you said that as we go inside the earth the value of'g' decreases but why?since this value is inversely proportional to the distance,then as the distance between the centre of earth and particle decreases the g-force must increase.
Asked by architsrivastava02
| 6th Nov, 2010,
12:00: AM
Expert Answer:
Dear student
The g force at a point inside the sphere is due to mass lying within the point & not lying outside.
Consider a point mass m at a depth d below the surface of the earth, so that its distance from the center of the earth is (RE − d). The earth can be thought of as being composed of a smaller
sphere of radius (RE – d ) and a spherical shell of thickness d. The force on m due to the outer
shell of thickness d is zero. As far as the smaller sphere of radius ( RE – d ) is concerned, the point mass is outside it and hence the force due to this smaller sphere is just as if the entire mass of
the smaller sphere is concentrated at the centre.
If Ms is the mass of the smaller sphere, then it will depeend only on the radius (RE – d ).
then,
Ms / ME = ( RE – d )3 / RE3
Since mass of a sphere is proportional to be cube of its radius.
We hope that clarifies your query.
Regards
Team
TopperLearning
sphere of radius (RE – d ) and a spherical shell of thickness d. The force on m due to the outer
shell of thickness d is zero. As far as the smaller sphere of radius ( RE – d ) is concerned, the point mass is outside it and hence the force due to this smaller sphere is just as if the entire mass of
the smaller sphere is concentrated at the centre.
If Ms is the mass of the smaller sphere, then it will depeend only on the radius (RE – d ).
Ms / ME = ( RE – d )3 / RE3
Since mass of a sphere is proportional to be cube of its radius.

Answered by
| 7th Nov, 2010,
06:53: AM
Kindly Sign up for a personalised experience
- Ask Study Doubts
- Sample Papers
- Past Year Papers
- Textbook Solutions
Sign Up
Verify mobile number
Enter the OTP sent to your number
Change