Question

A small point of mass \(m\) is placed at a distance \(2R\) from the center \(O\) of a big uniform solid sphere of mass \(M\) and radius \(R\). The gravitational force on \(m\) due to \(M\) is \(F_1\). A spherical part of radius \(R/3\) is removed from the big sphere as shown in the figure, and the gravitational force on \(m\) due to the remaining part of \(M\) is found to be \(F_2\). The value of the ratio \( F_1 : F_2 \) is:

 

• A. 16 : 9
• B. 11 : 10
• C. 12 : 11
• D. 12 : 9

Hint:

Use the inverse square law for gravitational force and break the mass into parts when necessary.

Answer 1

The Correct Option is C

The gravitational force on \(m\) due to the whole sphere is: \[ F_1 = \frac{GMm}{(2R)^2} \quad \text{(1)} \] The gravitational force due to the remaining mass after removing the spherical part of radius \(R/3\) is: \[ F_2 = \frac{GMm}{(2R)^2} \times \left(\frac{M}{27} \times \frac{4R}{3} \right)^2 = \frac{11 GMm}{48R^2} \quad \text{(2)} \] Thus, the ratio is: \[ \frac{F_1}{F_2} = 12 : 11 \] Thus, the answer is \( \boxed{12 : 11} \).

Answer 2

Force before removal. For a point outside a uniform sphere the whole mass acts as if concentrated at the centre. So \[ F_1=\frac{G M m}{(2R)^2}=\frac{G M m}{4R^2}. \]

1. Replace "removal" by superposition. The gravitational field of the remaining body = field of the original full sphere \(+\) field of a negative sphere (the removed part) located at the removed sphere's centre. Thus the net force on \(m\) is \[ F_2=\frac{G M m}{(2R)^2}-\frac{G m\,M_{\text{removed}}}{d^2}, \] where \(M_{\text{removed}}\) is the mass of the removed small sphere and \(d\) is the distance from the point mass \(m\) to the centre of the removed sphere.
2. Mass of removed small sphere. Same density \(\rho\) as the big sphere, so \[ M_{\text{removed}} = M\cdot\frac{\text{vol}_{\text{small}}}{\text{vol}_{\text{big}}} = M\cdot\frac{(R/3)^3}{R^3} = \frac{M}{27}. \]
3. Geometry — position of the removed sphere's centre. The small sphere of radius \(R/3\) is cut out from the big sphere so that it lies on the radius toward the external point. Its centre is at distance \[ R - \frac{R}{3} = \frac{2R}{3} \] from \(O\). Therefore the distance from the external point (at \(2R\) from \(O\)) to the removed-centre is \[ d = 2R - \frac{2R}{3} = \frac{4R}{3}. \]
4. Compute \(F_2\). \[ F_2 = \frac{G M m}{4R^2} \;-\; \frac{G m (M/27)}{(4R/3)^2}. \] Compute the second term denominator: \[ (4R/3)^2=\frac{16R^2}{9}, \] so \[ \frac{M/27}{(4R/3)^2}=\frac{1}{27}\cdot\frac{9}{16}\cdot\frac{M}{R^2}=\frac{1}{48}\cdot\frac{M}{R^2}. \] Hence \[ F_2 = \frac{G M m}{R^2}\Big(\frac{1}{4}-\frac{1}{48}\Big) = \frac{G M m}{R^2}\cdot\frac{11}{48}. \]
5. Find ratio \(F_1:F_2\). Express \(F_1\) with denominator \(48\): \[ F_1=\frac{G M m}{4R^2}=\frac{G M m}{R^2}\cdot\frac{12}{48}. \] Therefore \[ F_1:F_2 = \frac{12/48}{11/48} = 12:11. \]

Answer

\(\boxed{F_1:F_2 = 12:11}\) (Option 3)