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A non-satellite body fulfilling only the first two of the above criteria is classified as “dwarf planet.”

In 2006, Pluto was demoted to a ‘dwarf planet’ after scientists revised their definition of what constitutes a “true” planet.

Orbital data and kepler’s third law
Parent Satellite Average orbital radius r (km) Period T(y) r 3 / T 2 (km 3 / y 2 )
Earth Moon 3.84 × 10 5 size 12{3 "." "84" times "10" rSup { size 8{5} } } {} 0.07481 1 . 01 × 10 19 size 12{1 "." "01" times times "10" rSup { size 8{"18"} } } {}
Sun Mercury 5 . 79 × 10 7 size 12{5 "." "79" times "10" rSup { size 8{7} } } {} 0.2409 3 . 34 × 10 24 size 12{3 "." "34" times "10" rSup { size 8{"24"} } } {}
Venus 1 . 082 × 10 8 size 12{1 "." "082" times "10" rSup { size 8{8} } } {} 0.6150 3 . 35 × 10 24 size 12{3 "." "35" times "10" rSup { size 8{"24"} } } {}
Earth 1 . 496 × 10 8 size 12{1 "." "496" times "10" rSup { size 8{8} } } {} 1.000 3 . 35 × 10 24 size 12{3 "." "35" times "10" rSup { size 8{"24"} } } {}
Mars 2 . 279 × 10 8 size 12{2 "." "279" times "10" rSup { size 8{8} } } {} 1.881 3 . 35 × 10 24 size 12{3 "." "35" times "10" rSup { size 8{"24"} } } {}
Jupiter 7 . 783 × 10 8 size 12{7 "." "783" times "10" rSup { size 8{8} } } {} 11.86 3 . 35 × 10 24 size 12{3 "." "35" times "10" rSup { size 8{"24"} } } {}
Saturn 1 . 427 × 10 9 size 12{1 "." "427" times "10" rSup { size 8{9} } } {} 29.46 3 . 35 × 10 24 size 12{3 "." "35" times "10" rSup { size 8{"24"} } } {}
Neptune 4 . 497 × 10 9 size 12{4 "." "497" times "10" rSup { size 8{9} } } {} 164.8 3 . 35 × 10 24 size 12{3 "." "35" times "10" rSup { size 8{"24"} } } {}
Pluto 5 . 90 × 10 9 size 12{5 "." "90" times "10" rSup { size 8{9} } } {} 248.3 3 . 33 × 10 24 size 12{3 "." "33" times "10" rSup { size 8{"24"} } } {}
Jupiter Io 4 . 22 × 10 5 size 12{4 "." "22" times "10" rSup { size 8{5} } } {} 0.00485 (1.77 d) 3 . 19 × 10 21 size 12{3 "." "19" times "10" rSup { size 8{"21"} } } {}
Europa 6 . 71 × 10 5 size 12{6 "." "71" times "10" rSup { size 8{5} } } {} 0.00972 (3.55 d) 3 . 20 × 10 21 size 12{3 "." "20" times "10" rSup { size 8{"21"} } } {}
Ganymede 1 . 07 × 10 6 size 12{1 "." "07" times "10" rSup { size 8{6} } } {} 0.0196 (7.16 d) 3 . 19 × 10 21 size 12{3 "." "19" times "10" rSup { size 8{"21"} } } {}
Callisto 1 . 88 × 10 6 size 12{1 "." "88" times "10" rSup { size 8{6} } } {} 0.0457 (16.19 d) 3 . 20 × 10 21 size 12{3 "." "20" times "10" rSup { size 8{"21"} } } {}

The universal law of gravitation is a good example of a physical principle that is very broadly applicable. That single equation for the gravitational force describes all situations in which gravity acts. It gives a cause for a vast number of effects, such as the orbits of the planets and moons in the solar system. It epitomizes the underlying unity and simplicity of physics.

Before the discoveries of Kepler, Copernicus, Galileo, Newton, and others, the solar system was thought to revolve around Earth as shown in [link] (a). This is called the Ptolemaic view, for the Greek philosopher who lived in the second century AD. This model is characterized by a list of facts for the motions of planets with no cause and effect explanation. There tended to be a different rule for each heavenly body and a general lack of simplicity.

[link] (b) represents the modern or Copernican model. In this model, a small set of rules and a single underlying force explain not only all motions in the solar system, but all other situations involving gravity. The breadth and simplicity of the laws of physics are compelling. As our knowledge of nature has grown, the basic simplicity of its laws has become ever more evident.

In figure a the paths of the different planets are shown in the forms of dotted concentric circles with the Earth at the center with its Moon. The Sun is also shown revolving around the Earth. Each planet is labeled with its name. On the planets Mercury, Venus, Mars, Jupiter and Saturn green colored epicycles are shown. In the figure b Copernican view of planet is shown. The Sun is shown at the center of the solar system. The planets are shown moving around the Sun.
(a) The Ptolemaic model of the universe has Earth at the center with the Moon, the planets, the Sun, and the stars revolving about it in complex superpositions of circular paths. This geocentric model, which can be made progressively more accurate by adding more circles, is purely descriptive, containing no hints as to what are the causes of these motions. (b) The Copernican model has the Sun at the center of the solar system. It is fully explained by a small number of laws of physics, including Newton’s universal law of gravitation.

Section summary

  • Kepler’s laws are stated for a small mass m size 12{m} {} orbiting a larger mass M size 12{M} {} in near-isolation. Kepler’s laws of planetary motion are then as follows:

    Kepler’s first law

    The orbit of each planet about the Sun is an ellipse with the Sun at one focus.

    Kepler’s second law

    Each planet moves so that an imaginary line drawn from the Sun to the planet sweeps out equal areas in equal times.

    Kepler’s third law

    The ratio of the squares of the periods of any two planets about the Sun is equal to the ratio of the cubes of their average distances from the Sun:

    T 1  2 T 2  2 = r 1  3 r 2  3 , size 12{ { {T rSub { size 8{1} rSup { size 8{2} } } } over {T rSub { size 8{2} rSup { size 8{2} } } } } = { {r rSub { size 8{1} rSup { size 8{3} } } } over {r rSub { size 8{2} rSup { size 8{3} } } } } } {}

    where T size 12{m} {} is the period (time for one orbit) and r size 12{m} {} is the average radius of the orbit.

  • The period and radius of a satellite’s orbit about a larger body M size 12{m} {} are related by
    T 2 = 2 GM r 3 size 12{T rSup { size 8{2} } = { {4π rSup { size 8{2} } } over { ital "GM"} } r rSup { size 8{3} } } {}

    or

    r 3 T 2 = G 2 M . size 12{ { {r rSup { size 8{3} } } over {T rSup { size 8{2} } } } = { {G} over {4π rSup { size 8{2} } } } M} {}

Questions & Answers

how did you get 1640
Noor Reply
If auger is pair are the roots of equation x2+5x-3=0
Peter Reply
Wayne and Dennis like to ride the bike path from Riverside Park to the beach. Dennis’s speed is seven miles per hour faster than Wayne’s speed, so it takes Wayne 2 hours to ride to the beach while it takes Dennis 1.5 hours for the ride. Find the speed of both bikers.
MATTHEW Reply
420
Sharon
from theory: distance [miles] = speed [mph] × time [hours] info #1 speed_Dennis × 1.5 = speed_Wayne × 2 => speed_Wayne = 0.75 × speed_Dennis (i) info #2 speed_Dennis = speed_Wayne + 7 [mph] (ii) use (i) in (ii) => [...] speed_Dennis = 28 mph speed_Wayne = 21 mph
George
Let W be Wayne's speed in miles per hour and D be Dennis's speed in miles per hour. We know that W + 7 = D and W * 2 = D * 1.5. Substituting the first equation into the second: W * 2 = (W + 7) * 1.5 W * 2 = W * 1.5 + 7 * 1.5 0.5 * W = 7 * 1.5 W = 7 * 3 or 21 W is 21 D = W + 7 D = 21 + 7 D = 28
Salma
Devon is 32 32​​ years older than his son, Milan. The sum of both their ages is 54 54​. Using the variables d d​ and m m​ to represent the ages of Devon and Milan, respectively, write a system of equations to describe this situation. Enter the equations below, separated by a comma.
Aaron Reply
find product (-6m+6) ( 3m²+4m-3)
SIMRAN Reply
-42m²+60m-18
Salma
what is the solution
bill
how did you arrive at this answer?
bill
-24m+3+3mÁ^2
Susan
i really want to learn
Amira
I only got 42 the rest i don't know how to solve it. Please i need help from anyone to help me improve my solving mathematics please
Amira
Hw did u arrive to this answer.
Aphelele
hi
Bajemah
-6m(3mA²+4m-3)+6(3mA²+4m-3) =-18m²A²-24m²+18m+18mA²+24m-18 Rearrange like items -18m²A²-24m²+42m+18A²-18
Salma
complete the table of valuesfor each given equatio then graph. 1.x+2y=3
Jovelyn Reply
x=3-2y
Salma
y=x+3/2
Salma
Hi
Enock
given that (7x-5):(2+4x)=8:7find the value of x
Nandala
3x-12y=18
Kelvin
please why isn't that the 0is in ten thousand place
Grace Reply
please why is it that the 0is in the place of ten thousand
Grace
Send the example to me here and let me see
Stephen
A meditation garden is in the shape of a right triangle, with one leg 7 feet. The length of the hypotenuse is one more than the length of one of the other legs. Find the lengths of the hypotenuse and the other leg
Marry Reply
how far
Abubakar
cool u
Enock
state in which quadrant or on which axis each of the following angles given measure. in standard position would lie 89°
Abegail Reply
hello
BenJay
hi
Method
I am eliacin, I need your help in maths
Rood
how can I help
Sir
hmm can we speak here?
Amoon
however, may I ask you some questions about Algarba?
Amoon
hi
Enock
what the last part of the problem mean?
Roger
The Jones family took a 15 mile canoe ride down the Indian River in three hours. After lunch, the return trip back up the river took five hours. Find the rate, in mph, of the canoe in still water and the rate of the current.
cameron Reply
Shakir works at a computer store. His weekly pay will be either a fixed amount, $925, or $500 plus 12% of his total sales. How much should his total sales be for his variable pay option to exceed the fixed amount of $925.
mahnoor Reply
I'm guessing, but it's somewhere around $4335.00 I think
Lewis
12% of sales will need to exceed 925 - 500, or 425 to exceed fixed amount option. What amount of sales does that equal? 425 ÷ (12÷100) = 3541.67. So the answer is sales greater than 3541.67. Check: Sales = 3542 Commission 12%=425.04 Pay = 500 + 425.04 = 925.04. 925.04 > 925.00
Munster
difference between rational and irrational numbers
Arundhati Reply
When traveling to Great Britain, Bethany exchanged $602 US dollars into £515 British pounds. How many pounds did she receive for each US dollar?
Jakoiya Reply
how to reduced echelon form
Solomon Reply
Jazmine trained for 3 hours on Saturday. She ran 8 miles and then biked 24 miles. Her biking speed is 4 mph faster than her running speed. What is her running speed?
Zack Reply
d=r×t the equation would be 8/r+24/r+4=3 worked out
Sheirtina
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Source:  OpenStax, College physics. OpenStax CNX. Jul 27, 2015 Download for free at http://legacy.cnx.org/content/col11406/1.9
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