Here,
S
and
S
′ are the foci and
P
any point on the ellipse. The sun is at S and planet at P. The distance of each focus from the centre of ellipse is
ea
, where e is the dimensionless number between 0 to 1 called the eccentricity. If e = 0, the ellipse is a circle. The actual orbits of the planets are nearly circular, their eccentricities range from 0.007 for Venus to 0.248 for Pluto.
For earth e = 0.017. The point in the planet’s orbit closest to the sun is the perihelion and the point most distant from the sun is aphelion.
Law of orbits
Newton was able to show that for a body acted on by an attractive force proportional to 1/r
2
the only possible closed orbits are a circle or an ellipse. The open orbits must be parabolas or hyperbolas. He also showed that if total energy E is negative the orbit is an ellipse (or circle), if it is zero the orbit is a parabola and if E is positive the orbit is a hyperbola.
Further, it was also shown that the orbits under the attractive force
F
=
k
/
r
n
are stable for n < 3. Therefore, it follows that circular orbits will be stable for a force varying inversely as the distance or the square of the distance and will be unstable for the inverse cube (or a higher power) law
Explanation for law of Areas
PP
’ =
v
dt
P
’
M
= (
PP
’) sin (180
o
– θ) =
PP
’ sin θ
= (
v
sin θ)
dt
Kepler's second law is shown in figures. In a small time, interval
dt
, the line from the sun
S
to the planet
P
turns through an angle
d
θ. The area swept out in this time interval is,
dA
= area of triangle shown in figure
=
(base) (height)
=
(
SP
)(
P
,
M
)
=
(
r
) (
v
sin θ)
dT
∴
Areal velocity
rv
sin θ
……. (i)
Now, rv sin θ is the magnitude of the vector product
r
×
v
which in turn is
times the angular momentum
L
=
r
×
mv
of the planet with respect to the sum. So, we have,
…… (ii)
Or
= constant
Thus, Kepler's second law, that areal velocity is constant, means that angular momentum is constant. It is easy to see why the angular momentum of the planet must be constant. According to Newton's law the rate of change of L equals the torque of the gravitational force F acting on the planet,
Here, r is the radius vector of planet from the centre of the sun and the force F is directed from the planet towards the centre of the sun. So, these vectors always lie along the same line and their vector product r × F is zero. Hence,
or L = constant. Thus, from Eq. (ii) we can see that
= constant if L = constant. Thus, second law is actually the law of conservation of angular momentum.
Explanation of Law of periods
We have already derived Kepler’s third law for the particular case of circular orbits. T
2
is proportional to r
3
Newton was able to show that the same relationship holds for an elliptical orbit, with the orbit radius r replaced by semimajor axis a. Thus,
(elliptical orbit)
Here, M
s
is the mass of the sum.
Important Points To Remember
1.
The areal velocity of a planet is constant (Kepler’s second law) and is given by
dA/dt
=
L
/ 2
m
Here, L is the angular momentum of the planet about sun.
2.
Most of the problems of gravitation are solved by two conservation laws: (i) Law of Conservation of Angular Momentum about sun and (ii) Law of Conservation of Mechanical (potential + kinetic) Energy Hence, the following two equations are used in most of the cases,
Here, the following two equations are used in most of the cases,
mv r
sin θ = constant
….. (1)
mv
2
–
= constant
….. (2)
At aphelion and perihelion positions θ = 90
o
Hence, equation (1) can be written as,
mv
r sin90
o
= constant
Further, since mass of the planet (
m
) also remains constant, equation (1) can also be written as
vr
sinθ = constant
…. (4)
⇒
v
1
r
1
=
v
2
r
2
{θ = 90
o
}
Since
r
1
>
r
2
⇒
v
1
<
v
2
3.
Applying the above-mentioned conservation Laws at the aphelion and perihelion positions along with the knowledge that
r
1
=
a
phelion
=
a
(1 +
e
) and
r
2
=
p
erihelion
=
a
(1 –
e
)
where
e
is the eccentricity of the elliptical orbit. Then, we can show that.
v
min
=
a
phelion
=
v
1
=
v
max
=
v
perhilion
=
v
2
=
and the total energy of the planet is calculated by.
E = U + K = –
⇒
E = –
4.
Trajectory of a body projected from point A in the direction AB with different initial velocities: Let a body be projected from point A with velocity v in the direction AB. For different values of v the paths are different. Here are the possible cases.
Trajectory of a body projected from point A in the direction AB with different initial velocities:
Let a body be projected from A with velocity v in the direction AB. For different values of v the paths are different. Here are the possible cases.
(i)
If v = 0, path is a straight line from A to O.
(ii)
If 0 <
v
<
v
0
, path is an ellipse with centre O of the earth as a focus.
(iii)
If
v
=
v
0
, path is a circle with O as the centre.
(iv)
If
v
0
<
v
<
v
e
, path is again an ellipse with O as a focus.
(v)
If
v
=
v
e
, body escapes from the gravitational pull of the earth and path is a parabola.
(vi)
If v > v
e
, body again escapes but now the
Path is hyperbola. Here,
v
0
= orbital speed
at A and
v
e
= escape velocity at A.
Important Points Related To Kepler's Laws
(a)
From case (i) to (iv) total energy of the body is negative. Hence, these are the closed orbits and are also called Bound Trajectories. For case (v) total energy is zero and for case (vi) total energy is positive. In these two cases orbits are open and are called Unbound Trajectories.
(b)
If v is not very large the elliptical orbit will intersect the earth and the body will fall back to earth.
