Suppose you decide to go ice fishing. You use powerful strong Auger drill to cut

Question

Suppose you decide to go ice fishing. You use powerful strong Auger drill to cut through the ice. You drill a hole with smooth sides straight through. As a physics student, suddenly you want to find out whether one can drill straight through the center of the earth a hole into the earth as a tube to find (assuming If the air is removed from this tube and the tube doesn’t fill up with water, liquid rock or iron from the core), whether an object dropped into one end of the tube will have enough energy to just exit the other end and if so after an interval of time.Make the assumption that the earth has uniform mass density. While calculating the time for go through the steps (a) What is the gravitational force as a function of the distance r from the center of the earth? Express your answer in terms of the gravitational acceleration at the surface of the earth g and re Note: (neglect the amount of mass you drilled out.) (b) Use your result of part a) to explain why the object of mass m should oscillate (analogous to an object attached to a spring like SHM). In particular, what is the spring constant and how long would it take for this object to reach the other side of the earth? (c) What is the potential energy inside the earth as a function of r for the object-earth system? Can you think of a natural point to choose a zero point for the potential energy? Be careful because you will need relation between work and potential energy when the object moves inside the earth and with appropriate gravitation force when the object is inside the earth. (d) Use energy considerations to find the velocity of the object when it passes through the center of the earth

Explanation / Answer

Answer:

PART A:

Consider the object is at a distance "r" from the centre of earth in the tube for any instant "t".

Leu us draw a sphere of radius "r" with its centre at the centre of earth. Only this part of sphere will exert a force of attraction on the particle. Then the mass of the this part of sphere is

M' = (4/3) * pi *r3 * M / (4/3) * pi *Re3 = r3M/ Re3 Where M and Re are mass and radius of earth

The force of attraction on object of mass "m" is

F = ( Gr3Mm/ Re3 ) / r2  = GMmr / Re3

PART B

The force acts towards the centre of the earth. The resultant force on particle is opposite to displacement from the centre from the earth and proportional to it. therefore particle would execute SHM in the tube.

The force of constant (spring constant) k = GMm/Re3

Time period T = 2 * pi* ( Re3 / Gm )1/2

PART C

Potential energy = integratio (0, r) ( GMmr / Re3 ) dr = inte (0, r) k.r dr = (1/2) kr2   

We can not choose natural point as zero for potential energy because the point is not the part of spring that is executing SHM

Part D

  Velocity of the object when passes through centre

V = w (Re2 - x2)1/2 = Re w at the center x = 0

V = Re (GM/Re3)1/2

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