Elliptical pizza also called " Kepler's pizza" .
All slices have the same areas.
source http://www.till.com/blog/index.html
LAB KEPLER : polar paper, orbit of Mars
1)
ohannes Kepler was German and fond of Astronomy. He was an amazing mathematician too.
During the 30 years war in Europe (catholic were persecuting protestants and protestants were fighting each others) , he had to flee Germany
and moved to Prague. In Prague, he met Tycho Brahe who had collected, with the naked eye, and amazingly very accurately, positions of planets around the Sun.
Tycho Brahe hired Kepler to work on the data in search of patterns. The 2 men didn't get along and Brahe would only give Kepler the precious data a few at a time. Kepler got lucky when Brahe suddenly died (broken bladder) and stole the data from the family.
With the data (positions of Mars relative to the Sun) he was able to infer the famous 3 Kepler laws:
A) Planets move along ellipses around the Sun (most of them are almost circles). (the sun is at one focus)
B) When moving along their orbit, planets sweep the same area during the same time.
(meaning they race when they are close to the Sun and slows down when they are away from it)
C) For all the planets in our solar system, the ratio between (the orbital period)2 and (the distance from the Sun)3 is the same.
or (the orbital period of planet)2 / (the distance from the Sun of planet )3 = K K is a constant.
That was an major discovery. Only based on complex computations and on very accurate data. Next, Newton invents calculus
and is able to derive these laws and explain , using calculus and his universal gravitational law, why .
The third law is about ratio. If we use Earth as a reference we can write:
(orbital period planet)2 / (distance planet-Sun)3 = (orbital period earth)2 / (distance earth-Sun)3 = K (constant)
Let's take (distance Earth-Sun ) is 1 AU (1 astronomical unit is about 150 million km but don't use 150 million km in your computations)
(distance of planets will be in AUs)
Let's take the orbital period of the Earth is 1 year (period of planets will be in earth years)
We know that it takes MArs 1.88 earth years to go around the sun (orbital period planet)
Use the proportion to find the (distance mars-sun)3 = ____________ . (hint: solve the proportion above).
So solve for (distance mars-sun) , with your TI, raise the previous number to (1/3) (that will undo the third power)
(distance mars-sun) =_______________ AUs.
Do the same thing for Jupiter. Find in the Internet, the orbital period of Jupiter in Earth years = _______ years.
(distance Jupiter-Sun) = ___________ AUS.
2) Thanks to Kepler's laws we can find out the relative distances of planets from the Sun.
The distance Sun- Earth is taken as being 1 AU. The relative distance Sun-Venus is then 0.7 AU.
A) When The Sun, Venus and Earth are lined up, find the distance between the EArth and Venus in AUs.
distance EArth-VEnus = ______ AUs. (draw, it will help you)
B) To find the absolute distances from Sun to planets, radio waves were sent to VEnus. The radio wave bounced back from Venus
to Earth. It took 300s for the waves to cover the round trip. Find the distance Earth-Venus in km .
(radio waves are electromagnetic waves that travel at the speed of light in space (vacuum). The speed
of light is 300,000 km/s ). distance VEnus-Earth = __________________ km (scientific notation, keep 2 decimals)
(hint: you have 1s= 300,000km find 300s = ? proportion again, don't forget to divide the distance by 2, it is a round trip )
C) Use the distance Venus- EArth you found to find 1AU (distance EArth-Sun) in km. ( VEnus-Earth = 0.3AU )
(you can set a proportion again. 0.3 AU = ... km so 1 AU = ? km )
3) So Kepler derived a simple law from the data collected by the gifted astronomer Tycho Brahe:
(orbital period of a planet around the Sun)2 / (distance of the planet from Sun)3 = K K is a constant
or for short: P2 / R3 = K (constant)
A) In the formula P is the ___________________________ and R is the ______________.
B) The value of K was not known until Newton introduces the gravitational law ( 2 masses attract each other, the attraction depends on the
distance between the masses), invents calculus and states his 3 laws of motion.
Newton found that, if the mass of the Sun is really big compared to Earth's then :
K = 4 (Pi)2 / G M G is the gravitational constant G = 6.67 10-11 M is the mass of the Sun in Kg
You are going to find the mass of the Sun using Kepler formula and Newton's expression for K :
P2 / R3 = 4 (Pi)2 / G M
C) advanced - optional : In the previous algebraic expression, solve for M = ___________ (use P2, R3, pi and G)
D) If M = (4 R3 (Pi)2 ) / ( P2 G ) with G = 6.67 10-11 P is the orbital period (1 year) and R is the distance Earth- Sun in meters: 150,000,000,000 m
Find the mass of the Sun in kg. You need to convert P in seconds first. (1 year = 365x24x3600 seconds).
M = _____________kg (scientific notation)
Check with the Internet.
E) you can use the same formula to find the mass of the EArth using the Moon.
R is the distance Moon-Earth R = 384, 000,000 m . Find The time P in takes the Moon to orbit the EArth in seconds
using Internet. P = _______ days = ______________ seconds.
Find M using the formula and the TI.
3) Galileo discovered the moons of Jupiter. He could measure their orbital sizes (orbital distance) only by using by diameter of Jupiter as a unit of measure.
He found tat Io, which had a period of 1.8 days, was 4.2 units from the center of Jupiter. Callisto, Jupiter's fourth moon, had a period of 16.7 days.
Using the same unit that Galileo used, predict CAllisto's distance from Jupiter. use the steps:
A) sketch the orbits of Io and Callisto.
B) Find the orbital size of Callisto using Kepler's 1st law. Tc2 / Ti2 = rc3 /ri3
Tc = 16.7, Ti=1.8, rc = ? , ri = 4.2 units
4) P2 / R3 = 4 (Pi)2 / G M
What happen to the ratio P2 / R3 if the mass of the sun is doubled. (that is if M becomes 2 M)
5) A conic has an equation given by:
Ax2 + Bxy + Cy2 + Dx + Ey + F = 0
if B2 - 4AC <0, B=0 and A =C it is a circle
if B2 - 4AC <0, and either B not 0 or A not C it is an ellipse
if B2 - 4AC = 0 it is a parabola
if B2 - 4AC > 0 it is a hyperbola
Tell what type of path each comet follows. Which comet (s) will pass the sun more than once.
A) 3550x2 + 14,200x + 7100y - 13, 050 = 0
B) 2200x2 + 4600y2 - 13200x - 18,400 y + 12,900 =0
C) 5000x2 -6500y2 + 20,000x -52,000y - 695,000 = 0
6) Kepler found the relation: T2 / R3 = constant or T2 / R3 = K
T is the period in earth years and R is the average distance from the Sun in AUs. (1 AU is the distance EArth-Sun)
consider the data:
A) consider the equation T2 / R3 = K now T is x and R is y . so the equation comes : x2 / y3 = K
Isolate y3 , y3 = __________. Solve for y = ___________ (raise to power 1/3 on both side, call K1/3 K' just another constant ).
find the expression for lny , ln y = ______ + ______
If the data obey this model the graph of lny is a line with a slope = ______. lets see if this is true.
Fill the table:
Using your TI plot these data and find the slope. Do you get the right number ? slope = _________.
10)
The orbits of the planets are closed to a circle except for plutos. its orbit crosses Neptune's orbit.
Some comets orbit the Sun along very elongated ellipses. (eccentricity close to 1). Like Halley comet.
(it crosses our orbit every 75 years about)
11) ANy object in the heaven will move along a conic: parabola, circle, ellipse or a piece of hyperbola.
These curves are called conics because you get them by cutting through 2 double-napped cones (on top of each other) as follow: (see picture)
(depending on the plane that cuts the cones).
The curve the object will move along, depends on the bonding energy to a planet/star/black hole (potential energy)
of the object versus its kinetic energy.
circle or ellipse:
If the object has more potential energy than kinetic energy, it is bonded to another massive object (planet/star ..)
For example : artificial satellites and the Moon are bonded to EArth, Earth is bonded to the Sun, the Sun is bonded
to the center of the galaxy, binary stars are bonded to each other .. In that case the total energy is negative :
Total energy = kinetic energy + potential energy < 0 with potential energy <0 and |potential energy| > | kinetic energy|
The energy can switch back and forth from kinetic to potential, but the total energy stays negative.
The trajectory is an ellipse with the object orbited (like Sun) at one focus, as predicted but Kepler/ A circle is an ellipse.
(special case).
parabola or hyperbola
In that case, the object is not bonded to a planet or a star. It is attracted by the object (star/planet) and its
trajectory is curved but its kinetic energy is large enough to counterbalance the bonding energy.
So a comet can pass by our Sun and never come back, or a space ship can pass by Jupiter but can overcome
its attractive force. For that to happen, the kinetic energy of the space ship needs to be equal (parabola) or
greater (hyperbola) to the bonding energy (potential energy). The total is either 0 (parabola) or
greater than 0 (hyperbola).
Total energy = kinetic energy + potential energy with potential energy >= 0 and |potential energy| < | kinetic energy|
--------------------------------------
The total energy E = K + U (U is the potential energy and is negative / K is the kinetic energy and is positive)
If the kinetic energy can't fill the potential well, the planet can't escape.
Along the orbit, the energy switch back and for from potential energy U and kinetic energy K.
But the total energy K+U stays constanct. When U goes down, K goes up.
So when the planets gets close to the Sun, U reaches its minimum and K its maximum. so the speed is large.
When the planets is far form the Sun, U goes up and K goes down. the speed decreases.
The orbit can be a circle if e= 0 (total energy is the lowest/minimum for a given momentum and negative).
The orbit can be an ellipse if 0 <e < 1 (total energy is negative )
the orbit can be a parabola e =1 (kinetic energy = potential energy is total energy = 0 )
the orbit can be a hyperbola e >1 (kinetic energy > potential energy)
Function of energy versus the eccentricity e. It is a parabola shifted 1 unit down.
The minimum is when e = 0, E < 0.
of e =1, E = 0.
source: http://www.sparknotes.com/math/precalc/conicsections/section1.rhtml
ohannes Kepler was German and fond of Astronomy. He was an amazing mathematician too.
During the 30 years war in Europe (catholic were persecuting protestants and protestants were fighting each others) , he had to flee Germany
and moved to Prague. In Prague, he met Tycho Brahe who had collected, with the naked eye, and amazingly very accurately, positions of planets around the Sun.
Tycho Brahe hired Kepler to work on the data in search of patterns. The 2 men didn't get along and Brahe would only give Kepler the precious data a few at a time. Kepler got lucky when Brahe suddenly died (broken bladder) and stole the data from the family.
With the data (positions of Mars relative to the Sun) he was able to infer the famous 3 Kepler laws:
A) Planets move along ellipses around the Sun (most of them are almost circles). (the sun is at one focus)
B) When moving along their orbit, planets sweep the same area during the same time.
(meaning they race when they are close to the Sun and slows down when they are away from it)
C) For all the planets in our solar system, the ratio between (the orbital period)2 and (the distance from the Sun)3 is the same.
or (the orbital period of planet)2 / (the distance from the Sun of planet )3 = K K is a constant.
That was an major discovery. Only based on complex computations and on very accurate data. Next, Newton invents calculus
and is able to derive these laws and explain , using calculus and his universal gravitational law, why .
The third law is about ratio. If we use Earth as a reference we can write:
(orbital period planet)2 / (distance planet-Sun)3 = (orbital period earth)2 / (distance earth-Sun)3 = K (constant)
Let's take (distance Earth-Sun ) is 1 AU (1 astronomical unit is about 150 million km but don't use 150 million km in your computations)
(distance of planets will be in AUs)
Let's take the orbital period of the Earth is 1 year (period of planets will be in earth years)
We know that it takes MArs 1.88 earth years to go around the sun (orbital period planet)
Use the proportion to find the (distance mars-sun)3 = ____________ . (hint: solve the proportion above).
So solve for (distance mars-sun) , with your TI, raise the previous number to (1/3) (that will undo the third power)
(distance mars-sun) =_______________ AUs.
Do the same thing for Jupiter. Find in the Internet, the orbital period of Jupiter in Earth years = _______ years.
(distance Jupiter-Sun) = ___________ AUS.
2) Thanks to Kepler's laws we can find out the relative distances of planets from the Sun.
The distance Sun- Earth is taken as being 1 AU. The relative distance Sun-Venus is then 0.7 AU.
A) When The Sun, Venus and Earth are lined up, find the distance between the EArth and Venus in AUs.
distance EArth-VEnus = ______ AUs. (draw, it will help you)
B) To find the absolute distances from Sun to planets, radio waves were sent to VEnus. The radio wave bounced back from Venus
to Earth. It took 300s for the waves to cover the round trip. Find the distance Earth-Venus in km .
(radio waves are electromagnetic waves that travel at the speed of light in space (vacuum). The speed
of light is 300,000 km/s ). distance VEnus-Earth = __________________ km (scientific notation, keep 2 decimals)
(hint: you have 1s= 300,000km find 300s = ? proportion again, don't forget to divide the distance by 2, it is a round trip )
C) Use the distance Venus- EArth you found to find 1AU (distance EArth-Sun) in km. ( VEnus-Earth = 0.3AU )
(you can set a proportion again. 0.3 AU = ... km so 1 AU = ? km )
3) So Kepler derived a simple law from the data collected by the gifted astronomer Tycho Brahe:
(orbital period of a planet around the Sun)2 / (distance of the planet from Sun)3 = K K is a constant
or for short: P2 / R3 = K (constant)
A) In the formula P is the ___________________________ and R is the ______________.
B) The value of K was not known until Newton introduces the gravitational law ( 2 masses attract each other, the attraction depends on the
distance between the masses), invents calculus and states his 3 laws of motion.
Newton found that, if the mass of the Sun is really big compared to Earth's then :
K = 4 (Pi)2 / G M G is the gravitational constant G = 6.67 10-11 M is the mass of the Sun in Kg
You are going to find the mass of the Sun using Kepler formula and Newton's expression for K :
P2 / R3 = 4 (Pi)2 / G M
C) advanced - optional : In the previous algebraic expression, solve for M = ___________ (use P2, R3, pi and G)
D) If M = (4 R3 (Pi)2 ) / ( P2 G ) with G = 6.67 10-11 P is the orbital period (1 year) and R is the distance Earth- Sun in meters: 150,000,000,000 m
Find the mass of the Sun in kg. You need to convert P in seconds first. (1 year = 365x24x3600 seconds).
M = _____________kg (scientific notation)
Check with the Internet.
E) you can use the same formula to find the mass of the EArth using the Moon.
R is the distance Moon-Earth R = 384, 000,000 m . Find The time P in takes the Moon to orbit the EArth in seconds
using Internet. P = _______ days = ______________ seconds.
Find M using the formula and the TI.
3) Galileo discovered the moons of Jupiter. He could measure their orbital sizes (orbital distance) only by using by diameter of Jupiter as a unit of measure.
He found tat Io, which had a period of 1.8 days, was 4.2 units from the center of Jupiter. Callisto, Jupiter's fourth moon, had a period of 16.7 days.
Using the same unit that Galileo used, predict CAllisto's distance from Jupiter. use the steps:
A) sketch the orbits of Io and Callisto.
B) Find the orbital size of Callisto using Kepler's 1st law. Tc2 / Ti2 = rc3 /ri3
Tc = 16.7, Ti=1.8, rc = ? , ri = 4.2 units
4) P2 / R3 = 4 (Pi)2 / G M
What happen to the ratio P2 / R3 if the mass of the sun is doubled. (that is if M becomes 2 M)
5) A conic has an equation given by:
Ax2 + Bxy + Cy2 + Dx + Ey + F = 0
if B2 - 4AC <0, B=0 and A =C it is a circle
if B2 - 4AC <0, and either B not 0 or A not C it is an ellipse
if B2 - 4AC = 0 it is a parabola
if B2 - 4AC > 0 it is a hyperbola
Tell what type of path each comet follows. Which comet (s) will pass the sun more than once.
A) 3550x2 + 14,200x + 7100y - 13, 050 = 0
B) 2200x2 + 4600y2 - 13200x - 18,400 y + 12,900 =0
C) 5000x2 -6500y2 + 20,000x -52,000y - 695,000 = 0
6) Kepler found the relation: T2 / R3 = constant or T2 / R3 = K
T is the period in earth years and R is the average distance from the Sun in AUs. (1 AU is the distance EArth-Sun)
consider the data:
| planet | Mercury | VEnus | EArth | Mars | Jupiter | Saturn |
| x (period in earth's year) | 0.387 | 0.723 | 1.000 | 1.524 | 5.203 | 9.539 |
| y (distance from the Sun) | 0.241 | 0.615 | 1.000 | 1.881 | 11.862 | 29.458 |
A) consider the equation T2 / R3 = K now T is x and R is y . so the equation comes : x2 / y3 = K
Isolate y3 , y3 = __________. Solve for y = ___________ (raise to power 1/3 on both side, call K1/3 K' just another constant ).
find the expression for lny , ln y = ______ + ______
If the data obey this model the graph of lny is a line with a slope = ______. lets see if this is true.
Fill the table:
| planet | Mercury | VEnus | EArth | Mars | Jupiter | Saturn |
| ln(x) | ||||||
| ln(y) |
Using your TI plot these data and find the slope. Do you get the right number ? slope = _________.
10)
The orbits of the planets are closed to a circle except for plutos. its orbit crosses Neptune's orbit.
Some comets orbit the Sun along very elongated ellipses. (eccentricity close to 1). Like Halley comet.
(it crosses our orbit every 75 years about)
11) ANy object in the heaven will move along a conic: parabola, circle, ellipse or a piece of hyperbola.
These curves are called conics because you get them by cutting through 2 double-napped cones (on top of each other) as follow: (see picture)
(depending on the plane that cuts the cones).
The curve the object will move along, depends on the bonding energy to a planet/star/black hole (potential energy)
of the object versus its kinetic energy.
circle or ellipse:
If the object has more potential energy than kinetic energy, it is bonded to another massive object (planet/star ..)
For example : artificial satellites and the Moon are bonded to EArth, Earth is bonded to the Sun, the Sun is bonded
to the center of the galaxy, binary stars are bonded to each other .. In that case the total energy is negative :
Total energy = kinetic energy + potential energy < 0 with potential energy <0 and |potential energy| > | kinetic energy|
The energy can switch back and forth from kinetic to potential, but the total energy stays negative.
The trajectory is an ellipse with the object orbited (like Sun) at one focus, as predicted but Kepler/ A circle is an ellipse.
(special case).
parabola or hyperbola
In that case, the object is not bonded to a planet or a star. It is attracted by the object (star/planet) and its
trajectory is curved but its kinetic energy is large enough to counterbalance the bonding energy.
So a comet can pass by our Sun and never come back, or a space ship can pass by Jupiter but can overcome
its attractive force. For that to happen, the kinetic energy of the space ship needs to be equal (parabola) or
greater (hyperbola) to the bonding energy (potential energy). The total is either 0 (parabola) or
greater than 0 (hyperbola).
Total energy = kinetic energy + potential energy with potential energy >= 0 and |potential energy| < | kinetic energy|
--------------------------------------
The total energy E = K + U (U is the potential energy and is negative / K is the kinetic energy and is positive)
If the kinetic energy can't fill the potential well, the planet can't escape.
Along the orbit, the energy switch back and for from potential energy U and kinetic energy K.
But the total energy K+U stays constanct. When U goes down, K goes up.
So when the planets gets close to the Sun, U reaches its minimum and K its maximum. so the speed is large.
When the planets is far form the Sun, U goes up and K goes down. the speed decreases.
The orbit can be a circle if e= 0 (total energy is the lowest/minimum for a given momentum and negative).
The orbit can be an ellipse if 0 <e < 1 (total energy is negative )
the orbit can be a parabola e =1 (kinetic energy = potential energy is total energy = 0 )
the orbit can be a hyperbola e >1 (kinetic energy > potential energy)
Function of energy versus the eccentricity e. It is a parabola shifted 1 unit down.
The minimum is when e = 0, E < 0.
of e =1, E = 0.
source: http://www.sparknotes.com/math/precalc/conicsections/section1.rhtml