanning tunneling microscope
Sahil
how nano science is used for hydrophobicity
Santosh
Do u think that Graphene and Fullrene fiber can be used to make Air Plane body structure the lightest and strongest. Rafiq
Rafiq
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Mahi
what is simplest way to understand the applications of nano robots used to detect the cancer affected cell of human body.? How this robot is carried to required site of body cell.? what will be the carrier material and how can be detected that correct delivery of drug is done Rafiq
Rafiq
if virus is killing to make ARTIFICIAL DNA OF GRAPHENE FOR KILLED THE VIRUS .THIS IS OUR ASSUMPTION
Anam
analytical skills graphene is prepared to kill any type viruses .
Anam
Any one who tell me about Preparation and application of Nanomaterial for drug Delivery
Hafiz
what is Nano technology ?
Bob Reply
write examples of Nano molecule?
Bob
The nanotechnology is as new science, to scale nanometric
brayan
nanotechnology is the study, desing, synthesis, manipulation and application of materials and functional systems through control of matter at nanoscale
Damian
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Damian Reply
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What fields keep nano created devices from performing or assimulating ? Magnetic fields ? Are do they assimilate ?
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Kyle
yes I'm doing my masters in nanotechnology, we are being studying all these domains as well..
Adin
why?
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Kyle
biomolecules are e building blocks of every organics and inorganic materials.
Joe
how did you get the value of 2000N.What calculations are needed to arrive at it
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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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