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Step-by-Step Solution
Step 1: Express the mass of the planet in terms of density
For a spherical planet of radius $R$ and constant density $\rho$, its mass $M$ is given by
$M = \frac{4}{3} \pi R^3 \rho$.
Step 2: Relate the new planet’s radius to Earth’s radius
Let the new planet have mass $2M$ (twice Earth’s mass) but the same density $\rho$. Then
$2M = \frac{4}{3} \pi R'^3 \rho.$
Dividing by $M = \frac{4}{3} \pi R^3 \rho$, we get
$\frac{2M}{M} = \frac{\frac{4}{3} \pi R'^3 \rho}{\frac{4}{3} \pi R^3 \rho} \quad \Longrightarrow \quad 2 = \frac{R'^3}{R^3} \quad \Longrightarrow \quad R' = \sqrt[3]{2}\, R.$
Step 3: Write the expression for weight on Earth
The weight of an object of mass $m$ on Earth ($W$) can be written as
$W = \frac{GMm}{R^2},$
where $G$ is the universal gravitational constant, $M$ is Earth’s mass, and $R$ is Earth’s radius.
Step 4: Derive the weight on the new planet
Let $W_2$ be the weight of the same object on the new planet. Its mass is $2M$, and its radius is $\sqrt[3]{2}\, R$. Thus,
$W_2 = \frac{G \,(2M)\, m}{(\sqrt[3]{2}\, R)^2} = \frac{2\,GM\,m}{2^{\frac{2}{3}} \,R^2} = 2^{1 - \frac{2}{3}} \,\frac{GMm}{R^2} = 2^{\frac{1}{3}} \,W.$
Step 5: Conclude the ratio of the weights
Therefore, the weight on the new planet is $2^{\frac{1}{3}}$ times the weight on Earth. Hence,
$\displaystyle W_2 = 2^{\frac{1}{3}} \, W.$
