Question

In an imaginary gravitational field, the gravitational potential energy of a particle of mass m at a distance r from the centre of the field force is given by u=(kr^2)/2 where k is a positive constant of appropriate dimensions. If the particle is moving in a circular orbit of radius r about the centre, choose the correct options.


Total energy of the particle is


Total energy of the particle is


The orbital time period of the particle is proportional to


The angular momentum of the particle about the centre is proportional to

Answer 1

Answer:

Given potential energy V=

2

kr

2

F=−kr (towards centre) [F=−

dr

dV

]

At r=R,

kR=

R

mv

2

[Centripetal force]

where v is the speed of the particle

⇒ velocity v=

m

kR

2

=

m

k

R

L=mvR=m

m

k

R

2

solution

Given potential energy V=

2

kr

2

F=−kr (towards centre) [F=−

dr

dV

]

At r=R,

kR=

R

mv

2

[Centripetal force]

where v is the speed of the particle

ok follow me and mark me brilliant

Answer 2

Given:

In an imaginary gravitational field, the gravitational potential energy of a particle of mass m at a distance r from the centre of the field force is given by u=(kr^2)/2 where k is a positive constant of appropriate dimensions. The particle is moving in a circular orbit of radius r about the centre.

To find:

Proportionality of :

• Time period

• Angular momentum

Solution:

$\therefore \: U = \dfrac{k {r}^{2} }{2}$

$\implies \: F = - \dfrac{dU}{dr}$

$\implies \: F = - \dfrac{d( \frac{k {r}^{2} }{2} )}{dr}$

$\implies \: F = - 2kr \: \: \: \: \: \: ..........(1)$

Now, this force will provide the centripetal component :

$\therefore \: | F| = \dfrac{m {v}^{2} }{r}$

$\implies \: 2kr = \dfrac{m {v}^{2} }{r}$

$\implies \: m{v}^{2} = 2k {r}^{2}$

$\implies \: {v}^{2} = \dfrac{2k {r}^{2} }{m}$

$\implies \: v= \sqrt{ \dfrac{2k {r}^{2} }{m}}$

$\implies \: v= \sqrt{ \dfrac{2k }{m}} \times r$

$\implies \: v \propto r \: \: \: \: \: \: .......(2)$

Now, time period be T :

$\therefore \: T = \dfrac{2\pi r}{v}$

$\implies\: T \propto \: \dfrac{r}{v}$

$\implies\: T \propto \: \dfrac{r}{r}$

$\boxed{ \implies\: T \propto \: {r}^{0} }$

So, time period is independent of the radius of circular trajectory.

Let angular momentum be L :

$\therefore \: L = mvr$

$\implies\: L \propto \: (v \times r)$

$\implies\: L \propto \: (r\times r)$

$\boxed{ \implies\: L \propto \: {r}^{2} }$

So, angular momentum is directly proportional to square of the radius of the circular trajectory.