Physics, asked by Ketanjiwane4341, 1 year ago

derive kepler's second law of planetary motion from Newton's law of gravitation

Answers

Answered by BiswajitBiswas
5
Of course, Kepler’s Laws originated from
observations of the solar system, but
Newton’s great achievement was to
establish that they follow mathematically
from his Law of Universal Gravitation
and his Laws of Motion. We present
here a calculus-based derivation of
Kepler’s Laws. You should be familiar
with the results, but need not worry
about the details of the derivation—it’s
just here in case you’re curious.
The standard approach in analyzing
planetary motion is to use (r , q )
coordinates, where r is the distance from
the origin—which we take to be the
center of the Sun—and q is the angle
between the x -axis and the line from the
origin to the point in question.
In the picture above, in which I have
greatly exaggerated the ellipticity of the
orbit, suppose the planet goes from A to
B , a distance Ds, in a short time Dt, so
its speed in orbit is Ds/D t. Notice that
the velocity can be resolved into vector
components in the radial direction Dr /D t
(in this case negative) and in the
direction perpendicular to the radius,
r Dq /D t . The short line BC in the diagram
above is perpendicular to SB (S being the
center of the Sun), and therefore
becoming perpendicular to SC as well in
the limit of AB becoming an
infinitesimally small distance.
In the limit of small Ds, then, we have
where the
angular velocity .
Kepler’s Second Law
We shall consider Kepler’s Second Law
(that the planet sweeps out equal areas
in equal times) first, because it has a
simple physical interpretation.
Looking at the above picture, in the time
Dt during which the planet moves from
A to B , the area swept out is the
approximately triangular area ABS,
where S is the center of the Sun. For the
distance AB sufficiently small, this area
tends to that of the long thin triangle
BSC , which has a base of length rDq and
a height r . Using area of a triangle = ½
base´height, it follows immediately that
Now, the angular momentum L of the
planet in its orbit is given by
so the rate of sweeping out of area is
proportional to the angular momentum,
and equal to L/2m .
Newton’s Laws tell us that the rate of
change of angular momentum is equal to
the torque of the forces acting on the
system . For the planet orbiting the Sun,
this torque is zero : the only force acting
is gravity, and that force acts in a line
from the planet towards the center of the
Sun, and therefore has zero torque—no
leverage—about that central point, so the
planet’s angular momentum about that
point must remain constant.
Consequently, the rate at which area is
swept out is also constant, and Kepler’s
Second Law follows: equal areas are
swept out in equal times .
Ellipses
We’re now ready for Kepler’s First Law:
each planet moves in an elliptical orbit
with the Sun at one focus of the ellipse.
Let us begin by reviewing some basic
facts about ellipses.
An ellipse is essentially a circle scaled
shorter in one direction, in (x , y )
coordinates it is described by the
equation
,
a circle being given by a = b . The
lengths a and b are termed the
semimajor axis and the semiminor axis
respectively.
An ellipse has two foci, shown F 1 and F 2
on the diagram, which have the optical
property that if a point source of light is
placed at F 1 , and the ellipse is a mirror,
it will reflect—and therefore focus—all
the light to F 2 . One way to draw an
ellipse is to take a piece of string of
length 2 a , tie one end to F 1 and the
other to F 2 , and hold the string taut with
a pencil point. Anywhere this happens
on a flat piece of paper is a point on the
ellipse. In other words, the ellipse is the
set of points P such that PF1 + PF 2 = 2 a .
The ellipticity of the ellipse e is defined
by writing the distance from the center
of the ellipse to a focus OF 1 = ea.
In fact, in analyzing planetary motion, it
is more natural to take the origin of
coordinates at the center of the Sun
rather than the center of the elliptical
orbit. It is also more convenient to take
(r , q ) coordinates instead of (x , y )
coordinates, because the strength of the
gravitational force depends only on r.
Therefore, the relevant equation
describing a planetary orbit is the (r , q )
equation with the origin at one focus.
For an ellipse of semi major axis a and
eccentricity e the equation is:
.
It is not difficult to prove that this is
equivalent to the traditional equation in
terms of x , y presented above.
Answered by Anonymous
16

\textbf{Answer is in Attachment !!}
Attachments:
Similar questions