Two concentric shells of masses m_1 and m_2 h | vedclass

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(C) According to the shell theorem,the gravitational force inside a uniform spherical shell is zero. At position $A$ (outside both shells): The partic

Vedclass · Accepted Answer

$\frac{G(m_1 + m_2)m}{r_1^2}, \frac{G m_1 m}{r_2^2}, 0$

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(C) According to the shell theorem,the gravitational force inside a uniform spherical shell is zero. At position $A$ (outside both shells): The particle is at a distance $r_1$ from the center. Both shells act as point masses at the center. The total force is $F_A = \frac{G(m_1 + m_2)m}{r_1^2}$. At position $B$ (between the two shells): The particle is at a distance $r_2$ from the center. It is outside the inner shell $(m_1)$ but inside the outer shell $(m_2)$. The force due to the outer shell is $0$. The force due to the inner shell is $F_B = \frac{G m_1 m}{r_2^2}$. At position $C$ (inside both shells): The particle is at a distance $r_3$ from the center. It is inside both shells. Therefore,the gravitational force due to both shells is $0$. Thus,$F_C = 0$. Therefore,the forces at $A, B,$ and $C$ are $\frac{G(m_1 + m_2)m}{r_1^2}, \frac{G m_1 m}{r_2^2}, 0$.

Two concentric shells of masses $m_1$ and $m_2$ have radii $r_2$ and $r_1$ respectively (where $r_2 < r_1$). $A$ particle of mass $m$ is placed at positions $A, B,$ and $C$ as shown in the figure. The gravitational forces on the particle at these positions are:

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