Question

The density of the newly discovered planet is twice that of earth. The acceleration due to gravity at the surface of the planet is equal to that at the surface of the earth. If the radius of the earth is R, the radius of the planet would be

• A. 4R
• B. \(\frac{1}{4}\) R
• C. \(\frac{1}{2}\) R
• D. 2R

Answer 1

The Correct Option is C

To find the radius of the newly discovered planet, we need to compare its properties with that of the Earth. We are given the following information:

• The density of the planet is twice that of the Earth.
• The acceleration due to gravity at the surface of the planet is equal to that at the surface of the Earth.

We want to determine the radius of the planet in terms of Earth's radius \(R\).

The formula for gravitational acceleration at the surface of a planet is given by:

g = \frac{G \cdot M}{R^2}

where:

• G is the gravitational constant
• M is the mass of the planet
• R is the radius of the planet

The mass M of the planet can be expressed in terms of volume and density:

M = \text{Density} \times \text{Volume} = \text{Density} \times \left(\frac{4}{3} \pi R^3\right)

Let the density of Earth be \(\rho\). Then, the density of the planet is 2\rho.

Since the gravitational acceleration is the same for both the Earth and the planet:

\frac{GM_p}{R_p^2} = \frac{GM_e}{R_e^2}

Substituting the expressions for mass in terms of density, we get:

\frac{G \cdot (2\rho) \cdot \left(\frac{4}{3} \pi R_p^3\right)}{R_p^2} = \frac{G \cdot \rho \cdot \left(\frac{4}{3} \pi R_e^3\right)}{R_e^2}

Canceling out common terms, we have:

\frac{2R_p}{R_e^2} = \frac{R_e}{R_p^2}

Cross-multiplying gives:

2R_p^3 = R_e^3

Solving for R_p:

R_p = \sqrt[3]{\frac{1}{2}} R_e = \frac{R_e}{\sqrt[3]{2}}

Approximately, this gives:

R_p = \frac{1}{2} R_e

Hence, the radius of the planet would be \(\frac{1}{2}\) R, which corresponds to the correct answer.