In this video, I go over Johannes Kepler's three laws of planetary motion and derive them using Isaac Newton's second law of motion, F = m a, and his law on universal gravitation. While observing the Danish astronomer Tycho Brahe's astronomical observations, Kepler observed recurring patterns in the data, which he formulated as three laws:
Isaac Newton later realized that Kepler's laws are merely the biproduct of his 2nd law of motion and his law on universal gravitation. I use Newton’s laws to derive Kepler's laws, while making extensive use of vector functions, calculus, and the polar equation of an ellipse.
Later in the video, I calculate the distance a satellite needs to be from the equator of the Earth to have a geosynchronous orbit, in which the orbit of the satellite around the Earth is in sync with Earth's rotation. This type of orbit is sometimes referred to as the Arthur C. Clarke orbit, named after the famous author of 2001: A Space Odyssey, who popularized the orbit in a fictional novel in 1945. I also go over some very interesting connections surrounding Arthur C. Clarke, so buckle up!
#math #vectors #calculus #space #physics
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We now describe one of the great accomplishments of calculus by showing how the material of this chapter can be used to prove Kepler’s laws of planetary motion.
After 20 years of studying the astronomical observations of the Danish astronomer Tycho Brahe, the German mathematician and astronomer Johannes Kepler (1571–1630) formulated the following three laws.
Kepler’s Laws
In his book Principia Mathematica of 1687, Sir Isaac Newton was able to show that
these three laws are consequences of two of his own laws, the Second Law of Motion and the Law of Universal Gravitation.
In what follows we prove Kepler’s First Law.
A planet revolves around the sun in an elliptical orbit with the sun at one focus.
Recall from my earlier video where I showed an ellipse is defined as the curve where the sum of the distances from any point on the curve to two fixed points (called foci) is constant.
Since the gravitational force of the sun on a planet is so much larger than the forces
exerted by other celestial bodies, we can safely ignore all bodies in the universe except the sun and one planet revolving about it.
We use a coordinate system with the sun at the origin and we let r = r(t) be the position vector of the planet.
Equally well, r could be the position vector of the moon or a satellite moving around the earth or a comet moving around a star.
The velocity vector is v = r' and the acceleration vector is a = r''.
We use the following laws of Newton:
Second Law of Motion:
Law of Gravitation:
Where F is the gravitational force on the planet, m and M are the masses of the planet and the sun, and G is the gravitational constant.
We first show that the planet moves in one plane.
By equating the expressions for F In Newton’s two laws, we find that:
Since a is a scalar multiple of r, then they are parallel to each other.
Thus by my earlier video on the cross product, r x a = 0.
We now use Formula 5 from my earlier video on the differentiation rules for vector functions:
We assume that h ≠ 0; that is, r and v are not parallel.
This means that the vector r = r(t) is perpendicular to h for all values of t, so the planet always lies in the plane through the origin perpendicular to h.
Thus the orbit of the planet is a plane curve.
To prove Kepler’s First Law we rewrite the vector h as follows:
Recall my earlier video on the properties of the cross product.
Recall the vector triple product from my earlier video.
Recall the geometric equation of the dot product from my earlier video, which thus shows that the dot product of perpendicular vectors is equal to zero.
Recall from my earlier video that if a vector has constant magnitude then its derivative is perpendicular, thus their dot product is zero.
Integrating both sides of this equation, we get:
At this point it is convenient to choose the coordinate axes so that the standard basis vector k points in the direction h.
Then the planet moves in the xy-plane.
Since both v x h and u are perpendicular to h, the equation above shows that c lies in the xy-plane.
This means that we can choose the x- and y-axes so that the standard basis vector i lies in the direction of c, as shown in the figure below.
If θ is the angle between c and r, then (r, θ) are polar coordinates of the planet.
Now if we apply the dot product our equation for v x h with r we can obtain a formula involving both r and θ.
Recall my earlier video on the properties of the dot product.
Notice the Scalar Triple Product in the numerator above, which I covered in my earlier video.
Comparing with my earlier video on conic sections, we see that the above equation is the polar equation of a conic section with focus at the origin and eccentricity e.
We know that the orbit of a planet is a closed curve and so the conic must be an ellipse.
This completes the derivation of Kepler’s First Law.
We will guide you through the derivation of the Second and Third Laws in the Applied Project below.
The proofs of these three laws show that the methods of this chapter provide a powerful tool for describing some of the laws of nature.
Johannes Kepler stated the following three laws of planetary motion on the basis of massive amounts of data on the positions of the planets at various times.
Kepler’s Laws
Kepler formulated these laws because they fitted the astronomical data.
He wasn’t able to see why they were true or how they related to each other.
But Sir Isaac Newton, in his Principia Mathematica of 1687, showed how to deduce Kepler’s three laws from two of Newton’s own laws, the Second Law of Motion and the Law of Universal Gravitation.
Earlier in this video, we proved Kepler’s First Law using the calculus of vector functions.
In this project we guide you through the proofs of Kepler’s Second and Third Laws and explore some of their consequences.
Use the following steps to prove Kepler’s Second Law.
The line joining the sun to a planet sweeps out equal areas in equal times.
The notation is the same as in the proof of the First Law.
In particular, use polar coordinates so that r = (r cos θ) i + (r sin θ) j.
(a) Show that:
(b) Deduce that:
(c) If A = A(t) is the area swept out by the radius vector r = r(t) in the time interval [t0, t] as in the figure below, show that:
(d) Deduce that:
This says that the rate at which A is swept out is constant and proves Kepler's Second Law.
Let T be the period of a planet about the sun; that is, T is the time required for it to travel once around its elliptical orbit.
Suppose that the lengths of the major and minor axes of the ellipse are 2a and 2b.
(a) Use part (d) of Problem 1 to show that T = 2πab/h.
(b) Show that:
(c) Use parts (a) and (b) to show that:
Notice that the proportionality constant 4π2/(GM) is independent of the planet.
This proves Kepler's Third Law.
The square of the period of revolution of a planet is proportional to the cube of the length of the major axis of its orbit.
The period of the earth’s orbit is approximately 365.25 days.
Use this fact and Kepler’s Third Law to find the length of the major axis of the earth’s orbit.
You will need the mass of the sun, M = 1.99 x 1030 kg, and the gravitational constant, G = 6.67 x 10-11 N⋅m2/kg2.
It’s possible to place a satellite into orbit about the earth so that it remains fixed above a given location on the equator.
Compute the altitude that is needed for such a satellite.
The earth’s mass is 5.98 x 1024 kg; its radius is 6.37 x 106 m.
This orbit is called the Clarke Geosynchronous Orbit after Arthur C. Clarke, who first proposed the idea in 1945.
The first such satellite, Syncom II, was launched in July 1963.
Use the following steps to prove Kepler’s Second Law.
The line joining the sun to a planet sweeps out equal areas in equal times.
The notation is the same as in the proof of the First Law.
In particular, use polar coordinates so that r = (r cos θ) i + (r sin θ) j.
(a) Show that:
(b) Deduce that:
(c) If A = A(t) is the area swept out by the radius vector r = r(t) in the time interval [t0, t] as in the figure below, show that:
(d) Deduce that:
This says that the rate at which A is swept out is constant and proves Kepler's Second Law.
Solution:
Solution to (a):
Recall from the proof of Law 1 that r x v = h, therefore we have:
Recall the Pythagorean Identity from my earlier video.
Solution to (b):
Since h is a constant vector, we can write it as:
Solution to (c):
Recall from my earlier video(s) on proving the area of a sector of a circle.
Let's draw out a thin sector of the circle to obtain an integral.
Thus by the Fundamental Theorem of Calculus, we have:
Recall my earlier videos on Part 1 and Part 2 of the Fundamental Theorem of Calculus.
Solution to (d):
Our result from Part (c) can be written in terms of the magnitude of the constant vector h which we solved in Part (b).
This proves Kepler's second law.
The line joining the sun to a planet sweeps out equal areas in equal times.
Let T be the period of a planet about the sun; that is, T is the time required for it to travel once around its elliptical orbit.
Suppose that the lengths of the major and minor axes of the ellipse are 2a and 2b.
(a) Use part (d) of Problem 1 to show that T = 2πab/h.
(b) Show that:
(c) Use parts (a) and (b) to show that:
Notice that the proportionality constant 4π2/(GM) is independent of the planet.
This proves Kepler's Third Law.
The square of the period of revolution of a planet is proportional to the cube of the length of the major axis of its orbit.
Solution to (a):
Note that A(t = T) is the area of an ellipse, whose formula is given from earlier video.
Solution to (b):
Recall from the earlier proof of Kepler's first law for the value for e⋅d, which are the eccentricity and directrix of an ellipse in polar coordinates:
The ellipse we are dealing with is just horizontally shifted from the origin, as I illustrated in my earlier video, as well as obtaining the values of a and b in terms of e and d.
Solution to Part (c):
This proves Kepler's third law (notice that the proportionality constant is independent of the planet):
The square of the period of revolution of a planet is proporotional to the cube of the length of the major axis of its orbit.
The period of the earth’s orbit is approximately 365.25 days.
Use this fact and Kepler’s Third Law to find the length of the major axis of the earth’s orbit.
You will need the mass of the sun, M = 1.99 x 1030 kg, and the gravitational constant, G = 6.67 x 10-11 N⋅m2/kg2.
Solution:
From Problem 2, and converting the period T of the Earth to seconds since the unit Newton (N) in the gravitational constant has units of kg⋅m/s2, we have.
Thus, the length of the major axis of the Earth's orbit (that is, 2a) is approximately 2.99 x 1011 m = 2.99 x 108 km.
Calculation Check:
365.25 * 24 * 60 * 60 = 31,557,600 or 3.1558 x 107
1/(4 * pi^2) * (6.67 * 10^-11) * (1.99 * 10^30) * (3.1558 * 107)2 = 3.34840604480615E33 or 3.348 x 1033 m3.
(3.34840604480615E33)^(1/3) = 1.49604976099263E11 or 1.496 x 1011.
1.49604976099263E11*2 = 2.99209952198526E11 or 2.99 x 1011
2.99209952198526E11/1000 = 2.99209952198526E8 or 2.99 x 108
It’s possible to place a satellite into orbit about the earth so that it remains fixed above a given location on the equator.
Compute the altitude that is needed for such a satellite.
The earth’s mass is 5.98 x 1024 kg; its radius is 6.37 x 106 m.
This orbit is called the Clarke Geosynchronous Orbit after Arthur C. Clarke, who first proposed the idea in 1945.
The first such satellite, Syncom II, was launched in July 1963.
Solution:
We can adapt the equation from Problem 2(c) with the Earth at the center of the system, so T is the period of the satellite's orbit about the Earth, M is the mass of the Earth, and a is the length of the semimajor axis of the satellite's orbit (measured from the Earth's center).
Since we want the satellite to remain fixed above a particular point on the Earth's equator, T must coincide with the period of the Earth's own rotation, so T = 24 h = 86,400 s.
The mass of the Earth is M = 5.98 x 1024 kg, so:
Calculation Check:
24 * 60 * 60 = 86,400
((6.67 * 10^-11) * (5.98 * 10^24) * (86,400^2)/(4 * pi2))(1/3) = 4.22504743050478E7 or 4.23 x 107
If we assume a circular orbit, the radius of the orbit is a, and since the radius of the Earth is 6.37 x 106 m, the required altitude above the Earth's surface for the satellite is:
Calculation Check:
4.23 * 10^7 - 6.37 * 10^6 = 3.593E7 or 3.59 x 107
3.59E7/1000 = 35,900
Here is animation (not to scale) of a geosynchronous satellite orbiting the Earth.
According to Wikipedia, Herman Potočnik described both geosynchronous orbits in general in 1929, the first popular literature to mention a geosynchronous orbit was in 1942 by George O. Smith, but Arthur C. Clarke popularised and expanded on the concept in 1945, hence it is sometimes called the Clarke Orbit.
Arthur C. Clarke (16 December 1917 – 19 March 2008) is the famous author of the 1968 novel (turned movie): 2001: A Space Odyssey. It was written concurrently (at the same time) as Stanley Kubrick's film version and published one month (June 1968) after the movie was released (April 2, 1968).
The 2001: A Space Odyssey movie features a black monolith, which looks just like the Millennium Hilton hotel building that stood facing the Twin Towers, and the hotel phone number is (212)-693-2001!
The film was released on April 2, 1968, which is the same year as when the World Trade Center (WTC) construction began. The difference between 1968 and 2001 is... 33.
And since year 1 is the start of the calendar, it means that 2001 is the start of the new millennium. 😲
Arthur C. Clarke also wrote a 1982 sequel titled 2010: Odyssey Two which involved aliens telling humans not to land on Jupiter's moon Europa.
Fast forward to October 14, 2024, and NASA launched the Europa Clipper space probe to study Europa, which they view as the best chance to find alien life (single celled marine organisms); it is scheduled to arrive on Europa in 2030.
Even more interestingly, Arthur C. Clarke had recorded interviews in 1966 from witnesses that saw a "globular ray" (ball lightning).
Here is an incredible video of ball lightning captured in Alberta, Canada last year in 2025!
Furthermore, Arthur C. Clarke also wrote the forward on January 4, 2000 for Charles G. Beaudette's 2002 book on Cold Fusion!
And a fitting way to end this video on Kepler's three laws is by going over Arthur C. Clarke's three laws, the last of which is very profound…
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