Time for going from one point of earth surface to another

[Pushoam]

Homework Statement

Homework Equations

The Attempt at a Solution

[/B]
The thing to remember is that inside the Earth surface, the gravitational field does not remain inversely proportional to $r^2$.

The equation of motion is

$\ddot r = \frac { -gr} R$ ....(1)

This means that the particle will oscillate between the two points on the surface with angular frquency $\omega = \sqrt{ \frac { g} R}$. ...(2)

$T= \frac { 2\pi } {\omega }$...(3)

The required time is $\frac { T} 2 = 42.3 ~ min$...(4)

Hence, the correct option is (c).

The time taken for going from one point to another point of the Earth surface is independent of the distance between the two points.

Then, when I go into an underground room, I need to move for going from one point to another. Is it because of friction force?

 

[PeroK]

Homework Statement

View attachment 216656

Homework Equations

The Attempt at a Solution

[/B]
The thing to remember is that inside the Earth surface, the gravitational field does not remain inversely proportional to $r^2$.

The equation of motion is

$\ddot r = \frac { -gr} R$

This means that the particle will oscillate between the two points on the surface with angular frquency $\omega = \sqrt{ \frac { g} R$.

$T= \frac { 2\pi } {\omega }$

The required time is $\frac { T} 2 = 42.3 ~ min$.

Hence, the correct option is (c).

The time taken for going from one point to another point of the Earth surface is independent of the distance between the two points.

Then, when I go into an underground room, I need to move for going from one point to another. Is it because of friction force?

If the tunnel was thru the centre, then that formula comes out immediately. How do you justify the SHM for a tunnel near the surface?

Otherwise, what's your question?

 

[Pushoam]

If the tunnel was thru the centre, then that formula comes out immediately. How do you justify the SHM for a tunnel near the surface?

I don't know about which formula are you talking?
In writing the following equation,

The equation of motion is

$\ddot r = \frac { -gr} R$....(1)

I din't assume that the tunnel is passing through the center.
Even for the tunnel not passing through the center, I think the eqn (1) is valid. Is this wrong?

 

[PeroK]

I don't know about which formula are you talking?
In writing the following equation,I din't assume that the tunnel is passing through the center.
Even for the tunnel not passing through the center, I think the eqn (1) is valid. Is this wrong?

What does $r$ represent? For SHM it should be the displacement along the tunnel.

Only when the tunnel is thru the centre does this displacement coincide with the displacement from the centre.

 

[Pushoam]

What does rrr represent? For SHM it should be the displacement along the tunnel.

I took it as the displacement from the center.

For SHM it should be the displacement along the tunnel.

I am not getting why it should be so.
It seems as if I assumed that the tunnel is passing through the center unknowingly while writing equation (1).

 

[PeroK]

I took it as the displacement from the center.

I am not getting why it should be so.
It seems as if I assumed that the tunnel is passing through the center unknowingly while writing equation (1).

When the object is at the midpoint of the tunnel, according to SHM, the force and acceleration should be zero.

But, at the midpoint of the tunnel $r \ne 0$, so the formula cannot be correct in the form you have.

Also, for SHM, the displacement at the midpoint should be 0. In your equation $r$ is always positive, so additionally the force would not change sign.

 

[Pushoam]

When the object is at the midpoint of the tunnel, according to SHM, the force and acceleration should be zero.

But, at the midpoint of the tunnel r≠0r≠0r \ne 0, so the formula cannot be correct in the form you have.

Also, for SHM, the displacement at the midpoint should be 0. In your equation rrr is always positive, so additionally the force would not change sign.

I will reply to you after digesting it. Thanks for it.