Sec.13.4 - Kepler’s Laws

謝承修, Te-Sheng Lin (林得勝)
Last updated on Jan 6, 2021 2 min read Calculus

Applied Project in Sec.13.4, Calculus by Stewart
Chinese version 克卜勒定律

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

  1. A planet revolves around the sun in an elliptical orbit with the sun at one focus.
  2. The line joining the sun to a planet sweeps out equal areas in equal times.
  3. 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.

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. In Section 13.4 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.

Question 1： Use the following steps to prove Kepler’s Second Law. The notation is the same as in the proof of the First Law in Section 13.4. In particular, use polar coordinates so that $\vec{r} = r\cos{\theta} \vec{i} + r\sin{\theta} \vec{j}$.

(a) Show that $\vec{h}=r^2 \frac{d\theta}{dt} \vec{k}$.

(b) Deduce that $r^2 \frac{d\theta}{dt}=h$.

(c) If $A=A(t)$ is the area swept out by the radius vector $r=r(t)$ in the time interval $[t_0, t]$. Show that $$ \frac{dA}{dt}=\frac{1}{2}r^2\frac{d\theta}{dt} $$

(d) Deduce that $$ \frac{dA}{dt}=\frac{1}{2}h= c, $$ where $c$ is a constant. This says that the rate at which $A$ is swept out is constant and proves Kepler’s Second Law.

Question 2： 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 $T =2\pi ab h$.

(b) Show that $$ \frac{h^2}{GM}=ed=\frac{b^2}{a} $$

(c) Use parts (a) and (b) to show that $$ T^2=\frac{4\pi^2}{GM}a^3 $$

This proves Kepler’s Third Law. [Notice that the proportionality constant $\frac{4\pi^2}{GM}$ is independent of the planet.]

Quesion 3： 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, $M=1.99\times 10^{30}kg$, and the gravitational constant, $G=6.67\times10^{-11}N-m^2/kg^2$.

Solution ：

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

$$ a=(\frac{GMT^2}{4\pi^2})^{\frac{1}{3}}\approx149600000km $$

Question 4： 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 $5.98\times 10^{24}kg$; its radius is $6.37\times 10^{6}m$.

Solution： $$ h=(\frac{GM_eT^2}{4\pi^2})^{\frac{1}{3}}-R\approx42.25\times10^6-6.37\times10^6=35880km $$

applied project

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