Journey through the center of the earth
Congratulations! You just dug a tunnel to the opposite end of the Earth. To celebrate, you drop a rock straight down into the hole. If the tunnel passes directly through the center of the Earth, then what will happen to the rock after you drop it?
Assumptions
- There is no air resistance and the ball is thrown straight toward the center of the Earth.
- The Earth is a perfect sphere, the interior is a uniform solid, the rock is electrically neutral, and the tunnel is drilled along the axis of rotation of the Earth's spin. Neglect any thermal effects.
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1 solution
Discussions for this problem are now closed
It would halt at the center of the earth . I totally disagree with the solution. The point when it passes through the center of the earth the point just Above the center of the earth will have a very strong gravity. Making the object halt at the center
That's not right. At the center of the Earth, it will feel no gravitational pull, and around the center of the Earth, it will very a very weak pull.
Thank you Josh Silverman . I agree with the fact that I was wrong, later I researched about it and came to know about this fact
Will this equation remain valid while something travels to the center ? Momentum is momentum due to relative position of an object and that relative position is where the pull or push force is being exerted. while an object is at the position where the pull or push force is acting towards; the object has no relative momentum. The object should stop at the center of earth
I think the confusion is coming from the fact that the gravitational force changes based on where you are in the tunnel. As you travel down the tunnel, some mass will be above you while the rest of the mass will be below you. Both masses will try to pull you towards them. As you travel further towards the center, the two masses will become nearly equal is mass such that the force that pulls you up will be equal to the force that pulls you down such that there is no net force on you at the center of the earth, which is reflected in the above solution.
However, as you are being pulled down to the center, you will also be increasing speed, because F = m a = − R E a r t h 2 G m M E a r t h such that your kinetic energy in the core will equal your initial Gravitational potential energy, K c o r e = 1 / 2 m v c o r e 2 = r 0 G M m = U 0 . If we let r 0 = R E a r t h we get v c o r e = R e a r t h 2 G M ≈ 1 1 , 0 0 0 m / s , this velocity will then carry you to the exact same height on the other side where you will have no velocity and then you repeat the journey over and over again always reaching the same height and having the same velocity in the core during each oscillation.
To me gravitational pull is not because of some mass is over you or under you but it is because of movement of earth round its axis. Am I wrong here ?
@Fred Burger – Simply stated, yes you are incorrect. Gravitational pull at least in this regard has nothing to do with the rotation of the earth or its movement. The amount of gravitational pull has nothing to do with the rotation of the earth or movement of earth in this context.
F ( r ) = − r 2 G m M ( r ) , the force due to gravity, has no rotational or movement parameters.
Possibly you are getting gravity confused with centripetal acceleration which is a force generated by rotation. However, even at the equator, a c = − r v 2 ≈ − 0 . 0 3 4 m / s 2 , is several times weaker than the gravitational force, g = − 9 . 8 m / s 2 . At the poles there is no centripetal acceleration since it is aligned with the axis of rotation. The centripetal acceleration and force are often talked about with artificial gravity, but has nothing to do with this problem.
Please do a google search on Newtonian gravity and possibly centripetal acceleration to further help with any confusion.
To determine the motion, we'll need to determine the force and then solve Newton's Second Law , F = m a .
For gravitational forces , F ( r ) = − r 2 G m M ( r ) , where G is Newton's gravitational constant, m=mass of the particle, r=distance from the center of the earth, and M ( r ) = 3 4 ρ π r 3 , where ρ is density.
We assume that the Earth has a relatively uniform density such that ρ = 4 / 3 π R E a r t h 3 M E a r t h .
Combining the equations gives F = − G m 4 / 3 π R E a r t h 3 r 2 4 / 3 π M E a r t h r 3 = − R E a r t h 3 G m M E a r t h r . This is analogous to Hooke's Law , F = − k r , where k is a constant, since R E a r t h 3 G m M E a r t h = constant = k , which means that the motion of the particle will undergo simple harmonic motion just like a spring.
It should also be mentioned this is more readily recognized as F = − m g R E a r t h r , where g = R E a r t h 2 G M E a r t h = 9 . 8 m / s 2 showing it is similar to gravity as the surface of earth , F = − m g , but with a scalar term to account for how much of the Earth is pulling on the rock towards the center.
Lots of assumptions are made in this problem including that there is no air resistance or Earth's rotation can be ignored otherwise answer 1 would be valid because the equation of motion would be something like F = − k r − α v , where alpha is a constant and v is speed. This equation of motion describes a particle that would constantly lose speed and energy due the air resistance and would eventually get stuck in the center.