How Long Does It Take for a Particle to Travel Through an Earth-Crossing Tunnel?

  • Thread starter Avi1995
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In summary, the particle goes inside the tunnel with a speed of \sqrt{gR} and then it executes a SHM with a circular frequency ω2=GM/R3.
  • #1
Avi1995
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Problem: Assume that a tunnel is dug across the earth(radius=R) passing through its centre. Find the time a particle takes to cover the length of tunnel if it is projected into the tunnel with a speed of [itex]\sqrt{gR}[/itex]
2. Homework Equations :
Basic SHM equations:
1.F=-kx
2.T=(2*pi)/(omega)
Gravitation:
[itex]F=GMm/r^2[/itex]

Attempt:
V(potential due to earth)=GMx/R(R+x)
1/2*m*g*R=GMx/R(R+x)
x=R
So particle will first go inside the tunnel accelerating till centre then decelerating till surface and further till it reaches distance R from surface.
I am confused now As this is not a SHM(Outside surface the field is non linear) So how do I find the time?
 
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  • #2
g`=GMx/R3
Thats the acceleration at a distance x from the center.
So the acceleration of the particle is directly proportional to the (-ve) of the distance from the center.
So the particle will execute a Simple Harmonic Motion with its circular frequency, ω2=GM/R3
Now could you work it out?
 
  • #3
But outside the surface the field is non linear, so it won't be a shm.
 
  • #4
Avi1995 said:
But outside the surface the field is non linear, so it won't be a shm.

The particle won't be able to go outside! The force of gravity will keep on increasing just enought to make its velocity 0 at the surface! Think about it
 
  • #5
deep838 said:
The particle won't be able to go outside! The force of gravity will keep on increasing just enought to make its velocity 0 at the surface! Think about it

But there is no change in potential energy from one end to other end as both are on surface. So shouldn't K.E. remain same?
 
  • #6
Read the problem carefully: it asks how long the particle is inside the tunnel.

In the tunnel, the particle performs SHM. Why? What force acts on the particle at distance r<R from the centre of Earth? Assume that the density of the Earth is constant. You have to know that the force exerted by a homogeneous sphere is the same as if all mass enclosed in the sphere of radius r concentrated in the centre.

ehild
 
  • #7
deep838 said:
g`=GMx/R3
Thats the acceleration at a distance x from the center.
No. At radius r from the centre of a solid sphere radius R>r (assuming each concentric shell is in itself uniform), the gravitational pull from the portion of the sphere at radius > r exactly cancels itself. So only consider the pull from the part of the Earth at radius < r.
 
  • #8
haruspex said:
No. At radius r from the centre of a solid sphere radius R>r (assuming each concentric shell is in itself uniform), the gravitational pull from the portion of the sphere at radius > r exactly cancels itself. So only consider the pull from the part of the Earth at radius < r.

yeah that's what i did to get the equation!
 
  • #9
deep838 said:
yeah that's what i did to get the equation!

Sorry - misread it. Too hasty.
 
  • #10
ehild said:
Read the problem carefully: it asks how long the particle is inside the tunnel.

In the tunnel, the particle performs SHM. Why? What force acts on the particle at distance r<R from the centre of Earth? Assume that the density of the Earth is constant. You have to know that the force exerted by a homogeneous sphere is the same as if all mass enclosed in the sphere of radius r concentrated in the centre.

ehild

Understood. But the little doubt that remains in my mind is that particle was projected with a speed in the tunnel, can the equations of SHM be used?
 
  • #11
The initial velocity and initial position determines the amplitude and phase constant of the SHM, but the equation ma=-kx does not change.

ehild
 

Related to How Long Does It Take for a Particle to Travel Through an Earth-Crossing Tunnel?

What is SHM (Simple Harmonic Motion)?

Simple Harmonic Motion (SHM) is a type of motion in which an object moves back and forth in a regular pattern, around a central equilibrium point. It is characterized by a sinusoidal or circular motion and is caused by a restoring force that is directly proportional to the displacement from the equilibrium point.

What is the relationship between SHM and Gravitation?

The relationship between SHM and Gravitation is that SHM can occur in systems where there is a restoring force acting on an object, such as a mass on a spring or a pendulum. In these systems, the gravitational force acts as the restoring force, causing the object to oscillate back and forth.

How do you solve SHM + Gravitation problems?

To solve SHM + Gravitation problems, you need to first identify the system and the forces acting on the object. Then, you can use equations such as Hooke's Law (F = -kx) and Newton's Second Law (F = ma) to set up and solve the equations of motion for the system. It is also important to consider energy conservation when solving these problems.

What are some real-life examples of SHM + Gravitation problems?

Some real-life examples of SHM + Gravitation problems include a swing, a pendulum clock, a mass attached to a spring, and a bouncing ball. In all of these systems, the object experiences SHM due to the gravitational force acting as the restoring force.

What are the applications of solving SHM + Gravitation problems?

Solving SHM + Gravitation problems has many practical applications in fields such as engineering, physics, and astronomy. It can help us understand and predict the behavior of systems such as springs, pendulums, and celestial bodies. This knowledge is crucial for designing and building structures, machines, and devices that rely on SHM and gravitational forces.

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