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F net = ma c = m v 2 r . size 12{F rSub { size 8{ ital "net"} } = ital "ma" rSub { size 8{c} } =m { {v rSup { size 8{2} } } over {r} } } {}

The net external force on mass m size 12{m} {} is gravity, and so we substitute the force of gravity for F net size 12{F rSub { size 8{ ital "net"} } } {} :

G mM r 2 = m v 2 r . size 12{G { { ital "mM"} over {r rSup { size 8{2} } } } =m { {v rSup { size 8{2} } } over {r} } } {}

The mass m size 12{m} {} cancels, yielding

G M r = v 2 . size 12{G { {M} over {r} } =v rSup { size 8{2} } } {}

The fact that m size 12{m} {} cancels out is another aspect of the oft-noted fact that at a given location all masses fall with the same acceleration. Here we see that at a given orbital radius r size 12{r} {} , all masses orbit at the same speed. (This was implied by the result of the preceding worked example.) Now, to get at Kepler’s third law, we must get the period T size 12{T} {} into the equation. By definition, period T size 12{T} {} is the time for one complete orbit. Now the average speed v size 12{v} {} is the circumference divided by the period—that is,

v = r T . size 12{v= { {2π`r} over {T} } } {}

Substituting this into the previous equation gives

G M r = 2 r 2 T 2 . size 12{G { { ital "mM"} over {r rSup { size 8{2} } } } =m { {v rSup { size 8{2} } } over {r} } } {}

Solving for T 2 size 12{T rSup { size 8{2} } } {} yields

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} } } {}

Using subscripts 1 and 2 to denote two different satellites, and taking the ratio of the last equation for satellite 1 to satellite 2 yields

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} } } } } {}

This is Kepler’s third law. Note that Kepler’s third law is valid only for comparing satellites of the same parent body, because only then does the mass of the parent body M size 12{M} {} cancel.

Now consider what we get if we solve T 2 = 2 GM r 3 for the ratio r 3 / T 2 size 12{r rSup { size 8{3} } /T rSup { size 8{2} } } {} . We obtain a relationship that can be used to determine the mass M size 12{M} {} of a parent body from the orbits of its satellites:

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} {}

If r size 12{r} {} and T size 12{T} {} are known for a satellite, then the mass M size 12{M} {} of the parent can be calculated. This principle has been used extensively to find the masses of heavenly bodies that have satellites. Furthermore, the ratio r 3 / T 2 size 12{r rSup { size 8{3} } /T rSup { size 8{2} } } {} should be a constant for all satellites of the same parent body (because r 3 / T 2 = GM / 2 size 12{r rSup { size 8{3} } /T rSup { size 8{2} } = ital "GM"/4π rSup { size 8{2} } } {} ). (See [link] ).

It is clear from [link] that the ratio of r 3 / T 2 size 12{r rSup { size 8{3} } /T rSup { size 8{2} } } {} is constant, at least to the third digit, for all listed satellites of the Sun, and for those of Jupiter. Small variations in that ratio have two causes—uncertainties in the r size 12{r} {} and T size 12{T} {} data, and perturbations of the orbits due to other bodies. Interestingly, those perturbations can be—and have been—used to predict the location of new planets and moons. This is another verification of Newton’s universal law of gravitation.

Making connections

Newton’s universal law of gravitation is modified by Einstein’s general theory of relativity, as we shall see in Particle Physics . Newton’s gravity is not seriously in error—it was and still is an extremely good approximation for most situations. Einstein’s modification is most noticeable in extremely large gravitational fields, such as near black holes. However, general relativity also explains such phenomena as small but long-known deviations of the orbit of the planet Mercury from classical predictions.

The case for simplicity

The development of the universal law of gravitation by Newton played a pivotal role in the history of ideas. While it is beyond the scope of this text to cover that history in any detail, we note some important points. The definition of planet set in 2006 by the International Astronomical Union (IAU) states that in the solar system, a planet is a celestial body that:

  1. is in orbit around the Sun,
  2. has sufficient mass to assume hydrostatic equilibrium and
  3. has cleared the neighborhood around its orbit.

Questions & Answers

show that the set of all natural number form semi group under the composition of addition
Nikhil Reply
what is the meaning
Dominic
explain and give four Example hyperbolic function
Lukman Reply
_3_2_1
felecia
⅗ ⅔½
felecia
_½+⅔-¾
felecia
The denominator of a certain fraction is 9 more than the numerator. If 6 is added to both terms of the fraction, the value of the fraction becomes 2/3. Find the original fraction. 2. The sum of the least and greatest of 3 consecutive integers is 60. What are the valu
SABAL Reply
1. x + 6 2 -------------- = _ x + 9 + 6 3 x + 6 3 ----------- x -- (cross multiply) x + 15 2 3(x + 6) = 2(x + 15) 3x + 18 = 2x + 30 (-2x from both) x + 18 = 30 (-18 from both) x = 12 Test: 12 + 6 18 2 -------------- = --- = --- 12 + 9 + 6 27 3
Pawel
2. (x) + (x + 2) = 60 2x + 2 = 60 2x = 58 x = 29 29, 30, & 31
Pawel
ok
Ifeanyi
on number 2 question How did you got 2x +2
Ifeanyi
combine like terms. x + x + 2 is same as 2x + 2
Pawel
x*x=2
felecia
2+2x=
felecia
×/×+9+6/1
Debbie
Q2 x+(x+2)+(x+4)=60 3x+6=60 3x+6-6=60-6 3x=54 3x/3=54/3 x=18 :. The numbers are 18,20 and 22
Naagmenkoma
Mark and Don are planning to sell each of their marble collections at a garage sale. If Don has 1 more than 3 times the number of marbles Mark has, how many does each boy have to sell if the total number of marbles is 113?
mariel Reply
Mark = x,. Don = 3x + 1 x + 3x + 1 = 113 4x = 112, x = 28 Mark = 28, Don = 85, 28 + 85 = 113
Pawel
how do I set up the problem?
Harshika Reply
what is a solution set?
Harshika
find the subring of gaussian integers?
Rofiqul
hello, I am happy to help!
Shirley Reply
please can go further on polynomials quadratic
Abdullahi
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Mark
I need quadratic equation link to Alpa Beta
Abdullahi Reply
find the value of 2x=32
Felix Reply
divide by 2 on each side of the equal sign to solve for x
corri
X=16
Michael
Want to review on complex number 1.What are complex number 2.How to solve complex number problems.
Beyan
yes i wantt to review
Mark
16
Makan
x=16
Makan
use the y -intercept and slope to sketch the graph of the equation y=6x
Only Reply
how do we prove the quadratic formular
Seidu Reply
please help me prove quadratic formula
Darius
hello, if you have a question about Algebra 2. I may be able to help. I am an Algebra 2 Teacher
Shirley Reply
thank you help me with how to prove the quadratic equation
Seidu
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Opoku
what is math number
Tric Reply
4
Trista
x-2y+3z=-3 2x-y+z=7 -x+3y-z=6
Sidiki Reply
can you teacch how to solve that🙏
Mark
Solve for the first variable in one of the equations, then substitute the result into the other equation. Point For: (6111,4111,−411)(6111,4111,-411) Equation Form: x=6111,y=4111,z=−411x=6111,y=4111,z=-411
Brenna
(61/11,41/11,−4/11)
Brenna
x=61/11 y=41/11 z=−4/11 x=61/11 y=41/11 z=-4/11
Brenna
Need help solving this problem (2/7)^-2
Simone Reply
x+2y-z=7
Sidiki
what is the coefficient of -4×
Mehri Reply
-1
Shedrak
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Source:  OpenStax, Une: physics for the health professions. OpenStax CNX. Aug 20, 2014 Download for free at http://legacy.cnx.org/content/col11697/1.1
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