Fixed Attractor Point

## **Problem**

A particle of mass

m

{\displaystyle m}

 is attracted toward a a fixed point with a force inversely proportional to its instantaneous distance from the point. If the particle is released from rest, find the time for it to reach the fixed point.

## **Hint**

(1) Use the Gamma function

Γ
(
n
# )

∫

0

∞

t

n
−
1

e

−
t

d
t

{\displaystyle \Gamma (n)=\int \_{0}^{\infty }t^{n-1}e^{-t}\,dt}

 and the value

Γ
(
1

/

2
# )

π

{\displaystyle \Gamma(1/2) = \sqrt \pi}

.

## **Solution**

Let the fixed point be the origin. At time

# t
0

{\displaystyle t=0}

, let the particle be located on the x-axis at

# x
a

{\displaystyle x=a}

 where

a
>
0

{\displaystyle a>0}

. Then by Newton's second law, the equation of motion is

m

d

2

x

d

t

2

=
−

k
x

{\displaystyle m{\frac {d^{2}x}{dt^{2}}}=-{\frac {k}{x}}}

where

m

{\displaystyle m}

 is the mass of the particle and

k
>
0

{\displaystyle k>0}

 is the constant of proportionality.

Let

d
x

d
t

=
v

{\displaystyle {\frac {dx}{dt}}=v}

 be the velocity of the particle. Subsequently, by the chain rule

d

2

x

d

t

2

=

d
v

d
t

=

d
v

d
x

⋅

d
x

d
t

=
v
⋅

d
v

d
x

{\displaystyle {\begin{aligned}{\frac {d^{2}x}{dt^{2}}}&={\frac {dv}{dt}}\\[5pt]&={\frac {dv}{dx}}\cdot {\frac {dx}{dt}}\\[5pt]&=v\cdot {\frac {dv}{dx}}\end{aligned}}}

and the original equation of motion becomes

m
v

d
v

d
x

=
−

k
x

.

{\displaystyle mv{\frac {dv}{dx}}=-{\frac {k}{x}}.}

Integrating both sides yields:

m
v
∫
d
v

=
−
∫

k
x

d
x

1
2

m

v

2

=
−
k
ln
⁡
x
C
.

{\displaystyle {\begin{aligned}mv\int dv&=-\int {\frac {k}{x}}\,dx\\[5pt]{\frac {1}{2}}mv^{2}&=-k\ln x+C.\end{aligned}}}

Since

# v
0

{\displaystyle v=0}

 (at rest) at

# x
a

{\displaystyle x=a}

, we find

# c
k
ln
⁡
a

{\displaystyle c=k\ln a}

. Thus,

1
2

m

v

2

=
k
ln
⁡

(

a
x

)

v

=
−

2
k

m

ln
⁡

(

a
x

)

.

{\displaystyle {\begin{aligned}{\frac {1}{2}}mv^{2}&=k\ln \left({\frac {a}{x}}\right)\\[5pt]v&=-{\sqrt {\frac {2k}{m}}}{\sqrt {\ln \left({\frac {a}{x}}\right)}}.\end{aligned}}}

Note that the negative sign is chosen because the distance

x

{\displaystyle x}

 is decreasing as across time.

Since

# v

d
x

d
t

{\displaystyle v = \frac{dx}{dt} }

, we get

d
x

d
t

=
−

2
k

m

ln
⁡

(

a
x

)

.

{\displaystyle {\frac {dx}{dt}}=-{\sqrt {\frac {2k}{m}}}{\sqrt {\ln \left({\frac {a}{x}}\right)}}.}

To solve for the time

T

{\displaystyle T}

 taken for the particle to travel from

# x
a

{\displaystyle x=a}

 to

# x
0

{\displaystyle x = 0}

, rearrange the differential equation and integrate both sides

∫

0

T

d
# t

∫

a

0

d
x

−

2
k

m

ln
⁡

(

a
x

)

{\displaystyle \int \_{0}^{T}dt=\int \_{a}^{0}{\frac {dx}{-{\sqrt {\frac {2k}{m}}}{\sqrt {\ln \left({\frac {a}{x}}\right)}}}}}

∫

0

T

d
# t

m

2
k

∫

0

a

d
x

ln
⁡

(

a
x

)

{\displaystyle \int \_{0}^{T}dt={\sqrt {\frac {m}{2k}}}\int \_{0}^{a}{\frac {dx}{\sqrt {\ln \left({\frac {a}{x}}\right)}}}}

# T

m

2
k

∫

0

a

d
x

ln
⁡

(

a
x

)

.

{\displaystyle T={\sqrt {\frac {m}{2k}}}\int \_{0}^{a}{\frac {dx}{\sqrt {\ln \left({\frac {a}{x}}\right)}}}.}

Letting

# u
ln
⁡

(

a
x

)

{\displaystyle u=\ln \left({\frac {a}{x}}\right)}

 or

# x
a

e

−
u

{\displaystyle x=ae^{-u}}

, we get

T

=
a

m

2
k

∫

0

∞

u

−
1

/

2

e

−
u

d
u

=
a

m

2
k

Γ
(
1

/

2
)

=
a

π
m

2
k

.

{\displaystyle {\begin{aligned}T&=a{\sqrt {\frac {m}{2k}}}\int \_{0}^{\infty }u^{-1/2}e^{-u}\,du\\[5pt]&=a{\sqrt {\frac {m}{2k}}}\,\Gamma (1/2)\\[5pt]&=a{\sqrt {\frac {\pi m}{2k}}}.\end{aligned}}}