For attractive inverse forces the shape of orbit will be
Answers
Answer:
Motion in an Inverse-Square Central Force
Field
1 Central Forces and Angular Momentum
Suppose we have a central force, that is, a force that depends only on the
distance to the origin and is directed either toward or away from the origin.
Then we can write the force as
F~ = f(r)~r (1)
where f(r) is a scalar function (We write r =
p
x
2 + y
2 + z
2
for the length of
the vector ~r: of course this is just the distance to the origin). For a particle
of mass m whose position as a function of time is given by ~r(t), we define its
angular momentum by
L~ (t) = m~r(t) × ~r0
(t)
Fact: If a particle moves subject to a central force, its angular momentum
is constant.
Proof: Using the product rule, we compute the derivative of L~ (t) as
L~ 0
(t) = m~r0
(t) × ~r0
(t) + m~r(t) × ~r00(t) = m~r(t) × ~r00(t)
since ~r0
(t) crossed with itself is the zero vector. But Newton’s second law
says that
F~ = m~r00(t),
so our equation is
L~ 0
(t) = ~r(t) × F . ~
But then, using equation (1), we have
L~ 0
(t) = ~r(t) × f(r)~r(t) = 0
1and so L~ (t) must be a constant vector.
The constancy of the vector L~ has two important consequences.
Consequence 1: The motion takes place in a plane containing the origin.
This is because the vector ~r must always be perpendicular to the constant
vector L~ .
Consequence 2: The radius vector ~r sweeps out areas at a constant rate.
(This is Kepler’s first law in the context of planetary motion.)
Let ∆A be the area swept out by r(t) in the time interval ∆t. Then
∆A =
1
2
|~r(t) × ~r(t + ∆t)| ≈ 1
2
|~r(t) × (~r(t) + ~r0
(t)∆t| =
1
2
|~r(t) × ~r0
(t)|∆t
so that
∆A
∆t
≈
1
2
|~r(t) × ~r0
(t)| =
L
2m
,
where L = |L~ | (which is constant). In the limit,
dA
dt =
L
2m
.
2 Inverse-Square Force
Suppose now that f(r) = −K/r3
, where K is a constant. (Note that the
force is attractive if K > 0 and repulsive if K < 0.) In this case F~ is called
an inverse-square force since
F~ = −
K
r
3
~r = −
K
r
3
rˆr = −
K
r
2
ˆr,
where ˆr is the unit vector in the direction of ~r. In the case of planetary
motion, K = GM, where G is the universal gravitational constant and M
is the mass of the sun. In the case of motion in the electric field of a static
charged particle, K = Cq, where C is a universal constant and q is the
particle charge: in this case K can be negative.
From the product rule
L~ = m~r × (rˆr)
0 = mrˆr × (r
0
ˆr + rˆr
0
) = mr2
ˆr × ˆr
0
,
and since F~ = −Kr−2ˆr, we have
F~ × L~ = −Kr−2
ˆr × (mr2
ˆr × ˆr
0
) = −Kmˆr × (ˆr × ˆr
0
).
2Now use the vector identity
a × (b × c) = (a · c)b − (a · b)c (2)
to write this as
F~ × L~ = −Km((ˆr · ˆr
0
)ˆr − (ˆr · ˆr)ˆr
0
).
But ˆr · ˆr = 1, and it follows from differentiating this equation that ˆr · ˆr
0 = 0.
Hence
F~ × L~ = Kmˆr
0
. (3)
Since
d
dt
m~r0 × L~
= m~r00 × L~ = F~ × L, ~
equation (3) implies that
m~r0 × L~ = Kmˆr + ~c,
for some vector constant ~c. Dot both sides with ~r to get
~r · (m~r0 × L~ ) = Kmr + ~r · ~c. (4)
On the other hand, the vector identity (2) implies that
m~r0 × L~ = m2
~r0 × (~r × ~r0
) = m2
|~r0
|
2
~r − m2
(~r0
· ~r)~r0
and thus
~r · (m~r0 × L~ ) = m2
r
2
|~r0
|
2 − m2
(~r0
· ~r)
2
. (5)
Now
|a|
2
|b|
2 − (a · b)
2 = |a × b|
2
for any vectors a and b, so equation (5) gives
~r · (m~r0 × L~ ) = m2
|~r0 × ~r|
2 = L
2
.
Compare this with equation (4) to get
L
2 = Kmr + ~r · ~c. (6)
Now choose polar coodinates in the plane of motion so that ~c points in
the direction of the positive x-axis. Then equation (6) is
L
2 = Kmr + cr cos θ
3(where c = |~c|), or
r =
L
2
c cos θ + Km
. (7)
Equation (7) is always a conic section (ellipse, parabola or hyperbola). It is
the equation of an ellipse if c < Km (and a circle if c = 0). If Km = c it is
a parabola, and for Km < c it is an hyperbola (in particular, we always get
hyperbolic orbits for a repulsive force). If the path of a particle is a closed
orbit, then it must be an ellipse: this is Kepler’s second law.
QUESTION:-
For attractive inverse forces the shape of orbit will be
ANSWER:-
- For attractive inverse forces, the shape of the orbit will be elliptical
- Elliptical is the answer.
EXPLANATION:-
- In the Attractive Inverse Forces, the path of a particle is a closed orbit so it must be an ellipse according to Kepler’s second law.
- Kepler's second law states that a planet moves in its ellipse so that the line between it and the Sun positioned at a focus sweeps out equal areas at equal times.
- In attractive inverse forces the equation of an ellipse if c < Km (and a circle if c = 0).
- The elliptical orbits of planets were indicated by estimations of the orbit of Mars.
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