AP EAMCET · PHYSICS · Gravitation

A uniform solid sphere of radius $R$ produces a gravitational acceleration of $a_{o}$ on its surface. The distance of the point from the centre of the sphere where the gravitational acceleration becomes $\frac{a_{o}}{4}$ is,

A $4 R$
B $\frac{3}{2} R$
C $2 R$
D $3 R$

Verified Solution

Answer & Solution

Correct Answer

(C) $2 R$

Step-by-step Solution

Detailed explanation

On the surface of the uniform solid sphere, gravitational field(also called gravitational acceleration) is given by, -GMR2=a0. Gravitational field at any height h is given by, a=a01+hR-2⇒a04=a01+hR-2⇒1+hR2=4⇒h=R Hence, distance from the centre will be, =R+R=2R

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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A uniform solid sphere of radius R produces a gravitational acceleration of ao on its surface. The distance of the point from the centre of the sphere where the gravitational acceleration becomes ao4 is,

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A uniform solid sphere of radius $R$ produces a gravitational acceleration of $a_{o}$ on its surface. The distance of the point from the centre of the sphere where the gravitational acceleration becomes $\frac{a_{o}}{4}$ is,