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

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A particle projected into a tunnel through the Earth with a speed of √(gR) will experience Simple Harmonic Motion (SHM) due to the gravitational force acting on it. The acceleration at any distance x from the center is given by g' = GMx/R³, which leads to the conclusion that the force behaves as if all mass is concentrated at the center. The particle accelerates towards the center, decelerates as it moves back towards the surface, and will not exit the tunnel. The discussion emphasizes that while the gravitational field outside the surface is non-linear, inside the tunnel, the motion can be treated as SHM. The time taken for the particle to traverse the tunnel can be calculated using SHM equations, taking into account its initial conditions.
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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 \sqrt{gR}
2. Homework Equations :
Basic SHM equations:
1.F=-kx
2.T=(2*pi)/(omega)
Gravitation:
F=GMm/r^2

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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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?
 
But outside the surface the field is non linear, so it won't be a shm.
 
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
 
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?
 
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
 
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.
 
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!
 
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
 
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