Question:
There is a planet which is \(8\) times massive and \(27\) times denser than the earth. If \(g'\) and \(g\) are the accelerations due to gravity on the surfaces of the planet and the earth respectively, then:
There is a planet which is \(8\) times massive and \(27\) times denser than the earth. If \(g'\) and \(g\) are the accelerations due to gravity on the surfaces of the planet and the earth respectively, then:
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For a spherical planet, use \(\rho \propto \frac{M}{R^3}\) to first find the radius ratio, then apply \(g=\frac{GM}{R^2}\).
Updated On: Jun 26, 2026
- \(g'=8g\)
- \(g'=27g\)
- \(g'=18g\)
- \(g'=\dfrac{9}{4}g\)
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The Correct Option is C
Solution and Explanation
Step 1: Use the formula for acceleration due to gravity.
Acceleration due to gravity on the surface of a planet is \[ g=\frac{GM}{R^2} \] For the new planet, \[ g'=\frac{GM'}{R'^2} \] Given, \[ M'=8M \] Also, density of the new planet is \[ \rho'=27\rho \]
Step 2: Relate mass, density, and radius.
Density is given by \[ \rho=\frac{M}{V} \] For a spherical planet, \[ V=\frac{4}{3}\pi R^3 \] So, \[ \rho=\frac{M}{\frac{4}{3}\pi R^3} \] Hence, \[ \rho \propto \frac{M}{R^3} \] For the new planet, \[ \frac{\rho'}{\rho} = \frac{M'/R'^3}{M/R^3} \] Substituting the given ratios, \[ 27= \frac{8M}{R'^3}\cdot \frac{R^3}{M} \] \[ 27=8\frac{R^3}{R'^3} \] \[ \frac{R'^3}{R^3}=\frac{8}{27} \] Taking cube root, \[ \frac{R'}{R}=\frac{2}{3} \] Thus, \[ R'=\frac{2R}{3} \]
Step 3: Find the ratio of accelerations due to gravity.
Now, \[ \frac{g'}{g} = \frac{\frac{GM'}{R'^2}}{\frac{GM}{R^2}} \] \[ \frac{g'}{g} = \frac{M'}{M}\cdot \frac{R^2}{R'^2} \] Substituting, \[ \frac{M'}{M}=8 \] and \[ \frac{R'}{R}=\frac{2}{3} \] So, \[ \frac{R^2}{R'^2} = \left(\frac{R}{R'}\right)^2 = \left(\frac{3}{2}\right)^2 = \frac{9}{4} \] Therefore, \[ \frac{g'}{g}=8\times \frac{9}{4} \] \[ \frac{g'}{g}=18 \] Hence, \[ g'=18g \]
Step 4: Final conclusion.
Therefore, \[ \boxed{g'=18g} \]
Acceleration due to gravity on the surface of a planet is \[ g=\frac{GM}{R^2} \] For the new planet, \[ g'=\frac{GM'}{R'^2} \] Given, \[ M'=8M \] Also, density of the new planet is \[ \rho'=27\rho \]
Step 2: Relate mass, density, and radius.
Density is given by \[ \rho=\frac{M}{V} \] For a spherical planet, \[ V=\frac{4}{3}\pi R^3 \] So, \[ \rho=\frac{M}{\frac{4}{3}\pi R^3} \] Hence, \[ \rho \propto \frac{M}{R^3} \] For the new planet, \[ \frac{\rho'}{\rho} = \frac{M'/R'^3}{M/R^3} \] Substituting the given ratios, \[ 27= \frac{8M}{R'^3}\cdot \frac{R^3}{M} \] \[ 27=8\frac{R^3}{R'^3} \] \[ \frac{R'^3}{R^3}=\frac{8}{27} \] Taking cube root, \[ \frac{R'}{R}=\frac{2}{3} \] Thus, \[ R'=\frac{2R}{3} \]
Step 3: Find the ratio of accelerations due to gravity.
Now, \[ \frac{g'}{g} = \frac{\frac{GM'}{R'^2}}{\frac{GM}{R^2}} \] \[ \frac{g'}{g} = \frac{M'}{M}\cdot \frac{R^2}{R'^2} \] Substituting, \[ \frac{M'}{M}=8 \] and \[ \frac{R'}{R}=\frac{2}{3} \] So, \[ \frac{R^2}{R'^2} = \left(\frac{R}{R'}\right)^2 = \left(\frac{3}{2}\right)^2 = \frac{9}{4} \] Therefore, \[ \frac{g'}{g}=8\times \frac{9}{4} \] \[ \frac{g'}{g}=18 \] Hence, \[ g'=18g \]
Step 4: Final conclusion.
Therefore, \[ \boxed{g'=18g} \]
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