A uniform spherical planet (Radius R) has acc | vedclass

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(A) For a point $P$ inside the planet at a distance $r_1$ from the center, the acceleration due to gravity is given by $g_{in} = g(r_1/R)$.Given $g_{i

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$7R/4$

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(A) For a point $P$ inside the planet at a distance $r_1$ from the center, the acceleration due to gravity is given by $g_{in} = g(r_1/R)$.Given $g_{in} = g/4$, we have $r_1/R = 1/4$, so $r_1 = R/4$.The distance of point $P$ from the surface is $d = R - r_1 = R - R/4 = 3R/4$.For a point $Q$ outside the planet at a distance $r_2$ from the center, the acceleration due to gravity is given by $g_{out} = g(R/r_2)^2$.Given $g_{out} = g/4$, we have $(R/r_2)^2 = 1/4$, which implies $R/r_2 = 1/2$, so $r_2 = 2R$.The distance of point $Q$ from the surface is $h = r_2 - R = 2R - R = R$.The maximum separation between $P$ and $Q$ is the sum of their distances from the surface: $d + h = 3R/4 + R = 7R/4$.

$A$ uniform spherical planet (Radius $R$) has acceleration due to gravity at its surface $g$. Points $P$ and $Q$ located inside and outside the planet have acceleration due to gravity $g/4$. The maximum possible separation between $P$ and $Q$ is:

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