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(B) The acceleration due to gravity $g$ decreases as we move away from the earth's surface,given by $g(h) = g_0(1 - 2h/R_e)$.Let $C$ be the centre of

Vedclass · Accepted Answer

$A$ moves slowly upward and $B$ moves slowly downward relative to the cabin

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(B) The acceleration due to gravity $g$ decreases as we move away from the earth's surface,given by $g(h) = g_0(1 - 2h/R_e)$.Let $C$ be the centre of mass of the cabin. The acceleration of any object at a height $h$ is $a = g(h)$.Since $A$ is above $C$ and $B$ is below $C$,their heights are $h_A > h_C > h_B$.Consequently,the gravitational accelerations satisfy $a_B > a_C > a_A$.When viewed from the frame of the cabin (which has acceleration $a_C$),the relative accelerations are:$a_{A,rel} = a_A - a_C $a_{B,rel} = a_B - a_C > 0$ (meaning $B$ accelerates downward relative to the cabin).Thus,$A$ moves slowly upward and $B$ moves slowly downward relative to the cabin. Therefore,option $(b)$ is correct.

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