Physics, asked by manikdahiya8747, 11 months ago

For attractive inverse forces the shape of orbit will be

Answers

Answered by SabrezAlam
0

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.

Answered by Jaswindar9199
0

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.

#SPJ2

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