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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: general relativity and mercury

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:

Questions & Answers

A golfer on a fairway is 70 m away from the green, which sits below the level of the fairway by 20 m. If the golfer hits the ball at an angle of 40° with an initial speed of 20 m/s, how close to the green does she come?
Aislinn Reply
cm
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A mouse of mass 200 g falls 100 m down a vertical mine shaft and lands at the bottom with a speed of 8.0 m/s. During its fall, how much work is done on the mouse by air resistance
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what is inorganic
emma
Chemistry is a branch of science that deals with the study of matter,it composition,it structure and the changes it undergoes
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2. A sled plus passenger with total mass 50 kg is pulled 20 m across the snow (0.20) at constant velocity by a force directed 25° above the horizontal. Calculate (a) the work of the applied force, (b) the work of friction, and (c) the total work.
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you have been hired as an espert witness in a court case involving an automobile accident. the accident involved car A of mass 1500kg which crashed into stationary car B of mass 1100kg. the driver of car A applied his brakes 15 m before he skidded and crashed into car B. after the collision, car A s
Samuel Reply
can someone explain to me, an ignorant high school student, why the trend of the graph doesn't follow the fact that the higher frequency a sound wave is, the more power it is, hence, making me think the phons output would follow this general trend?
Joseph Reply
Nevermind i just realied that the graph is the phons output for a person with normal hearing and not just the phons output of the sound waves power, I should read the entire thing next time
Joseph
Follow up question, does anyone know where I can find a graph that accuretly depicts the actual relative "power" output of sound over its frequency instead of just humans hearing
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"Generation of electrical energy from sound energy | IEEE Conference Publication | IEEE Xplore" ***ieeexplore.ieee.org/document/7150687?reload=true
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progressive wave
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Mujahid
A string is 3.00 m long with a mass of 5.00 g. The string is held taut with a tension of 500.00 N applied to the string. A pulse is sent down the string. How long does it take the pulse to travel the 3.00 m of the string?
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Source:  OpenStax, College physics for ap® courses. OpenStax CNX. Nov 04, 2016 Download for free at https://legacy.cnx.org/content/col11844/1.14
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