A small ball of mass m is released at ht R above the surface of earth.If the maximum depth of the ball to which it goes is R/2 inside the earth through a marrow groove before coming to rest momentarily. The groovet, contains an ideal spring of spring constant k and natural length R find the value of k if R is the radius of earth and M mass of earth?

Ans: 7GMm/( R cube)

Question

A small ball of mass m is released at ht R above the surface of earth.If the maximum depth of the ball to which it goes is R/2 inside the earth through a marrow groove before coming to rest momentarily. The groovet, contains an ideal spring of spring constant k and natural length R find the value of k if R is the radius of earth and M mass of earth?

Ans: 7GMm/( R cube)

vikas askiitian expert · Answer

gravitational potential inside the earth is given by -3GM/2R [1-r²/3R²]

potential energy of ball when it was at a height of R above earth is-GMm/2R

kinetic energy at this instant is 0...

total energy = -GM/2R .................1

finally ball agian comes to rest , now ball is at a depth of R/2 from earth so

its gravitational energy is -3GMm/2R[1-r²/3R²]

=-3GMm/2R[11/12] (r=R/2)

=-11GMm/8R

potential energy stored in spring is KX²/2 = KR²/8 (compression is R/2)

total energy now is sum of potential energy of spring + gravitational potential energy

TE = -11GMm/8R + KR²/8 ................2

total energy remains same so eq1 = eq 2

KR²/8 -11GMm/8R = -GMm/2R

KR² /8 = 7GMm/8R or

K = 7GMm/R³

approve my ans if u like

Kushagra Madhukar · Answer

Dear student, Please find the solution to your problem. Gravitational potential inside the earth is given by – 3GM/2R [1 – r 2 /3R 2 ] Potential energy of ball when it was at a height of R above earth is-GMm/2R Kinetic energy at this instant is 0... Total energy = – GM/2R .................1 Finally ball agian comes to rest , now ball is at a depth of R/2 from earth so Its gravitational energy is – 3GMm/2R[1 – r 2 /3R 2 ] = – 3GMm/2R[11/12] (r = R/2) = – 11GMm/8R potential energy stored in spring is KX 2 /2 = KR 2 /8 (compression is R/2) total energy now is sum of potential energy of spring + gravitational potential energy TE = – 11GMm/8R + KR 2 /8 ................2 total energy remains same so eq1 = eq 2 KR 2 /8 – 11GMm/8R = – GMm/2R KR 2 /8 = 7GMm/8R or, K = 7GMm/R 3 Thanks and regards, Kushagra

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