Certainly! Let's go through this step-by-step.
1. Understanding the Problem:
- Original mass of the planet is [tex]\( M \)[/tex].
- The mass of the planet is reduced to [tex]\(\frac{1}{8}\)[/tex] of the original mass.
- Density of the planet remains constant.
- We need to determine the new acceleration due to gravity ([tex]\(g'\)[/tex]) of the planet.
2. Key Formulas:
- Acceleration due to gravity [tex]\( g \)[/tex] on a planet is given by:
[tex]\[
g = \frac{GM}{R^2}
\][/tex]
where [tex]\( G \)[/tex] is the gravitational constant, [tex]\( M \)[/tex] is the mass of the planet, and [tex]\( R \)[/tex] is the radius of the planet.
- Since density ([tex]\( \rho \)[/tex]) remains constant:
[tex]\[
\rho = \frac{M}{V} \quad \text{and} \quad V = \frac{4}{3} \pi R^3
\][/tex]
3. Mass Reduction and Density Constant:
- New mass, [tex]\( M' = \frac{M}{8} \)[/tex].
- Density remains constant, hence:
[tex]\[
\rho = \frac{M}{\frac{4}{3} \pi R^3} = \frac{M'}{\frac{4}{3} \pi {R'}^3}
\][/tex]
Since [tex]\(\rho\)[/tex] doesn't change:
[tex]\[
\frac{M}{R^3} = \frac{M'}{{R'}^3}
\][/tex]
4. Relating Volumes and Radii:
- Using [tex]\( M' = \frac{M}{8} \)[/tex]:
[tex]\[
\frac{M}{{R}^3} = \frac{\frac{M}{8}}{R'^3}
\][/tex]
[tex]\[
\frac{M}{{R}^3} = \frac{M}{8R'^3}
\][/tex]
[tex]\[
{R'^3} = \frac{R^3}{8}
\][/tex]
[tex]\[
R' = \left(\frac{R^3}{8}\right)^{\frac{1}{3}} = \frac{R}{2}
\][/tex]
5. Calculating the New Gravity ([tex]\( g' \)[/tex]):
- Using the formula for gravity:
[tex]\[
g' = \frac{G M'}{R'^2}
\][/tex]
- Substitute [tex]\( M' = \frac{M}{8} \)[/tex] and [tex]\( R' = \frac{R}{2} \)[/tex] into the gravity formula:
[tex]\[
g' = \frac{G \left(\frac{M}{8}\right)}{\left(\frac{R}{2}\right)^2}
\][/tex]
[tex]\[
g' = \frac{\frac{G M}{8}}{\frac{R^2}{4}}
\][/tex]
[tex]\[
g' = \frac{GM}{8} \cdot \frac{4}{R^2}
\][/tex]
[tex]\[
g' = \frac{GM}{R^2} \cdot \frac{4}{8}
\][/tex]
[tex]\[
g' = \frac{GM}{R^2} \cdot \frac{1}{2}
\][/tex]
[tex]\[
g' = \frac{g}{2}
\][/tex]
So, the new value of acceleration due to gravity of the planet when its mass is reduced to [tex]\(\frac{1}{8}\)[/tex] without change in density is:
[tex]\[
g' = \frac{g}{2}
\][/tex]
Thus, the correct answer is:
b) [tex]\(\frac{g}{2}\)[/tex]
