Hard Calc question

ccbc101 · #1 $Old$ December 1st, 2008, 11:37 AM

NO one seems to know how to do this.

According to Kepler's laws, the planets in our solar system move in
elliptical orbits around the sun. If a planet's closest approach to
the sun occurs at time t=0, then the distance r from the center of the
planet to the center of the sun at some later time t can be determined
from the equation

r=a(1-ecos(x))

where a is the average distance between centers, e is the positive
constant that measures the "flatness" of the elliptical orbit, and "x"
is the solution of Kepler's equation:

(2*pi*t)/T=x-esin(x)

in which T is the time it takes for one complete orbit of the planet.

a) Estimate the distance from the planet mars to the sun when t=1 year. a= 228 m km..e=.0934..t=1.88 years (Newton's Law)

b) calculate dr/dt. Show that r reaches a maximum value at t=T/2

vincisonfire · #2 $Old$ December 1st, 2008, 03:24 PM

I would say that you must use the approximation $sin(x) \simeq x$ .
It follows that $\frac{2\pi t}{T}=x(1-e) \implies x=\frac{2\pi t}{T(1-e)}$
Then $r=a(1-ecos(x)) =a(1-ecos(\frac{2\pi t}{T(1-e)}))$
and
$\frac{dr}{dt}= \frac{2e\pi t}{T(1-e)}) sin(\frac{2\pi t}{T(1-e)}))$
and we would have a minimum at $t= \frac{T(1-e)}{2}$