# Spring in a groove inside the Earth

1. Oct 31, 2014

### erisedk

1. The problem statement, all variables and given/known data
A small ball of mass 'm' is released at a height 'R' above the earth's surface. The maximum depth of the ball to which it goes is R/2 inside the earth through a narrow groove before coming to rest momentarily. The groove contains an ideal spring of spring constant K and natural length R. The value of K, if R is the radius of the earth and M is the mass of the earth is,

Ans: 7GMm/R^3

2. Relevant equations
PE= -GMm/R
KE= 1/2 m v^2
PE(of spring)= 1/2 kx^2

3. The attempt at a solution
Conserving energy of the spring- mass system at point of release (at a distance 2R from the centre of the earth) and when it comes to rest at a distance R/2 from the centre of the earth,
PE(of body initially)=PE(of body finally)+PE(of spring)
-GMm/2R = -2GMm/R + 1/2*k*(R^2/4)
k= 12GMm/R^3

I don't understand what's wrong with my answer.

2. Oct 31, 2014

### BvU

Must be the PE relationship. Does PE go to infinity if R goes to 0 ?

3. Oct 31, 2014

### erisedk

Yeah, it goes to negative infinity.

4. Oct 31, 2014

### BvU

Think again. not all the mass of the earth is located at the center

5. Mar 24, 2016

### manasi bandhaokar

actually the GPE inside and outside a solid sphere are different.
outside : -GMm/r (r=distance from centre of sphere)
inside: << Formula deleted by Mentor >>

Last edited by a moderator: Mar 24, 2016
6. Mar 24, 2016

### Staff: Mentor

Welcome to the PF.

Remember that the student must do the bulk of their homework themselves. It's fine to provide the hint that the equation for the GPE is different inside the Earth compared to outside, but the student is expected to figure out the equation by themselves Thanks..

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