## Lecture 22: Kepler's Laws - Elliptical Orbits - Satellites - Change of Orbits - Ham Sandwich

recorded by: Massachusetts Institute of Technology, MIT

published: Oct. 10, 2008, recorded: November 1999, views: 38223

released under terms of: Creative Commons Attribution Non-Commercial Share Alike (CC-BY-NC-SA)

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**1. Kepler's Laws and Elliptical Orbits:**

A review of equations for the period, velocity and mechanical energy of a circular orbit is given. Kepler's 3 Laws, and the planetary data that led him to his third law are introduced. The equations for elliptical orbits are discussed and compared with the equations for circular orbits.

**2. Elliptical Orbit from Initial Conditions:**

The elliptical orbit follows from the initial conditions. Using the conservation of mechanical energy, you can find the semimajor axis, and this, in combination with Kepler's 3rd law, enables you to calculate the orbital period. A numerical example for a high eccentricity orbit of an Earth orbiting satellite, is worked out. Using the conservation of angular momentum, one can determine the distances to apogee and perigee, and the satellite's velocity at apogee and perigee.

**3. Changing from Circular to Elliptical Orbits:**

By firing a rocket on board a spacecraft that is in a circular orbit, the spacecraft's velocity (vector) will change, and this leads to an elliptical orbit and a change in orbital period. The new elliptical orbits are sketched along with the original circular orbit, setting the stage for how astronauts in different spacecrafts can pass a sandwich.

**4. Astronauts Pass a Ham Sandwich:**

Peter and Mary are astronauts in different spacecrafts but in the same circular orbit. Peter wants to throw a ham sandwich to Mary. The question is: how to do that? There is a large family of solutions which are discussed.

**5. Simulations of the Passing of the Sandwich:**

A computer model for finding solutions to how astronauts Peter and Mary can pass a sandwich is introduced and exercised by its author, Dave Pooley. GREAT FUN!

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## Reviews and comments:

NECAT, April 29, 2010 at 10:15 p.m.:Hi,

Kepler himself has changed his reasoning on elliptical orbit by giving his third law about periods:(P1/P2)^2=(r1/r2)^3.He wanted to say that the elliptical orbits will be transformed to circular orbits with time.How a solid ellipse may be transformed to a circle? By being not an ellipse.Then what is the real shape of the orbits? They are spiraled.Newton's F*dt=m*dv,when solved, gives the equation of the motion: r=-a*t*(t-Tmax)+K.The polar graph of this equation shows billions of cardioidal looking spirals: like in the photos of the galaxies.So,orbits are not elliptical,the Sun is not at one of the focus of these ellipses,no aphelion,no perihelion,nor equality of swept out areas in equal interval of time.All these are old estimation of Kepler,proved by the mathematicians(!).Mathematicians should consider "spiraled orbits theory", and they will be surprised by the consequences they may discover: ligth year having no sense;years length different in the very beginning of the Earth (15 days a year for example) and ..etc .necattasdelen@ttmail.com

Eric Dong, July 2, 2010 at 6:03 a.m.:Don't say stupid things like that. Kepler's 3rd law is: (P1/P2)^2=(a1/a2)^3. There is no such thing as R in the equation; it is a, the semimajor axis. And your "derivation" of the spiral orbit is sheer nonsense. Please show me the t, Tmax, and r in Fdt=mdv! And you are just throwing technobabble after that. Kepler's laws were not proved by the mathematicians, they were proved by Kepler himself by observation. Newton derived Kepler's 3rd law HIMSELF. And your reference to light year making no sense makes no sense. Light year is as concrete as meter: it's the distance light goes in a year. Please do not play sham astrophysicist here. If you want my derivation (using PHYSICS!!!) of Kepler's 3rd law or a demonstration of the stableness of elliptical orbits, please e-mail me: ericcesium133@aol.com.

Eric Dong, July 2, 2010 at 6:04 a.m.:Don't say stupid things like that. Kepler's 3rd law is: (P1/P2)^2=(a1/a2)^3. There is no such thing as R in the equation; it is a, the semimajor axis. And your "derivation" of the spiral orbit is sheer nonsense. Please show me the t, Tmax, and r in Fdt=mdv! And you are just throwing technobabble after that. Kepler's laws were not proved by the mathematicians, they were proved by Kepler himself by observation. Newton derived Kepler's 3rd law HIMSELF. And your reference to light year making no sense makes no sense. Light year is as concrete as meter: it's the distance light goes in a year. Please do not play sham astrophysicist here. If you want my derivation (using PHYSICS!!!) of Kepler's 3rd law or a demonstration of the stableness of elliptical orbits, please e-mail me: ericcesium133@aol.com.

Peter L. Griffiths, May 13, 2011 at 5:37 p.m.:Galileo's law of falling bodies v^2=d is effectively the same as Kepler's distance law v^2=1/r. Let L indicate a small change then we have

v^2+Lv^2=d+Ld=1/(r-Ld). This is the usual way of measuring elocity. For the reciprocal method of measuring velocity we have

v^2+Lv^2=r+Lr=1/(d-Lr). There are only three variables involved distance, time and velocity. There are two methods of measuring the same velocity, distance per unit time and time per unit distance , one method is the reciprocal of the other. d+r equals the major axis of the elliptical orbit.

Peter L. Griffiths, October 31, 2012 at 4:26 p.m.:One of the simplest properties of all elliptical orbits is that in every elliptical orbit there is constant acceleration for half the orbit and constant deceleration for the other half. Velocity is a different matter but can be compared between planets in a different way.

Peter L. Griffiths, December 18, 2012 at 8:08 p.m.:Further to my previous comments, the connection between Galileo's v^2=d at the empty focus end of the elliptical orbit and Kepler's v^2=(1/r) at the Sun focus end is mathematically very interesting and not at all straight forward. Kepler's version can be adapted for further research purposes by including a constant V being the maximum velocity, then the variable velocities can be expressed as as V/#r where # is my notation for square root. In this way the same velocity arises on both the accelerating side as well as the decelerating side, but in opposite directions. As one of the properties of all perfect ellipses d is the distance from the curve to the empty focus, and r is the distance from the curve to the Sun focus, d+r equals the major axis of the elliptical orbit.

Peter L. Griffiths, January 24, 2013 at 3:55 p.m.:The time version of Kepler's distance law is t=r^(1.5). The time version of Galileo's law of falling bodies is t=d^(0.5). The time taken by Halley's comet to cover the decelerating half of the elliptical orbit is 38 years. Over half this decelerating distance, the comet will have taken 38/2#2 years which is 13.435 years decelerating. The time taken by Halley's comet to cover the accelerating half of the elliptical orbit is 38/#2 years which is 26.87 years. As previously # is my notation for square root.

Peter L. Griffiths, May 8, 2013 at 4:25 p.m.:In the second edition 1713 page 540 Cajori edition of his Principia, Isaac Newton incorrectly calculates the major axis of the orbit of Halley's Comet as being 35 AU. Newton does this by taking the ratio of the time of the Comet's orbit to the time of the Earth's orbit round the Sun as being 75:1, and then applying Kepler's distance law to this 75 namely 75^(2/3) which gives about 17 AU which Newton correctly thought was far too low so he doubled it to approximately 35 AU. In fact the correct figure for the major axis of the orbit of Halley's Comet is about 60 AU involving far more detailed calculations.

Peter L. Griffiths, June 8, 2013 at 9:16 p.m.:One important point about Kepler which does not seem to have been mentioned in the comments is the impact of Copernicus who showed that the Earth was not stationary but orbited round the Sun. Kepler recognised that with this information the existing recorded observations needed to be amended to convert these existing recorded observations from the orbiting starting point into observations from a stationary starting point. I think there is a name for this parallax.

NECAT TASDELEN, July 26, 2013 at 1:20 p.m.:Kepler's laws are wrong.Newton never confirm these laws,but agrees that "if the orbits are conical section"they may not be hyperbolic,or parabolic section as the body will get lost in the infinity.So it should be elliptical,he said.The condition "if conical section",may be turned out to "if the orbits are not of conical section nature".Then what should be shape of the orbits? Newton studied this case also and found spiraled orbits,wrongly commented.(PRINCIPIA by Andre Motte,page 206-209).He said the bodies will also go on "ad infinitum",due to the law of period.So he decided on elliptical shape.Without proving.But the bodies do not go on "ad infinitum",but on "ad finitum" as the equation of celestial motion is: r=-4*t^2+4*t*T-4*T^2/6; an expanding then compressing spiral.This equation is due to the conservation of energy.And the period law do not exist for celestial bodies but a time law Vp^2*r=CT is valid for all the bodies of the sun system even for Pluto or Halley .For our Earth VpEarth^2*rEarth=(29,786km/sec)^2*(149597890km) = 132 724 771 939, 26km3/sec2=CT for a given era(2009).

NECAT TASDELEN, July 27, 2013 at 9:33 p.m.:Hi,

I have to insist that celestial orbits are spiraled. Kepler's area law r*Vp=CT is wrong.To prove this I am using

work equation: W=F*L.The components in this equation are vectors.So when differentiated ,we write:

dWradial=dFradial*Lr+Fr*dLradial and similarly

dWperpendicular=dFp*Lp+Fp*dLp

From physics we know :the work done in the direction perpendicular to the attraction field direction equal zero.

So dWp=0.And for this we have to write m*dVp/dt=0.

That means,when integrated,Vp=CT.Then Kepler's area law is wrong.

On other hand,

The energy conservation equation which is:

Vr^2 + 2*a*r + 2Ct1 = 2*r*dVr/dt.

is a differential equation of the classical form

r'^2 + 2*a*r + K = 2*r*r'' where (K = 2Ct1 = Iw2/m).

with solution:

r=-a*t^2+a*t*T-a*T^2/6 where (t=real time,T=lifetime)

This is the equation of celestial motion.It shows no sign of ellipse,but a many layer cardioidal type spirals,placed on a parabolic volume surface along the trajectory of the Sun.No repetition of celestial values,no aphelion,no perihelion,no period.Just a time relation r*Vp^2=CT for all the planets.Even for Halley and Pluton:Just look the data.

VpEarth^2*rEarth = (29,786km/sec)^2*(149597890km) = 132 724 771 939, 26km3/sec2

VpMars^2*rMars = (24,131km/sec)^2 *(227939150km) = 132 724 771 939, 26km3/sec2

The spiraled orbits theory bring many novelties for studying:Definition of ligthyear,of AU,of year days,of I*w^2,of time law constant variation according the era,etc,etc.necattasdelen@ttmail.com

Peter L. Griffiths, August 13, 2013 at 8:33 p.m.:Kepler's area law is nearly but not quite right. It is time taken equals the area covered. If the time taken equals the area of a triangle being perpendicular times base times a half, we can see that this has a similarity with the correct distance law of time equals distance times distance to the power of one half. It is possible that Kepler needed to arrive at the area law before reaching the correct distance law, which is now accepted as applying throughout the whole universe.

Peter L. Griffiths, July 3, 2014 at 6:47 p.m.:I have come across a very important paper written by Kepler not mentioned in the list of his works. It has the title De Coni Sectionibus and is part of Ad Vitellionem paralipomena quibus Astronomiae Pars Optica traditur, published in Frankfurt in 1604. This paper shows Kepler trying to reconcile Apollonius's definition of a conic section with the cylindric section which Kepler needed. It also shows that Kepler had the right idea about constructing the cylindric section with pins and thread, the pins being stuck at the foci. Curiously enough Kepler fails to recognise that the distances from a point on the curve to the two foci add up to the major axis of the ellipse.

Necat Tasdelen, October 15, 2014 at 9:54 p.m.:Kepler laws start with the acceptance of elliptical orbits, without proving.Then a constant areal velocity is proved knowing that the orbit is elliptical.Spiraled orbit laws starts by proving: the cycling velocity Vp of the celestial body around the sun is constant.This is due to Newton's attraction force.Radial force is Fr and it exist Fr=G*M*m/r^2. A side attraction does not exist. This is Fp=m*dVp/dt=0=does not exist.This gives,when integrated, Vp=Ct .Using Vp=Ct in the energy conservation eequation we reach to r=-4*t^2+4*t*T*4*T^2/6 the movement equation in the space,in an attraction field.The attraction field is the sun for the planets or the planets for their moons.

If the cycling velocity was different at Kepler's aphelion and perihelion,we should feel the effects of this difference as acceleration on human bodies twice a year.But we feel nothing.Because Vp=Ct

Peter L. Griffiths, January 11, 2015 at 6:32 p.m.:Kepler's laws are based on recorded observations of velocities and distances from the Sun focus. The existence of this focus means that the shape of the orbits is cylindric sectional. In his little known 1603 paper on Cylindric sections Kepler recognised the pins and thread method of drawing the orbit.

Peter L. Griffiths, August 6, 2015 at 4:22 p.m.:Astronomers have failed to recognise that once you know the symmetrical orbit, you can then calculate the location of the foci and the varying velocity of the orbiting body.

Peter L. Griffiths, August 15, 2015 at 4:57 p.m.:Further to my comment of 6 August 2015, if the symmetrical ellipse is given then take the half of its major axis, this half will serve as the hypotenuse of a right angle triangle with 90 degrees at the centre of the major axis. The point at which this hypotenuse cuts the major axis will be the location of one of the foci. The same procedure applied to the other half of the symmetrical ellipse will locate the other focus.

Peter L. Griffiths, December 14, 2015 at 5:19 p.m.:Further to my comment of 15 August 2015, the focus of a symmetrical elliptical orbit is not the same as the focus of the parabola y= x^2, which is at y =1, x=0. I am not sure whether Kepler recognised this difference.

Necat Tasdelen, October 22, 2016 at 9:51 a.m.:When I write "planet's orbits are spiraled",I started with the area law of Kepler.Is r*Vp=Ct or not?.The radial attraction force of Newton,when the force vectors are considered,means that a side force does not exist.Then Fp=m*dVp/dt=0 means Vp=Ct.So Kepler's area law is wrong.Only Vp=Ct is correct.Using Vp=Ct in the energy conservation expression we reach to the celestial motion equation r=-4*t^2+4*t*T-4*T^2/6,where (r=distance planet sun,t=real time,T=life time of the planet).This equation do not show an ellipse but a parabolic vortex spiral.The spiral start from zero,goes to a max amplitude,then continue to go to zero.So the spiral have a closed shape,which Newton refused saying "if the orbits are spiraled the body should go "ad infinitum"".See PRINCIPIA page 290 translation of A.Motte.

Peter L. Griffiths, December 16, 2016 at 5:01 p.m.:There is a clear mathematical relationship between the focus (f) of an ellipse, the major axis (a) and the minor axis (b), it is f= a - [a^2 - b^2]^0.5. This also applies to a cylindric section

Peter L. Griffiths, September 4, 2017 at 4:26 p.m.:There is a small correction to my comment of 16 December 2016, a should be replaced by half the major axis, and b should be replaced by half the minor axis, so that

the focus f =a/2 -[ (a/2)^2 - (b/2)^2 ]^0.5 . Very few astronomers seem to know about this.

Davor form VideoLectures, December 18, 2017 at 10:30 a.m.:Hi all!

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