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sohan singh singh Grade: 12
        

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???         

6 years ago

Answers : (2)

Askiitians Expert Soumyajit IIT-Kharagpur
28 Points
										

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


radious of the earth=R


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)}


 


 


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All the best Amrit Pal !!!


 



Regards,


Askiitians Experts
Soumyajit Das IIT Kharagpur

6 years ago
Rishabh Dabral
32 Points
										

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.

6 years ago
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