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Somebody made a giant spherical cavity in the earth such that the earth's center and a point in the surface are diametrically opposite.The someone drops a small ball from a small opening at the surface into the cavity. In how many minutes does the ball reach the center of earth???  Askiitians Expert Soumyajit IIT-Kharagpur
28 Points
11 years ago

Dear Amrit Pal,

Ans:- Hi first of all I must say that you see the answer of the question. because it is a conceptual and if the solution is not correct then I will also have to attempt it in another way.

Let us consider some terms

Mass of the earth=M

Density of the earth=d

Universal gravitational constant=G

mass of the particle is=m

Now you must know that the value of the gravitational acc g′ at a depth h inside the earth is=(1- h/R)g

where g is it's value on the surface of the earth. g=GM/R² and hence  the total force is= 4/3 ∏GmRd(1 - h/R)

F=4/3 ∏Gmd(R-h)..............(1)

Now the value we are getting is due to the superposition of the cut out mass and the remaining mass.

HenceF=Fr+Fc..............(2)

Fr=force due to the remaining and Fc=force due to the cut out

Using eq 1 we can find out the force due to the cut out portion which is Fc=4/3∏ Gmd (R/2 - h)....................(3)

Putting these values of eq 1 and 3 in 2 we get,

Fr=F-Fc

=4/3 ∏Gmd{(R-h)-(R/2 -h)}

=2/3 ∏GmRd

Which is constant i.e h independent

then the acceleration due to this portion is a=Fr/m=2/3 ∏ GRd

So the time taken is t=√{(2R/2)/(2/3 ∏ GRd)}

=√{3/(2GR∏d)}

Please feel free to post as many doubts on our discussion forum as you can. If you find any question
Difficult to understand - post it here and we will get you the answer and detailed solution very quickly. We

All the best Amrit Pal !!!

Regards,

Soumyajit Das IIT Kharagpur

11 years ago

Dear Saumyajit,

we know that radius of earth is 6400km or 6400000m.

So, in the mouth of the cavity, the ball is at a potential=mgh.

and, at the centre, it will have the KE=mv2/2.

Equating both.

mgh=mv2/2, where h=6400000m.

by this, we get, v=11200m/s.

now, u may apply the first eq. of motion to find the time taken.