as the size of a planetâ€™s orbit increases, what happens to its period?

13 Gravitation

thirteen.5 Kepler'southward Laws of Planetary Motion

Learning Objectives

By the end of this section, y'all will be able to:

Describe the conic sections and how they relate to orbital motion
Draw how orbital velocity is related to conservation of athwart momentum
Determine the period of an elliptical orbit from its major centrality

Using the precise data collected past Tycho Brahe, Johannes Kepler carefully analyzed the positions in the sky of all the known planets and the Moon, plotting their positions at regular intervals of time. From this analysis, he formulated 3 laws, which we accost in this section.

Kepler'southward First Constabulary

The prevailing view during the time of Kepler was that all planetary orbits were circular. The data for Mars presented the greatest challenge to this view and that eventually encouraged Kepler to give upwards the popular idea. Kepler'south first police states that every planet moves along an ellipse, with the Sun located at a focus of the ellipse. An ellipse is defined as the fix of all points such that the sum of the distance from each bespeak to two foci is a constant. (Figure) shows an ellipse and describes a simple way to create it.

Figure a shows an x y coordinate system and an ellipse centered on the origin with foci f 1 on the left and f 2 on the right, both on the x axis. Focus f 1 is also labeled M. A point above focus f 2 is labeled m. The right triangle formed by f 1, f 2, and m is shown in red. Figure b shows a similar ellipse, with the sun shown and labeled as M and as Sun at f 1. A planet mass m is shown above f 1, at a vertical distance r from f 1. The location where the ellipse intersects the horizontal axis on the left is labeled as point A, and the location where the ellipse intersects the horizontal axis on the right is labeled as point B. — **Figure 13.16** (a) An ellipse is a curve in which the sum of the distances from a point on the bend to ii foci
$({f}_{1}\,\text{and}\,{f}_{2})$

is a constant. From this definition, you can come across that an ellipse tin be created in the following fashion. Place a pin at each focus, then place a loop of string effectually a pencil and the pins. Keeping the string taught, movement the pencil around in a complete circuit. If the two foci occupy the same place, the result is a circle—a special case of an ellipse. (b) For an elliptical orbit, if

$m\ll M$

, so yard follows an elliptical path with Thousand at one focus. More exactly, both 1000 and K move in their own ellipse about the common centre of mass.

\[({f}_{1}\,\text{and}\,{f}_{2})\] — **Figure 13.16** (a) An ellipse is a curve in which the sum of the distances from a point on the bend to ii foci
$({f}_{1}\,\text{and}\,{f}_{2})$

is a constant. From this definition, you can come across that an ellipse tin be created in the following fashion. Place a pin at each focus, then place a loop of string effectually a pencil and the pins. Keeping the string taught, movement the pencil around in a complete circuit. If the two foci occupy the same place, the result is a circle—a special case of an ellipse. (b) For an elliptical orbit, if

$m\ll M$

, so yard follows an elliptical path with Thousand at one focus. More exactly, both 1000 and K move in their own ellipse about the common centre of mass.

For elliptical orbits, the point of closest approach of a planet to the Sun is chosen the perihelion. It is labeled point A in (Figure). The farthest point is the aphelion and is labeled indicate B in the figure. For the Moon's orbit nearly Earth, those points are called the perigee and apogee, respectively.

An ellipse has several mathematical forms, but all are a specific case of the more general equation for conic sections. There are four different conic sections, all given past the equation

$\frac{\alpha }{r}=1+e\text{cos}\theta .$

The variables r and

$\theta$

are shown in (Figure) in the case of an ellipse. The constants

$\alpha$

and eastward are determined by the total free energy and angular momentum of the satellite at a given point. The constant eastward is chosen the eccentricity. The values of

$\alpha$

and e determine which of the four conic sections represents the path of the satellite.

An x y coordinate system and an ellipse centered on the origin with foci f 1 on the left and f 2 on the right, both on the x axis, are shown. Focus f 1 is also labeled M. A point on the ellipse in the first quadrant is labeled m. The horizontal segment connecting the foci f 1 and f 2, and the segment connecting f 1 and m are shown in red. The angle between those segments is labeled Theta. — **Figure 13.17** As before, the distance between the planet and the Sun is r, and the angle measured from the x-axis, which is along the major axis of the ellipse, is
$\theta$

.

1 of the real triumphs of Newton's law of universal gravitation, with the force proportional to the inverse of the altitude squared, is that when it is combined with his second law, the solution for the path of any satellite is a conic section. Every path taken by 1000 is one of the iv conic sections: a circle or an ellipse for leap or closed orbits, or a parabola or hyperbola for unbounded or open orbits. These conic sections are shown in (Figure).

A cone and its conic sections is shown. At the top a horizontal cut is shaded and a dashed line shown across the shading. This section is labeled circle. Below this a diagonal cut and line are shown. The line and cut intersect the sides of the cone. This section is labeled ellipse. Next is a diagonal cut and line that intersect the sides and the bottom of the cone and are labeled parabola. The last section is a vertical line and shaded cut labeled hyperbola — **Effigy thirteen.18** All motion caused past an inverse square force is ane of the four conic sections and is determined by the energy and management of the moving trunk.

If the total free energy is negative, then

$0\le e<1$

, and (Figure) represents a bound or closed orbit of either an ellipse or a circle, where

$e=0$

. [You can see from (Figure) that for

$e=0$

$r=\alpha$

, and hence the radius is constant.] For ellipses, the eccentricity is related to how ellipsoidal the ellipse appears. A circle has zero eccentricity, whereas a very long, drawn-out ellipse has an eccentricity virtually one.

If the total energy is exactly zero, then

$e=1$

and the path is a parabola. Retrieve that a satellite with zero total energy has exactly the escape velocity. (The parabola is formed merely by slicing the cone parallel to the tangent line forth the surface.) Finally, if the total energy is positive, then

$e>1$

and the path is a hyperbola. These last ii paths stand for unbounded orbits, where yard passes by M in one case and just in one case. This situation has been observed for several comets that approach the Sun and and so travel abroad, never to render.
We have confined ourselves to the case in which the smaller mass (planet) orbits a much larger, and hence stationary, mass (Lord's day), but (Effigy) also applies to any two gravitationally interacting masses. Each mass traces out the exact aforementioned-shaped conic section as the other. That shape is adamant by the total energy and angular momentum of the arrangement, with the center of mass of the organization located at the focus. The ratio of the dimensions of the 2 paths is the inverse of the ratio of their masses.

Y'all tin see an animation of ii interacting objects at the My Solar System page at Phet. Choose the Sunday and Planet preset option. Y'all tin can likewise view the more complicated multiple torso problems likewise. You may find the bodily path of the Moon quite surprising, yet is obeying Newton's simple laws of motion.

Orbital Transfers

People accept imagined traveling to the other planets of our solar system since they were discovered. But how can we all-time practice this? The most efficient method was discovered in 1925 past Walter Hohmann, inspired by a pop scientific discipline fiction novel of that time. The method is at present called a Hohmann transfer . For the case of traveling betwixt ii circular orbits, the transfer is along a "transfer" ellipse that perfectly intercepts those orbits at the aphelion and perihelion of the ellipse. (Effigy) shows the instance for a trip from Earth's orbit to that of Mars. Every bit before, the Sun is at the focus of the ellipse.

For any ellipse, the semi-major centrality is defined as half the sum of the perihelion and the aphelion. In (Effigy), the semi-major axis is the distance from the origin to either side of the ellipse along the ten-axis, or just one-half the longest axis (chosen the major axis). Hence, to travel from one circular orbit of radius

${r}_{1}$

to some other round orbit of radius

${r}_{2}$

, the aphelion of the transfer ellipse will be equal to the value of the larger orbit, while the perihelion volition exist the smaller orbit. The semi-major axis, denoted a, is therefore given by

$a=\frac{1}{2}({r}_{1}+{r}_{2})$

An illustration of the sun and three orbits around it are shown. All three orbits are circular. The innermost orbit is centered on the sun and is labeled Earth Orbit. The middle orbit is not centered on the sun. It coincides with the earth orbit at a point labeled — **Figure 13.19** The transfer ellipse has its perihelion at World'southward orbit and aphelion at Mars' orbit.

Allow'south take the case of traveling from Earth to Mars. For the moment, we ignore the planets and presume we are alone in Earth's orbit and wish to move to Mars' orbit. From (Figure), the expression for full free energy, we can see that the total energy for a spacecraft in the larger orbit (Mars) is greater (less negative) than that for the smaller orbit (Earth). To move onto the transfer ellipse from Globe's orbit, we will demand to increase our kinetic energy, that is, we need a velocity boost. The most efficient method is a very quick acceleration forth the circular orbital path, which is also along the path of the ellipse at that betoken. (In fact, the acceleration should be instantaneous, such that the circular and elliptical orbits are congruent during the acceleration. In practice, the finite acceleration is curt enough that the difference is non a significant consideration.) Once you take arrived at Mars orbit, yous volition need another velocity boost to move into that orbit, or y'all will stay on the elliptical orbit and only fall dorsum to perihelion where y'all started. For the return trip, y'all simply reverse the process with a retro-boost at each transfer point.

To make the move onto the transfer ellipse and and so off again, nosotros need to know each circular orbit velocity and the transfer orbit velocities at perihelion and aphelion. The velocity boost required is simply the difference between the circular orbit velocity and the elliptical orbit velocity at each signal. We can find the circular orbital velocities from (Figure). To decide the velocities for the ellipse, nosotros land without proof (as it is across the scope of this course) that total energy for an elliptical orbit is

$E=-\frac{Gm{M}_{\text{S}}}{2a}$

where

${M}_{\text{S}}$

is the mass of the Sun and a is the semi-major axis. Remarkably, this is the same as (Figure) for circular orbits, simply with the value of the semi-major axis replacing the orbital radius. Since nosotros know the potential energy from (Figure), nosotros can find the kinetic free energy and hence the velocity needed for each point on the ellipse. We go out it equally a challenge trouble to find those transfer velocities for an Earth-to-Mars trip.

We end this discussion by pointing out a few important details. Get-go, we have non accounted for the gravitational potential energy due to Earth and Mars, or the mechanics of landing on Mars. In practice, that must exist part of the calculations. Second, timing is everything. You practice non want to go far at the orbit of Mars to find out it isn't in that location. Nosotros must leave World at precisely the right fourth dimension such that Mars will be at the aphelion of our transfer ellipse just as we go far. That opportunity comes about every 2 years. And returning requires correct timing likewise. The total trip would take just under iii years! There are other options that provide for a faster transit, including a gravity assist flyby of Venus. But these other options come up with an additional cost in free energy and danger to the astronauts.

Visit this site for more than details well-nigh planning a trip to Mars.

Kepler's 2d Law

Kepler'south second police force states that a planet sweeps out equal areas in equal times, that is, the expanse divided past fourth dimension, chosen the areal velocity, is constant. Consider (Figure). The time information technology takes a planet to motion from position A to B, sweeping out area

${A}_{1}$

, is exactly the time taken to movement from position C to D, sweeping area

${A}_{2}$

, and to move from E to F, sweeping out expanse

${A}_{3}$

. These areas are the same:

${A}_{1}={A}_{2}={A}_{3}$

An x y coordinate system is shown with the sun, also labeled as M, on the x axis to the left of the origin and an unlabeled point to the right of the origin. A planet, labeled also as m, is shown in the second quadrant. An arrow, labeled v, extends from the planet and points down and left, tangent to the orbit. Points A, B, C, D, E, and F are labeled on the orbit. Points A and B are in the third quadrant. The area of the region defined by A B and the sun is labeled A 1. Points C and D are in on the orbit on either side of the – y axis. The area of the region defined by C D and the sun is labeled A 2. Points E and F are in the first quadrant. The area of the region defined by E F and the sun is labeled A 3. The pair of points A B have the largest distance between them and is closest to the sun. E F have the smallest distance between them and are farthest from the sun. — **Effigy xiii.20** The shaded regions shown have equal areas and represent the same time interval.

Comparing the areas in the figure and the distance traveled along the ellipse in each case, we tin can see that in social club for the areas to be equal, the planet must speed up as it gets closer to the Sun and slow down as it moves abroad. This behavior is completely consistent with our conservation equation, (Effigy). Only nosotros will prove that Kepler'due south second law is actually a issue of the conservation of angular momentum, which holds for any organisation with only radial forces.

Recall the definition of angular momentum from Angular Momentum,

$\overset{\to }{L}=\overset{\to }{r}\,×\,\overset{\to }{p}$

. For the example of orbiting motility,

$\overset{\to }{L}$

is the angular momentum of the planet about the Sun,

$\overset{\to }{r}$

is the position vector of the planet measured from the Sun, and

$\overset{\to }{p}=m\overset{\to }{v}$

is the instantaneous linear momentum at any indicate in the orbit. Since the planet moves forth the ellipse,

$\overset{\to }{p}$

is always tangent to the ellipse.

We can resolve the linear momentum into 2 components: a radial component

${\overset{\to }{p}}_{\text{rad}}$

forth the line to the Sun, and a component

${\overset{\to }{p}}_{\text{perp}}$

perpendicular to

$\overset{\to }{r}$

. The cross product for angular momentum can so exist written as

$\overset{\to }{L}=\overset{\to }{r}\,×\,\overset{\to }{p}=\overset{\to }{r}\,×\,({\overset{\to }{p}}_{\text{rad}}+{\overset{\to }{p}}_{\text{perp}})=\overset{\to }{r}\,×\,{\overset{\to }{p}}_{\text{rad}}+\overset{\to }{r}\,×\,{\overset{\to }{p}}_{\text{perp}}$

The first term on the correct is zero because

$\overset{\to }{r}$

is parallel to

${\overset{\to }{p}}_{\text{rad}}$

, and in the second term

$\overset{\to }{r}$

is perpendicular to

${\overset{\to }{p}}_{\text{perp}}$

, and then the magnitude of the cross product reduces to

$L=r{p}_{\text{perp}}=rm{v}_{\text{perp}}$

. Notation that the angular momentum does not depend upon

${p}_{\text{rad}}$

. Since the gravitational strength is only in the radial direction, information technology can alter only

${p}_{\text{rad}}$

and non

${p}_{\text{perp}}$

; hence, the angular momentum must remain abiding.

Now consider (Figure). A small triangular expanse

$\text{Δ}A$

is swept out in time

$\text{Δ}t$

. The velocity is along the path and it makes an angle

$\theta$

with the radial direction. Hence, the perpendicular velocity is given by

${v}_{\text{perp}}=v\text{sin}\theta$

. The planet moves a altitude

$\text{Δ}s=v\text{Δ}t\text{sin}\theta$

projected along the management perpendicular to r. Since the area of a triangle is i-half the base of operations (r) times the height

$(\text{Δ}s)$

, for a small-scale displacement, the area is given by

$\text{Δ}A=\frac{1}{2}r\text{Δ}s$

. Substituting for

$\text{Δ}s$

, multiplying past m in the numerator and denominator, and rearranging, we obtain

$\text{Δ}A=\frac{1}{2}r\text{Δ}s=\frac{1}{2}r(v\text{Δ}t\text{sin}\theta )=\frac{1}{2m}r(mv\text{sin}\theta \text{Δ}t)=\frac{1}{2m}r(m{v}_{\text{perp}}\text{Δ}t)=\frac{L}{2m}\text{Δ}t.$

A diagram showing the sun and a planet separated by a distance r. The velocity vector of the planet is shown as an arrow pointing at an obtuse angle to the distance r between the sun and planet. The line connecting the sun and planet is extended past the planet as a dashed line, and another dashed line is drawn from the tip of the velocity arrow to the dashed extension of r. The dashed lines meet at a right angle and form a triangle with the velocity arrow forming the hypotenuse and the planet at one vertex. The angle near the planet is labeled theta. The hypotenuse is also labeled v delta t, and the side opposite the planet labeled v delta t sin theta. The triangular region defined by the sun, planet and the tip of the velocity arrow is labeled Delta A, and the angle near the sun is labeled delta phi. — **Effigy 13.21** The chemical element of expanse
$\text{Δ}A$

swept out in time

$\text{Δ}t$

every bit the planet moves through angle

$\text{Δ}\varphi$

. The angle between the radial direction and

$\overset{\to }{v}$

is

$\theta$

.

The areal velocity is simply the rate of alter of surface area with time, so nosotros have

$\text{areal velocity}=\,\frac{\text{Δ}A}{\text{Δ}t}=\frac{L}{2m}.$

Since the athwart momentum is constant, the areal velocity must also be constant. This is exactly Kepler'southward second police. As with Kepler's first constabulary, Newton showed it was a natural upshot of his constabulary of gravitation.

You tin can view an animated version of (Figure), and many other interesting animations besides, at the School of Physics (Academy of New South Wales) site.

Kepler's Third Law

Kepler's third law states that the foursquare of the period is proportional to the cube of the semi-major centrality of the orbit. In Satellite Orbits and Energy, we derived Kepler's third law for the special case of a round orbit. (Figure) gives united states of america the menstruum of a circular orbit of radius r about Earth:

$T=2\pi \sqrt{\frac{{r}^{3}}{G{M}_{\text{E}}}}.$

For an ellipse, think that the semi-major axis is 1-half the sum of the perihelion and the aphelion. For a circular orbit, the semi-major axis (a) is the same as the radius for the orbit. In fact, (Figure) gives us Kepler'southward third law if we simply supplant r with a and square both sides.

${T}^{2}=\frac{4{\pi }^{2}}{GM}{a}^{3}$

Nosotros have changed the mass of World to the more general M, since this equation applies to satellites orbiting any large mass.

Instance

Orbit of Halley's Comet

Determine the semi-major axis of the orbit of Halley's comet, given that it arrives at perihelion every 75.3 years. If the perihelion is 0.586 AU, what is the aphelion?

Strategy

We are given the period, so we can rearrange (Figure), solving for the semi-major axis. Since we know the value for the perihelion, we can utilize the definition of the semi-major axis, given earlier in this department, to find the aphelion. Nosotros notation that 1 Astronomical Unit (AU) is the average radius of Earth's orbit and is defined to exist

$1\,\text{AU}=1.50\,×\,{10}^{11}\,\text{m}$

Solution

Rearranging (Figure) and inserting the values of the catamenia of Halley's comet and the mass of the Sun, nosotros have

$\begin{array}{}\\ \hfill a& ={(\frac{GM}{4{\pi }^{2}}{T}^{2})}^{1\text{/}3}\hfill \\ & ={(\frac{(6.67\,×\,{10}^{-11}\,\text{N}·{\text{m}}^{2}{\text{/kg}}^{2})(2.00\,×\,{10}^{30}\,\text{kg})}{4{\pi }^{2}}{(75.3\,\text{yr}\,×\,365\,\text{days/yr}\,×\,24\,\text{hr/day}\,×\,3600\,\text{s/hr})}^{2})}^{1\text{/}3}.\hfill \end{array}$

This yields a value of

$2.67\,×\,{10}^{12}\,\text{m}$

or 17.eight AU for the semi-major centrality.

The semi-major centrality is one-half the sum of the aphelion and perihelion, so we accept

$\begin{array}{}\\ \hfill a& =\hfill & \frac{1}{2}(\text{aphelion}+\text{perihelion})\hfill \\ \hfill \text{aphelion}& =\hfill & 2a-\text{perihelion}.\hfill \end{array}$

Substituting for the values, we plant for the semi-major centrality and the value given for the perihelion, we find the value of the aphelion to exist 35.0 AU.

Significance

Edmond Halley , a contemporary of Newton, first suspected that three comets, reported in 1531, 1607, and 1682, were really the aforementioned comet. Before Tycho Brahe made measurements of comets, it was believed that they were i-time events, perhaps disturbances in the atmosphere, and that they were not affected past the Sun. Halley used Newton'southward new mechanics to predict his namesake comet's render in 1758.

Check Your Understanding

The nearly round orbit of Saturn has an average radius of nigh 9.5 AU and has a period of 30 years, whereas Uranus averages nearly 19 AU and has a menstruation of 84 years. Is this consistent with our results for Halley's comet?

[reveal-answer q="fs-id1168326822487″]Prove Solution[/reveal-answer]

[hidden-answer a="fs-id1168326822487″]

The semi-major axis for the highly elliptical orbit of Halley'south comet is 17.viii AU and is the average of the perihelion and aphelion. This lies between the 9.five AU and 19 AU orbital radii for Saturn and Uranus, respectively. The radius for a circular orbit is the same as the semi-major axis, and since the menstruum increases with an increase of the semi-major axis, the fact that Halley's menstruation is betwixt the periods of Saturn and Uranus is expected.

[/hidden-answer]

Summary

All orbital motion follows the path of a conic section. Spring or closed orbits are either a circumvolve or an ellipse; unbounded or open orbits are either a parabola or a hyperbola.
The areal velocity of any orbit is constant, a reflection of the conservation of angular momentum.
The square of the menstruation of an elliptical orbit is proportional to the cube of the semi-major axis of that orbit.

Conceptual Questions

Are Kepler'due south laws purely descriptive, or exercise they contain causal information?

In the diagram below for a satellite in an elliptical orbit about a much larger mass, betoken where its speed is the greatest and where information technology is the least. What conservation law dictates this beliefs? Indicate the directions of the force, acceleration, and velocity at these points. Draw vectors for these aforementioned three quantities at the ii points where the y-axis intersects (along the semi-minor axis) and from this determine whether the speed is increasing decreasing, or at a max/min.

A diagram showing an x y coordinate system and an ellipse, centered on the origin with foci on the x axis. The focus on the left is labeled f 1 and M. The focus on the right is labeled f 2. A location labeled as m is shown above f 2. The right triangle defined by f 1, f 2, and m is shown in red. The clockwise direction tangent to the ellipse is indicated by blue arrows.

[reveal-respond q="103119″]Testify Solution[/reveal-answer]
[hidden-answer a="103119″]The speed is greatest where the satellite is closest to the large mass and least where farther away—at the periapsis and apoapsis, respectively. Information technology is conservation of angular momentum that governs this relationship. But it can also be gleaned from conservation of free energy, the kinetic energy must exist greatest where the gravitational potential energy is the to the lowest degree (well-nigh negative). The force, and hence acceleration, is e'er directed towards M in the diagram, and the velocity is ever tangent to the path at all points. The acceleration vector has a tangential component forth the direction of the velocity at the upper location on the y-centrality; hence, the satellite is speeding upwardly. Only the opposite is true at the lower position.[/hidden-answer]

Problems

Calculate the mass of the Sun based on data for average Earth's orbit and compare the value obtained with the Lord's day'south commonly listed value of

$1.989\,×\,{10}^{30}\,\text{kg}$

.
[reveal-answer q="114708″]Prove Solution[/reveal-answer]
[hidden-reply a="114708″]

$1.98\,×\,{10}^{30}\,\text{kg}$

; The values are the same within 0.05%.[/hidden-answer]

Io orbits Jupiter with an average radius of 421,700 km and a flow of i.769 days. Based upon these data, what is the mass of Jupiter?

The "mean" orbital radius listed for astronomical objects orbiting the Lord's day is typically not an integrated average but is calculated such that information technology gives the right period when applied to the equation for round orbits. Given that, what is the mean orbital radius in terms of aphelion and perihelion?

[reveal-respond q="fs-id1168329510821″]Bear witness Solution[/reveal-reply]

[hidden-reply a="fs-id1168329510821″]

Compare (Figure) and (Figure) to come across that they differ only in that the circular radius, r, is replaced past the semi-major axis, a. Therefore, the hateful radius is half the sum of the aphelion and perihelion, the aforementioned as the semi-major axis.

[/hidden-respond]

The perihelion of Halley's comet is 0.586 AU and the aphelion is 17.8 AU. Given that its speed at perihelion is 55 km/due south, what is the speed at aphelion (

$1\,\text{AU}=1.496\,×\,{10}^{11}\,\text{m}$

)? (Hint: Y'all may utilize either conservation of energy or angular momentum, but the latter is much easier.)

The perihelion of the comet Lagerkvist is ii.61 AU and information technology has a flow of 7.36 years. Show that the aphelion for this comet is 4.95 AU.

[reveal-answer q="fs-id1168329148430″]Show Solution[/reveal-answer]

[hidden-answer a="fs-id1168329148430″]

The semi-major axis, 3.78 AU is plant from the equation for the period. This is one-one-half the sum of the aphelion and perihelion, giving an aphelion distance of 4.95 AU.

[/hidden-answer]

What is the ratio of the speed at perihelion to that at aphelion for the comet Lagerkvist in the previous problem?

Eros has an elliptical orbit almost the Sun, with a perihelion altitude of i.xiii AU and aphelion distance of 1.78 AU. What is the menstruation of its orbit?

[reveal-answer q="fs-id1168329181708″]Show Solution[/reveal-respond]

[hidden-reply a="fs-id1168329181708″]

1.75 years

[/hidden-answer]

Glossary

aphelion: uttermost bespeak from the Sun of an orbiting trunk; the corresponding term for the Moon's farthest indicate from Earth is the apogee

Kepler's first law: law stating that every planet moves forth an ellipse, with the Sun located at a focus of the ellipse

Kepler's second law: law stating that a planet sweeps out equal areas in equal times, meaning it has a constant areal velocity

Kepler's third constabulary: law stating that the square of the period is proportional to the cube of the semi-major axis of the orbit

perihelion: point of closest approach to the Sun of an orbiting body; the corresponding term for the Moon's closest approach to Earth is the perigee

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Source: https://opentextbc.ca/universityphysicsv1openstax/chapter/13-5-keplers-laws-of-planetary-motion/

as the size of a planetâ€™s orbit increases, what happens to its period?

thirteen.5 Kepler'southward Laws of Planetary Motion

Learning Objectives

Kepler'southward First Constabulary

Orbital Transfers

Kepler's 2d Law

Kepler's Third Law

Instance

Orbit of Halley's Comet

Strategy

Solution

Significance

Check Your Understanding

Summary

Conceptual Questions

Problems

Glossary

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